3.4.68 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx\) [368]

Optimal. Leaf size=306 \[ -\frac {5 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}+\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {\text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{3/2} f}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{4 \sqrt {2} a^2 d^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.48, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3650, 3730, 3734, 12, 3557, 335, 217, 1179, 642, 1176, 631, 210, 3715, 65, 211} \begin {gather*} -\frac {5 \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}+\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{2 \sqrt {2} a^2 d^{3/2} 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^{3/2} 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^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2),x]

[Out]

(-5*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*d^(3/2)*f) + ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d
]]/(2*Sqrt[2]*a^2*d^(3/2)*f) - ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(2*Sqrt[2]*a^2*d^(3/2)*f) +
Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - Log[Sqrt[d] + S
qrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - 5/(2*a^2*d*f*Sqrt[d*Tan[e + f*
x]]) + 1/(2*d*f*Sqrt[d*Tan[e + f*x]]*(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 3650

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 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 3730

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*Ta
n[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 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 {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx &=\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\int \frac {\frac {5 a^2 d}{2}-a^2 d \tan (e+f x)+\frac {3}{2} a^2 d \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx}{2 a^3 d}\\ &=-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {\frac {7 a^3 d^3}{4}+\frac {1}{2} a^3 d^3 \tan (e+f x)+\frac {5}{4} a^3 d^3 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{a^4 d^4}\\ &=-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {a^4 d^3}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^6 d^4}-\frac {5 \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a d}\\ &=-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\int \frac {1}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d}-\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a d f}\\ &=-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\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 {5 \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a d^2 f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 d f}-\frac {\text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{2 a^2 d f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}+\frac {\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 d^{3/2} f}+\frac {\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 d^{3/2} f}-\frac {\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 d f}-\frac {\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 d f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} 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^{3/2} 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^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac {\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 d^{3/2} f}+\frac {\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 d^{3/2} f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} 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^{3/2} 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^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.38, size = 203, normalized size = 0.66 \begin {gather*} \frac {\left (\sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-\sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )-10 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right )+\frac {\log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )}{\sqrt {2}}-\frac {2 (4 \cos (e+f x)+5 \sin (e+f x))}{(\cos (e+f x)+\sin (e+f x)) \sqrt {\tan (e+f x)}}\right ) \tan ^{\frac {3}{2}}(e+f x)}{4 a^2 f (d \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(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]]] - 10*ArcTan[
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*(4*Cos[e + f*x] + 5*Sin[e + f*x]))/((Cos[e + f*x] + Sin[e + f*x])*Sqrt[Tan[e +
f*x]]))*Tan[e + f*x]^(3/2))/(4*a^2*f*(d*Tan[e + f*x])^(3/2))

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Maple [A]
time = 0.17, size = 212, normalized size = 0.69

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{4}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{5}}\right )}{f \,a^{2}}\)	\(212\)
default	\(\frac {2 d^{3} \left (-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{4}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{5}}\right )}{f \,a^{2}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*d^3*(-1/d^4/(d*tan(f*x+e))^(1/2)-1/2/d^4*(1/2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+5/2/d^(1/2)*arctan
((d*tan(f*x+e))^(1/2)/d^(1/2)))-1/16/d^5*(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*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))

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Maxima [A]
time = 0.52, size = 240, normalized size = 0.78 \begin {gather*} -\frac {\frac {4 \, {\left (5 \, d \tan \left (f x + e\right ) + 4 \, d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} + \sqrt {d \tan \left (f x + e\right )} a^{2} d} + \frac {\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}}}{a^{2}} + \frac {20 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} \sqrt {d}}}{8 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/8*(4*(5*d*tan(f*x + e) + 4*d)/((d*tan(f*x + e))^(3/2)*a^2 + sqrt(d*tan(f*x + e))*a^2*d) + (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*tan(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*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a^
2 + 20*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*sqrt(d)))/(d*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1151 vs. \(2 (248) = 496\).
time = 2.18, size = 2384, normalized size = 7.79 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/8*(5*(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)) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqr
t(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-
sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*
d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f
^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) - 1) - 4*(sqrt(2)*a^2*d^2*f*co
s(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f
*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4
))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d
*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f
^4))^(3/4) + 1) + (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3
- sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))
*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*s
in(f*x + e))/cos(f*x + e)) - (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(
f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^
8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x
+ e) + d*sin(f*x + e))/cos(f*x + e)) + 4*(9*cos(f*x + e)^4 - 9*cos(f*x + e)^2 + (cos(f*x + e)^3 - 5*cos(f*x +
e))*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^2*f*cos(f*x + e)^2 - a^2*d^2*f + 2*(a^2*d^2*f*cos
(f*x + e)^3 - a^2*d^2*f*cos(f*x + e))*sin(f*x + e)), -1/8*(20*(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)) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x
+ e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x +
e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3
/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(
f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^
(3/4) - 1) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - s
qrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*
x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos
(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f
*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + 1) + (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f +
2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*lo
g((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^
8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) - (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a
^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4
))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x +
e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) + 4*(9*cos(f*x + e)^4 - 9*cos(f*x +
e)^2 + (cos(f*x + e)^3 - 5*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^2*f*cos(f*x +
e)^2 - a^2*d^2*f + 2*(a^2*d^2*f*cos(f*x + e)^3 - a^2*d^2*f*cos(f*x + e))*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )} + 2 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(3/2)/(a+a*tan(f*x+e))**2,x)

