3.369 \(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=331 \[ \frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {\log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {\log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 d^{5/2} f}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) (d \tan (e+f x))^{3/2}}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}} \]

[Out]

7/2*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(5/2)/f-1/4*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/
d^(5/2)/f*2^(1/2)+1/4*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(5/2)/f*2^(1/2)+1/8*ln(d^(1/2)-2^(1
/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)-1/8*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)
+d^(1/2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)+9/2/a^2/d^2/f/(d*tan(f*x+e))^(1/2)-7/6/a^2/d/f/(d*tan(f*x+e))^(3/2)
+1/2/d/f/(d*tan(f*x+e))^(3/2)/(a^2+a^2*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.99, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3569, 3649, 3653, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {\log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {\log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 d^{5/2} f}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) (d \tan (e+f x))^{3/2}}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^2),x]

[Out]

(7*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*d^(5/2)*f) - ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]
]/(2*Sqrt[2]*a^2*d^(5/2)*f) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(2*Sqrt[2]*a^2*d^(5/2)*f) + L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(5/2)*f) - Log[Sqrt[d] + Sq
rt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(5/2)*f) - 7/(6*a^2*d*f*(d*Tan[e + f*x])^(
3/2)) + 9/(2*a^2*d^2*f*Sqrt[d*Tan[e + f*x]]) + 1/(2*d*f*(d*Tan[e + f*x])^(3/2)*(a^2 + a^2*Tan[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^2} \, dx &=\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {\frac {7 a^2 d}{2}-a^2 d \tan (e+f x)+\frac {5}{2} a^2 d \tan ^2(e+f x)}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx}{2 a^3 d}\\ &=-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {\frac {27 a^3 d^3}{4}+\frac {3}{2} a^3 d^3 \tan (e+f x)+\frac {21}{4} a^3 d^3 \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx}{3 a^4 d^4}\\ &=-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {2 \int \frac {\frac {21 a^4 d^5}{8}+\frac {3}{4} a^4 d^5 \tan (e+f x)+\frac {27}{8} a^4 d^5 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{3 a^5 d^7}\\ &=-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {3 a^5 d^5 \tan (e+f x)}{2 \sqrt {d \tan (e+f x)}} \, dx}{3 a^7 d^7}+\frac {7 \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a d^2}\\ &=-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d^2}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a d^2 f}\\ &=-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \sqrt {d \tan (e+f x)} \, dx}{2 a^2 d^3}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a d^3 f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{2 a^2 d^2 f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d^2 f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 d^2 f}+\frac {\operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 d^2 f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}+\frac {\operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^2 d^2 f}+\frac {\operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^2 d^2 f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}\\ &=\frac {7 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{5/2} f}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{5/2} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{5/2} f}-\frac {7}{6 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{2 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} \left (a^2+a^2 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 3.97, size = 203, normalized size = 0.61 \[ \frac {\tan ^{\frac {5}{2}}(e+f x) \left (-\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )+\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+14 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )+\frac {\log \left (-\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}+\frac {2 \left (20 \cot (e+f x)-4 \csc ^2(e+f x)+31\right )}{3 \sqrt {\tan (e+f x)} (\cot (e+f x)+1)}\right )}{4 a^2 f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^2),x]

[Out]

((-(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]) + Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 14*ArcT
an[Sqrt[Tan[e + f*x]]] + (Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[e + f
*x]] + Tan[e + f*x]])/Sqrt[2] + (2*(31 + 20*Cot[e + f*x] - 4*Csc[e + f*x]^2))/(3*(1 + Cot[e + f*x])*Sqrt[Tan[e
+ f*x]]))*Tan[e + f*x]^(5/2))/(4*a^2*f*(d*Tan[e + f*x])^(5/2))

