3.4.64 \(\int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx\) [364]

Optimal. Leaf size=281 \[ \frac {3 d^{5/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}+\frac {d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.34, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3646, 3734, 12, 3557, 335, 217, 1179, 642, 1176, 631, 210, 3715, 65, 211} \begin {gather*} \frac {3 d^{5/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}+\frac {d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 f}+\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f}-\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{4 \sqrt {2} a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*f) + (d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]]
)/Sqrt[d]])/(2*Sqrt[2]*a^2*f) - (d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(2*Sqrt[2]*a^2*f)
+ (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) - (d^(5/2)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(4*Sqrt[2]*a^2*f) - (d^2*Sqrt[d*Tan[e + f*x
]])/(2*f*(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 65

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 210

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

Rule 211

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

Rule 217

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

Rule 335

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 631

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 642

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 1176

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 1179

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 3557

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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3715

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 3734

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 {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx &=-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {\frac {a^2 d^3}{2}-a^2 d^3 \tan (e+f x)+\frac {3}{2} a^2 d^3 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{2 a^3}\\ &=-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int -\frac {2 a^3 d^3}{\sqrt {d \tan (e+f x)}} \, dx}{4 a^5}+\frac {\left (3 d^3\right ) \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a}\\ &=-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {d^3 \int \frac {1}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2}+\frac {\left (3 d^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\left (3 d^2\right ) \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a f}-\frac {d^4 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {d^4 \text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {d^3 \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 f}-\frac {d^3 \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 f}\\ &=\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}+\frac {d^{5/2} \text {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 f}+\frac {d^{5/2} \text {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 f}-\frac {d^3 \text {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 f}-\frac {d^3 \text {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 f}\\ &=\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}-\frac {d^{5/2} \text {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 f}+\frac {d^{5/2} \text {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 f}\\ &=\frac {3 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}+\frac {d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 3.55, size = 226, normalized size = 0.80 \begin {gather*} \frac {\csc (e+f x) (\cos (e+f x)+\sin (e+f x))^2 \left (-\frac {4 \cot (e+f x)}{\cos (e+f x)+\sin (e+f x)}+\frac {\left (2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+12 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right ) \sec (e+f x)}{\tan ^{\frac {3}{2}}(e+f x)}\right ) (d \tan (e+f x))^{5/2}}{8 a^2 f (1+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e + f*x]*(Cos[e + f*x] + Sin[e + f*x])^2*((-4*Cot[e + f*x])/(Cos[e + f*x] + Sin[e + f*x]) + ((2*Sqrt[2]*A
rcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 12*ArcTan[Sqrt[Tan[
e + f*x]]] + Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e
+ f*x]] + Tan[e + f*x]])*Sec[e + f*x])/Tan[e + f*x]^(3/2))*(d*Tan[e + f*x])^(5/2))/(8*a^2*f*(1 + Tan[e + f*x])
^2)

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 191, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (d \tan \left (f x +e \right )+d \right )}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}\right )}{f \,a^{2}}\)	\(191\)
default	\(\frac {2 d^{3} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (d \tan \left (f x +e \right )+d \right )}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}\right )}{f \,a^{2}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^2*d^3*(-1/4*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+3/4/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))-1/16/
d*(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*tan(f*x+e)-(d
^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*ar
ctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))

