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

Optimal. Leaf size=301 \[ \frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}+\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

((9/16 + (5*I)/16)*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/16 + (5*I
)/16)*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/32 - (5*I)/32)*d^(5/2)
*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + ((9/32 - (5*I)/32)*d^(5
/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + (((5*I)/8)*d^2*Sqrt[
d*Tan[e + f*x]])/(a^2*f*(1 + I*Tan[e + f*x])) - (d*(d*Tan[e + f*x])^(3/2))/(4*f*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.427365, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3558, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}+\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9/16 + (5*I)/16)*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/16 + (5*I
)/16)*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/32 - (5*I)/32)*d^(5/2)
*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + ((9/32 - (5*I)/32)*d^(5
/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + (((5*I)/8)*d^2*Sqrt[
d*Tan[e + f*x]])/(a^2*f*(1 + I*Tan[e + f*x])) - (d*(d*Tan[e + f*x])^(3/2))/(4*f*(a + I*a*Tan[e + f*x])^2)

Rule 3558

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

Rule 3595

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

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}-\frac{\int \frac{\sqrt{d \tan (e+f x)} \left (-\frac{3 a d^2}{2}+\frac{7}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int \frac{-\frac{5}{2} i a^2 d^3-\frac{9}{2} a^2 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4}\\ &=\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{5}{2} i a^2 d^4-\frac{9}{2} a^2 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 f}\\ &=\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}+\frac{\left (\left (\frac{9}{16}-\frac{5 i}{16}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2}\right ) \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 )}{\sqrt{2} a^2 f}+-\frac{\left (\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2}\right ) \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 )}{\sqrt{2} a^2 f}+-\frac{\left (\left (\frac{9}{32}+\frac{5 i}{32}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}+-\frac{\left (\left (\frac{9}{32}+\frac{5 i}{32}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}\\ &=\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.05158, size = 231, normalized size = 0.77 \[ -\frac{d^3 \sec ^3(e+f x) \left (5 i \sin (e+f x)+5 i \sin (3 (e+f x))-7 \cos (e+f x)+7 \cos (3 (e+f x))+(9+5 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(5+9 i) \sin ^{\frac{3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(9-5 i) \sqrt{\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 f (\tan (e+f x)-i)^2 \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*Sec[e + f*x]^3*(-7*Cos[e + f*x] + 7*Cos[3*(e + f*x)] + (5*I)*Sin[e + f*x] + (9 - 5*I)*Cos[2*(e + f*x)]*L
og[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (9 + 5*I)*ArcSin[Cos[e + f*x
] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sqrt[Sin[2*(e + f*x)]] + (5 + 9*I)*Log[Cos[e + f*x]
+ Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (5*I)*Sin[3*(e + f*x)]))/(32*a^2*f*Sqrt[d*Ta
n[e + f*x]]*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.049, size = 145, normalized size = 0.5 \begin{align*}{\frac{7\,{d}^{3}}{8\,f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\frac{5\,i}{8}}{d}^{4}}{f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{7\,{d}^{3}}{8\,f{a}^{2}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}-{\frac{{d}^{3}}{4\,f{a}^{2}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.43271, size = 1569, normalized size = 5.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log((-2*I*d^3*e^(2*I*f*x + 2*I*e) + (8*I*a^2*f*e
^(2*I*f*x + 2*I*e) + 8*I*a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I*d
^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log((-2*I*d^
3*e^(2*I*f*x + 2*I*e) + (-8*I*a^2*f*e^(2*I*f*x + 2*I*e) - 8*I*a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^
(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 4*a^2*f*sqrt(-49/64*I*d^5/(a^4
*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(7*d^3 + 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*
e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^2*f)) + 4*a^2*f*sqr
t(-49/64*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(7*d^3 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*
d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^
2*f)) - (6*I*d^2*e^(4*I*f*x + 4*I*e) + 5*I*d^2*e^(2*I*f*x + 2*I*e) - I*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I
*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.16374, size = 274, normalized size = 0.91 \begin{align*} -\frac{1}{8} \, d^{3}{\left (-\frac{7 i \, \sqrt{2} \arctan \left (-\frac{16 i \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a^{2} \sqrt{d} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 i \, \sqrt{2} \arctan \left (-\frac{16 i \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{-8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a^{2} \sqrt{d} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{7 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 5 i \, \sqrt{d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*d^3*(-7*I*sqrt(2)*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*
sqrt(d)))/(a^2*sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - 2*I*sqrt(2)*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*
sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) - (7*sqrt(d*tan(f*x + e))
*d*tan(f*x + e) - 5*I*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^2*a^2*f))