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fricas [B]  time = 1.10, size = 2269, normalized size = 6.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/24*(21*(cos(f*x + e)^2 + 2*(cos(f*x + e)^3 - cos(f*x + e))*sin(f*x + e) - 1)*sqrt(-d)*log(-(6*d^2*cos(f*x
+ e)*sin(f*x + e) - d^2 - 4*(d*cos(f*x + e)^2 - d*cos(f*x + e)*sin(f*x + e))*sqrt(-d)*sqrt(d*sin(f*x + e)/cos(
f*x + e)))/(2*cos(f*x + e)*sin(f*x + e) + 1)) + 12*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(
sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*arct
an(-sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) + sqrt(2)*a^2*d^2*f*sqrt((sqr
t(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(1/
(a^8*d^10*f^4))*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) - 1) + 12*(sqrt(2)*a^2*d
^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e)
)*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*arctan(-sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d
^10*f^4))^(1/4) + sqrt(2)*a^2*d^2*f*sqrt(-(sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*
f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(1/(a^8*d^10*f^4))*cos(f*x + e) - d*sin(f*x + e))/cos(f*x + e))*(1/
(a^8*d^10*f^4))^(1/4) + 1) + 3*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sqrt(2)*a^2*d^3*f*co
s(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*log((sqrt(2)*a^6*d^8*f^3
*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(1/(a^8*d^10*f^4))*
cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) - 3*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sq
rt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*log(-(
sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt
(1/(a^8*d^10*f^4))*cos(f*x + e) - d*sin(f*x + e))/cos(f*x + e)) - 4*(51*cos(f*x + e)^4 - 47*cos(f*x + e)^2 + (
11*cos(f*x + e)^3 - 27*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^3*f*cos(f*x + e)^
2 - a^2*d^3*f + 2*(a^2*d^3*f*cos(f*x + e)^3 - a^2*d^3*f*cos(f*x + e))*sin(f*x + e)), 1/24*(84*(cos(f*x + e)^2
+ 2*(cos(f*x + e)^3 - cos(f*x + e))*sin(f*x + e) - 1)*sqrt(d)*arctan(sqrt(d*sin(f*x + e)/cos(f*x + e))/sqrt(d)
) - 12*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a
^2*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*arctan(-sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/co
s(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) + sqrt(2)*a^2*d^2*f*sqrt((sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x
+ e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(1/(a^8*d^10*f^4))*cos(f*x + e) + d*sin(f*x + e
))/cos(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) - 1) - 12*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*
(sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*arc
tan(-sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) + sqrt(2)*a^2*d^2*f*sqrt(-(s
qrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(
1/(a^8*d^10*f^4))*cos(f*x + e) - d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^10*f^4))^(1/4) + 1) - 3*(sqrt(2)*a^2*
d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^3*f*cos(f*x + e
))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*log((sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d
^10*f^4))^(3/4)*cos(f*x + e) + a^4*d^6*f^2*sqrt(1/(a^8*d^10*f^4))*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))
+ 3*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^3*f + 2*(sqrt(2)*a^2*d^3*f*cos(f*x + e)^3 - sqrt(2)*a^2
*d^3*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^10*f^4))^(1/4)*log(-(sqrt(2)*a^6*d^8*f^3*sqrt(d*sin(f*x + e)/cos(
f*x + e))*(1/(a^8*d^10*f^4))^(3/4)*cos(f*x + e) - a^4*d^6*f^2*sqrt(1/(a^8*d^10*f^4))*cos(f*x + e) - d*sin(f*x
+ e))/cos(f*x + e)) + 4*(51*cos(f*x + e)^4 - 47*cos(f*x + e)^2 + (11*cos(f*x + e)^3 - 27*cos(f*x + e))*sin(f*x
+ e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^3*f*cos(f*x + e)^2 - a^2*d^3*f + 2*(a^2*d^3*f*cos(f*x + e)^3
- a^2*d^3*f*cos(f*x + e))*sin(f*x + e))]

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giac [A]  time = 2.16, size = 309, normalized size = 0.93 \[ \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{4 \, a^{2} d^{4} f} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{4 \, a^{2} d^{4} f} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{8 \, a^{2} d^{4} f} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{8 \, a^{2} d^{4} f} + \frac {7 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{2 \, a^{2} d^{\frac {5}{2}} f} + \frac {\sqrt {d \tan \left (f x + e\right )}}{2 \, {\left (d \tan \left (f x + e\right ) + d\right )} a^{2} d^{2} f} + \frac {2 \, {\left (6 \, d \tan \left (f x + e\right ) - d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f \tan \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/4*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^2
*d^4*f) + 1/4*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(ab
s(d)))/(a^2*d^4*f) - 1/8*sqrt(2)*abs(d)^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) +
abs(d))/(a^2*d^4*f) + 1/8*sqrt(2)*abs(d)^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d))
+ abs(d))/(a^2*d^4*f) + 7/2*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*d^(5/2)*f) + 1/2*sqrt(d*tan(f*x + e))/(
(d*tan(f*x + e) + d)*a^2*d^2*f) + 2/3*(6*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*a^2*d^3*f*tan(f*x + e))

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maple [A]  time = 0.32, size = 276, normalized size = 0.83 \[ \frac {\sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f \,a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f \,a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f \,a^{2} d^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {d \tan \left (f x +e \right )}}{2 f \,a^{2} d^{2} \left (d \tan \left (f x +e \right )+d \right )}+\frac {7 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 a^{2} d^{\frac {5}{2}} f}-\frac {2}{3 a^{2} d f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4}{a^{2} d^{2} f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^2,x)

[Out]