________________________________________________________________________________________

Maxima [A]
time = 0.53, size = 226, normalized size = 0.80 \begin {gather*} -\frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{4}}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} - \frac {12 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}} + \frac {2 \, \sqrt {2} d^{\frac {7}{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 ) + 2 \, \sqrt {2} d^{\frac {7}{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 {2} d^{\frac {7}{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 {2} d^{\frac {7}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{a^{2}}}{8 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*sqrt(d*tan(f*x + e))*d^4/(a^2*d*tan(f*x + e) + a^2*d) - 12*d^(7/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)
)/a^2 + (2*sqrt(2)*d^(7/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + 2*sqrt(2)*
d^(7/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + sqrt(2)*d^(7/2)*log(d*tan(f*
x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - sqrt(2)*d^(7/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f
*x + e))*sqrt(d) + d))/a^2)/(d*f)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 879 vs. \(2 (226) = 452\).
time = 1.50, size = 1840, normalized size = 6.55 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(2*d^2*cos(f*x + e)*sin(f*x + e) + d^2)*sqrt(-d)*log(-(6*d^2*cos(f*x + e)*sin(f*x + e) - d^2 + 4*(d*co
s(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)) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*arctan(-(s
qrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sq
rt(d^10/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*c
os(f*x + e) + d^5*sin(f*x + e))/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) + d^10)/d^10) + 4*(2*sqrt(2)*a^2*f*cos(f*
x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/c
os(f*x + e))*(d^10/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x + e) - sqrt(2
)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x + e))/cos(f*x
+ e))*(d^10/(a^8*f^4))^(3/4) - d^10)/d^10) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10
/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f
*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x + e))/cos(f*x + e)) + (2*sqrt(2)*a^2*f*cos(f*x + e)
*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x + e) - sqrt(2)
*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x + e))/cos(f*x +
e)) - 4*(d^2*cos(f*x + e)^2 + d^2*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(2*a^2*f*cos(
f*x + e)*sin(f*x + e) + a^2*f), 1/8*(12*(2*d^2*cos(f*x + e)*sin(f*x + e) + d^2)*sqrt(d)*arctan(sqrt(d*sin(f*x
+ e)/cos(f*x + e))/sqrt(d)) + 4*(2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(
1/4)*arctan(-(sqrt(2)*a^6*d^2*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*s
qrt((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^
8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x + e))/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) + d^10)/d^10) + 4*(2*sqrt(
2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*arctan(-(sqrt(2)*a^6*d^2*f^3*sqrt(d
*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) - sqrt(2)*a^6*f^3*sqrt((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*
x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x
+ e))/cos(f*x + e))*(d^10/(a^8*f^4))^(3/4) - d^10)/d^10) - (2*sqrt(2)*a^2*f*cos(f*x + e)*sin(f*x + e) + sqrt(
2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin
(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x + e))/cos(f*x + e)) + (2*sqrt(2)*a^2
*f*cos(f*x + e)*sin(f*x + e) + sqrt(2)*a^2*f)*(d^10/(a^8*f^4))^(1/4)*log((a^4*f^2*sqrt(d^10/(a^8*f^4))*cos(f*x
+ e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(d^10/(a^8*f^4))^(1/4)*cos(f*x + e) + d^5*sin(f*x
+ e))/cos(f*x + e)) - 4*(d^2*cos(f*x + e)^2 + d^2*cos(f*x + e)*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e))
)/(2*a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.78, size = 254, normalized size = 0.90 \begin {gather*} -\frac {1}{8} \, d^{2} {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \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 )}{a^{2} f} + \frac {2 \, \sqrt {2} \sqrt {{\left | d \right |}} \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 )}{a^{2} f} + \frac {\sqrt {2} \sqrt {{\left | d \right |}} \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 )}{a^{2} f} - \frac {\sqrt {2} \sqrt {{\left | d \right |}} \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 )}{a^{2} f} - \frac {12 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} f} + \frac {4 \, \sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) + d\right )} a^{2} f}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*d^2*(2*sqrt(2)*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d
)))/(a^2*f) + 2*sqrt(2)*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(
abs(d)))/(a^2*f) + sqrt(2)*sqrt(abs(d))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d
))/(a^2*f) - sqrt(2)*sqrt(abs(d))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(a^
2*f) - 12*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*f) + 4*sqrt(d*tan(f*x + e))*d/((d*tan(f*x + e) + d
)*a^2*f))

________________________________________________________________________________________

Mupad [B]
time = 4.69, size = 376, normalized size = 1.34 \begin {gather*} \frac {\mathrm {atan}\left (\frac {4\,d^{20}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{1/4}}{\frac {36\,d^{23}}{a^2\,f}-4\,a^2\,d^{18}\,f\,\sqrt {-\frac {d^{10}}{a^8\,f^4}}}+\frac {36\,d^{15}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{3/4}}{\frac {36\,d^{23}}{a^6\,f^3}-\frac {4\,d^{18}\,\sqrt {-\frac {d^{10}}{a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {d^{20}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {36\,d^{23}}{a^2\,f}+64\,a^2\,d^{18}\,f\,\sqrt {-\frac {d^{10}}{256\,a^8\,f^4}}}-\frac {d^{15}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{3/4}\,2304{}\mathrm {i}}{\frac {36\,d^{23}}{a^6\,f^3}+\frac {64\,d^{18}\,\sqrt {-\frac {d^{10}}{256\,a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (a^2\,d\,f+a^2\,d\,f\,\mathrm {tan}\left (e+f\,x\right )\right )}+\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}\,3{}\mathrm {i}}{2\,a^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((4*d^20*(d*tan(e + f*x))^(1/2)*(-d^10/(a^8*f^4))^(1/4))/((36*d^23)/(a^2*f) - 4*a^2*d^18*f*(-d^10/(a^8*f^
4))^(1/2)) + (36*d^15*(d*tan(e + f*x))^(1/2)*(-d^10/(a^8*f^4))^(3/4))/((36*d^23)/(a^6*f^3) - (4*d^18*(-d^10/(a
^8*f^4))^(1/2))/(a^2*f)))*(-d^10/(a^8*f^4))^(1/4))/2 + atan((d^20*(d*tan(e + f*x))^(1/2)*(-d^10/(256*a^8*f^4))
^(1/4)*16i)/((36*d^23)/(a^2*f) + 64*a^2*d^18*f*(-d^10/(256*a^8*f^4))^(1/2)) - (d^15*(d*tan(e + f*x))^(1/2)*(-d
^10/(256*a^8*f^4))^(3/4)*2304i)/((36*d^23)/(a^6*f^3) + (64*d^18*(-d^10/(256*a^8*f^4))^(1/2))/(a^2*f)))*(-d^10/
(256*a^8*f^4))^(1/4)*2i - (d^3*(d*tan(e + f*x))^(1/2))/(2*(a^2*d*f + a^2*d*f*tan(e + f*x))) + (atan(((d*tan(e
+ f*x))^(1/2)*(-d^5)^(1/2)*1i)/d^3)*(-d^5)^(1/2)*3i)/(2*a^2*f)

________________________________________________________________________________________