3.1113 \(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=216 \[ \frac{4 a^3 (-10 d+i c) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{8 a^3 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

((-8*I)*a^3*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((8*I)*a^3*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (8*a^3*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((8*I)/5)*a^3*(c + d*Tan[e + f*x
])^(5/2))/f + (4*a^3*(I*c - 10*d)*(c + d*Tan[e + f*x])^(7/2))/(63*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c +
d*Tan[e + f*x])^(7/2))/(9*d*f)

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Rubi [A]  time = 0.633928, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3556, 3592, 3528, 3537, 63, 208} \[ \frac{4 a^3 (-10 d+i c) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{8 a^3 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-8*I)*a^3*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((8*I)*a^3*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (8*a^3*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((8*I)/5)*a^3*(c + d*Tan[e + f*x
])^(5/2))/f + (4*a^3*(I*c - 10*d)*(c + d*Tan[e + f*x])^(7/2))/(63*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c +
d*Tan[e + f*x])^(7/2))/(9*d*f)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3592

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

Rule 3528

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

Rule 3537

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

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 208

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

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(2 a) \int (a+i a \tan (e+f x)) (a (i c+8 d)+a (c+10 i d) \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx}{9 d}\\ &=\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(2 a) \int (c+d \tan (e+f x))^{5/2} \left (18 a^2 d+18 i a^2 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(2 a) \int (c+d \tan (e+f x))^{3/2} \left (18 a^2 (c-i d) d+18 a^2 d (i c+d) \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(2 a) \int \sqrt{c+d \tan (e+f x)} \left (18 a^2 (c-i d)^2 d+18 i a^2 (c-i d)^2 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(2 a) \int \frac{18 a^2 (c-i d)^3 d-18 a^2 d (i c+d)^3 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{9 d}\\ &=\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{\left (72 i a^5 (c-i d)^6 d\right ) \operatorname{Subst}\left (\int \frac{1}{\left (324 a^4 d^2 (i c+d)^6+18 a^2 (c-i d)^3 d x\right ) \sqrt{c-\frac{x}{18 a^2 (i c+d)^3}}} \, dx,x,-18 a^2 d (i c+d)^3 \tan (e+f x)\right )}{f}\\ &=\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac{\left (2592 a^7 (c-i d)^9 d\right ) \operatorname{Subst}\left (\int \frac{1}{324 a^4 c (c-i d)^3 d (i c+d)^3+324 a^4 d^2 (i c+d)^6-324 a^4 (c-i d)^3 d (i c+d)^3 x^2} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{8 i a^3 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}\\ \end{align*}

Mathematica [B]  time = 11.0489, size = 528, normalized size = 2.44 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^3 \left (\sec (e) \left (-\frac{2}{315} \sin (3 e)-\frac{2}{315} i \cos (3 e)\right ) \sec ^2(e+f x) \left (75 c^2 \cos (e)+95 c d \sin (e)-405 i c d \cos (e)-135 i d^2 \sin (e)-322 d^2 \cos (e)\right )+\sec (e) \left (\frac{2 \cos (3 e)}{315 d}-\frac{2 i \sin (3 e)}{315 d}\right ) \sec (e+f x) \left (-405 c^2 d \sin (f x)-5 i c^3 \sin (f x)+1019 i c d^2 \sin (f x)+555 d^3 \sin (f x)\right )+\sec (e) \left (\frac{2 \cos (3 e)}{315 d^2}-\frac{2 i \sin (3 e)}{315 d^2}\right ) \left (-405 c^2 d^2 \sin (e)+2007 i c^2 d^2 \cos (e)-5 i c^3 d \sin (e)-135 c^3 d \cos (e)+10 i c^4 \cos (e)+1019 i c d^3 \sin (e)+3345 c d^3 \cos (e)+555 d^4 \sin (e)-1547 i d^4 \cos (e)\right )+\sec (e) \left (\frac{2}{63} \cos (3 e)-\frac{2}{63} i \sin (3 e)\right ) \sec ^3(e+f x) \left (-27 d^2 \sin (f x)-19 i c d \sin (f x)\right )+\left (-\frac{2}{9} d^2 \sin (3 e)-\frac{2}{9} i d^2 \cos (3 e)\right ) \sec ^4(e+f x)\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^3}-\frac{8 i e^{-3 i e} (c-i d)^{5/2} \cos ^3(e+f x) (a+i a \tan (e+f x))^3 \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-8*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c -
I*d]]*Cos[e + f*x]^3*(a + I*a*Tan[e + f*x])^3)/(E^((3*I)*e)*f*(Cos[f*x] + I*Sin[f*x])^3) + (Cos[e + f*x]^3*(Se
c[e]*Sec[e + f*x]^2*(75*c^2*Cos[e] - (405*I)*c*d*Cos[e] - 322*d^2*Cos[e] + 95*c*d*Sin[e] - (135*I)*d^2*Sin[e])
*(((-2*I)/315)*Cos[3*e] - (2*Sin[3*e])/315) + Sec[e]*((10*I)*c^4*Cos[e] - 135*c^3*d*Cos[e] + (2007*I)*c^2*d^2*
Cos[e] + 3345*c*d^3*Cos[e] - (1547*I)*d^4*Cos[e] - (5*I)*c^3*d*Sin[e] - 405*c^2*d^2*Sin[e] + (1019*I)*c*d^3*Si
n[e] + 555*d^4*Sin[e])*((2*Cos[3*e])/(315*d^2) - (((2*I)/315)*Sin[3*e])/d^2) + Sec[e + f*x]^4*(((-2*I)/9)*d^2*
Cos[3*e] - (2*d^2*Sin[3*e])/9) + Sec[e]*Sec[e + f*x]^3*((2*Cos[3*e])/63 - ((2*I)/63)*Sin[3*e])*((-19*I)*c*d*Si
n[f*x] - 27*d^2*Sin[f*x]) + Sec[e]*Sec[e + f*x]*((2*Cos[3*e])/(315*d) - (((2*I)/315)*Sin[3*e])/d)*((-5*I)*c^3*
Sin[f*x] - 405*c^2*d*Sin[f*x] + (1019*I)*c*d^2*Sin[f*x] + 555*d^3*Sin[f*x]))*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x]
+ d*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[f*x] + I*Sin[f*x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.04, size = 2936, normalized size = 13.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x)