1/8/f/a^2/d^2/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*ta
n(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/4/f/a^2/d^2/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f/a^2/d^2/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*
x+e))^(1/2)+1)+1/2/f/a^2/d^2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+7/2*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^
2/d^(5/2)/f-2/3/a^2/d/f/(d*tan(f*x+e))^(3/2)+4/a^2/d^2/f/(d*tan(f*x+e))^(1/2)

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maxima [A]  time = 0.54, size = 254, normalized size = 0.77 \[ \frac {\frac {4 \, {\left (27 \, d^{2} \tan \left (f x + e\right )^{2} + 20 \, d^{2} \tan \left (f x + e\right ) - 4 \, d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} d + \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} d^{2}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{2} d} + \frac {84 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} d^{\frac {3}{2}}}}{24 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

1/24*(4*(27*d^2*tan(f*x + e)^2 + 20*d^2*tan(f*x + e) - 4*d^2)/((d*tan(f*x + e))^(5/2)*a^2*d + (d*tan(f*x + e))
^(3/2)*a^2*d^2) + 3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d)
+ 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*ta
n(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*t
an(f*x + e))*sqrt(d) + d)/sqrt(d))/(a^2*d) + 84*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*d^(3/2)))/(d*f)

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mupad [B]  time = 4.98, size = 424, normalized size = 1.28 \[ \frac {\frac {9\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {10\,\mathrm {tan}\left (e+f\,x\right )}{3}-\frac {2}{3}}{a^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+a^2\,d\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {2048\,a^{10}\,d^{18}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^{10}\,f^4}\right )}^{1/4}}{2048\,a^8\,d^{16}\,f^4+100352\,a^{12}\,d^{21}\,f^6\,\sqrt {-\frac {1}{a^8\,d^{10}\,f^4}}}+\frac {100352\,a^{14}\,d^{23}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^{10}\,f^4}\right )}^{3/4}}{2048\,a^8\,d^{16}\,f^4+100352\,a^{12}\,d^{21}\,f^6\,\sqrt {-\frac {1}{a^8\,d^{10}\,f^4}}}\right )\,{\left (-\frac {1}{a^8\,d^{10}\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {a^{10}\,d^{18}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^{10}\,f^4}\right )}^{1/4}\,8192{}\mathrm {i}}{2048\,a^8\,d^{16}\,f^4-1605632\,a^{12}\,d^{21}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^{10}\,f^4}}}-\frac {a^{14}\,d^{23}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^{10}\,f^4}\right )}^{3/4}\,6422528{}\mathrm {i}}{2048\,a^8\,d^{16}\,f^4-1605632\,a^{12}\,d^{21}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^{10}\,f^4}}}\right )\,{\left (-\frac {1}{256\,a^8\,d^{10}\,f^4}\right )}^{1/4}\,2{}\mathrm {i}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-d^5}\,1{}\mathrm {i}}{d^3}\right )\,\sqrt {-d^5}\,7{}\mathrm {i}}{2\,a^2\,d^5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x))^2),x)

[Out]

((10*tan(e + f*x))/3 + (9*tan(e + f*x)^2)/2 - 2/3)/(a^2*f*(d*tan(e + f*x))^(5/2) + a^2*d*f*(d*tan(e + f*x))^(3
/2)) + (atan((2048*a^10*d^18*f^5*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^10*f^4))^(1/4))/(2048*a^8*d^16*f^4 + 100352
*a^12*d^21*f^6*(-1/(a^8*d^10*f^4))^(1/2)) + (100352*a^14*d^23*f^7*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^10*f^4))^(
3/4))/(2048*a^8*d^16*f^4 + 100352*a^12*d^21*f^6*(-1/(a^8*d^10*f^4))^(1/2)))*(-1/(a^8*d^10*f^4))^(1/4))/2 + ata
n((a^10*d^18*f^5*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^10*f^4))^(1/4)*8192i)/(2048*a^8*d^16*f^4 - 1605632*a^12
*d^21*f^6*(-1/(256*a^8*d^10*f^4))^(1/2)) - (a^14*d^23*f^7*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^10*f^4))^(3/4)
*6422528i)/(2048*a^8*d^16*f^4 - 1605632*a^12*d^21*f^6*(-1/(256*a^8*d^10*f^4))^(1/2)))*(-1/(256*a^8*d^10*f^4))^
(1/4)*2i + (atan(((d*tan(e + f*x))^(1/2)*(-d^5)^(1/2)*1i)/d^3)*(-d^5)^(1/2)*7i)/(2*a^2*d^5*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )} + 2 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(5/2)/(a+a*tan(f*x+e))**2,x)

[Out]

Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x)**2 + 2*(d*tan(e + f*x))**(5/2)*tan(e + f*x) + (d*tan(e + f*x)
)**(5/2)), x)/a**2

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