[Out]

Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x)**2 + 2*(d*tan(e + f*x))**(3/2)*tan(e + f*x) + (d*tan(e + f*x)
)**(3/2)), x)/a**2

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Giac [A]
time = 0.75, size = 291, normalized size = 0.95 \begin {gather*} -\frac {\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} d 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} d 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} d 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} d f} + \frac {20 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} \sqrt {d} f} + \frac {4 \, {\left (5 \, d \tan \left (f x + e\right ) + 4 \, d\right )}}{{\left (\sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + \sqrt {d \tan \left (f x + e\right )} d\right )} a^{2} f}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/8*(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*d*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(ab
s(d)))/(a^2*d*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*d*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*d*f) + 20*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*sqrt(d)*f) + 4*(5*d*tan(f*x + e) + 4*d)/((sqrt(d*tan(f
*x + e))*d*tan(f*x + e) + sqrt(d*tan(f*x + e))*d)*a^2*f))/d

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Mupad [B]
time = 4.72, size = 415, normalized size = 1.36 \begin {gather*} \frac {\mathrm {atan}\left (\frac {2048\,a^{10}\,d^{13}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{1/4}}{51200\,a^8\,d^{12}\,f^4-2048\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{a^8\,d^6\,f^4}}}+\frac {51200\,a^{14}\,d^{16}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{3/4}}{51200\,a^8\,d^{12}\,f^4-2048\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{a^8\,d^6\,f^4}}}\right )\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {a^{10}\,d^{13}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{1/4}\,8192{}\mathrm {i}}{51200\,a^8\,d^{12}\,f^4+32768\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^6\,f^4}}}-\frac {a^{14}\,d^{16}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{3/4}\,3276800{}\mathrm {i}}{51200\,a^8\,d^{12}\,f^4+32768\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^6\,f^4}}}\right )\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{2}+2}{a^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-d^3}\,1{}\mathrm {i}}{d^2}\right )\,\sqrt {-d^3}\,5{}\mathrm {i}}{2\,a^2\,d^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x))^2),x)

[Out]

(atan((2048*a^10*d^13*f^5*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^6*f^4))^(1/4))/(51200*a^8*d^12*f^4 - 2048*a^12*d^1
5*f^6*(-1/(a^8*d^6*f^4))^(1/2)) + (51200*a^14*d^16*f^7*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^6*f^4))^(3/4))/(51200
*a^8*d^12*f^4 - 2048*a^12*d^15*f^6*(-1/(a^8*d^6*f^4))^(1/2)))*(-1/(a^8*d^6*f^4))^(1/4))/2 + atan((a^10*d^13*f^
5*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^6*f^4))^(1/4)*8192i)/(51200*a^8*d^12*f^4 + 32768*a^12*d^15*f^6*(-1/(25
6*a^8*d^6*f^4))^(1/2)) - (a^14*d^16*f^7*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^6*f^4))^(3/4)*3276800i)/(51200*a
^8*d^12*f^4 + 32768*a^12*d^15*f^6*(-1/(256*a^8*d^6*f^4))^(1/2)))*(-1/(256*a^8*d^6*f^4))^(1/4)*2i - ((5*tan(e +
f*x))/2 + 2)/(a^2*f*(d*tan(e + f*x))^(3/2) + a^2*d*f*(d*tan(e + f*x))^(1/2)) - (atan(((d*tan(e + f*x))^(1/2)*
(-d^3)^(1/2)*1i)/d^2)*(-d^3)^(1/2)*5i)/(2*a^2*d^3*f)

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