[Out]

8/5*I*a^3*(c+d*tan(f*x+e))^(5/2)/f+2/7*I/f*a^3/d^2*(c+d*tan(f*x+e))^(7/2)*c+2*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^3+4*I/f*a^3/(2
*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)
-2*c)^(1/2))*c^3-4/f*a^3*d^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^
(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c+4/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)
*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^3-6*I/f*a^3*d^2/(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1
/2))*c-12*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(
1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+12*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)
+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+2*I/f*a^3*d^4/(2*(c^2+d^2)^(1/2)+2*c)^(
1/2)/(c^2+d^2)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+4
*I/f*a^3/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e)
)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^4-4*I/f*a^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)
+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-2*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^3+6/f*a^3*d/(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2
))*c^2-12/f*a^3*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2)
)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+12/f*a^3*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)
+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-6/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(
d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2-6/7/f*a^3/d*(c+d*tan(
f*x+e))^(7/2)+8/3/f*a^3*d*(c+d*tan(f*x+e))^(3/2)+16/f*a^3*d*c*(c+d*tan(f*x+e))^(1/2)-2/f*a^3*d^3/(2*(c^2+d^2)^
(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))-4/f*a
^3*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d
^2)^(1/2)-2*c)^(1/2))+4/f*a^3*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*t
an(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+2/f*a^3*d^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(
c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))-2/9*I/f*a^3/d^2*(c+d*tan(f*x+e))^(9/2)-8*
I/f*a^3*d^2*(c+d*tan(f*x+e))^(1/2)+8/3*I/f*a^3*(c+d*tan(f*x+e))^(3/2)*c+8*I/f*a^3*c^2*(c+d*tan(f*x+e))^(1/2)+6
*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)+(c^2+d^2)^(1/2))*c+4*I/f*a^3*d^4/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^
(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2*I/f*a^3*d^4/(2*(c^2+d^2)^(1/2)+2*c)^(1/2
)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))+4/f*
a^3*d^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-
d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c-4*I/f*a^3*d^4/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+
d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-8/f*a^3*d/(c^2+d^2)^(1/2)/(2*(c
^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*
c)^(1/2))*c^3-4*I/f*a^3/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2
+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^4-2*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(
1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^4+2*I/f*a^3/(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e
)-c-(c^2+d^2)^(1/2))*c^4-8/f*a^3*d^3/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^
(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+8/f*a^3*d/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(
1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))
*c^3+8/f*a^3*d^3/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*ta
n(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-4/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln(d
*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^3

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 7.90456, size = 2770, normalized size = 12.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/1260*(315*(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f*e^(6*I*f*x + 6*I*e) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*
e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(64*a^6*c^5 - 320*I*a^6*c^4*d - 640*a^6*c^3*d^2 + 640*I*a^6*c^2*d^3 + 320*a
^6*c*d^4 - 64*I*a^6*d^5)/f^2)*log((8*a^3*c^3 - 16*I*a^3*c^2*d - 8*a^3*c*d^2 - (I*f*e^(2*I*f*x + 2*I*e) + I*f)*
sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(64*a^6*c^5 - 320*I*a^6*c^4*d
- 640*a^6*c^3*d^2 + 640*I*a^6*c^2*d^3 + 320*a^6*c*d^4 - 64*I*a^6*d^5)/f^2) + (8*a^3*c^3 - 24*I*a^3*c^2*d - 24*
a^3*c*d^2 + 8*I*a^3*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(4*a^3*c^2 - 8*I*a^3*c*d - 4*a^3*d^2)) - 31
5*(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f*e^(6*I*f*x + 6*I*e) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*e^(2*I*f*x
+ 2*I*e) + d^2*f)*sqrt(-(64*a^6*c^5 - 320*I*a^6*c^4*d - 640*a^6*c^3*d^2 + 640*I*a^6*c^2*d^3 + 320*a^6*c*d^4 -
64*I*a^6*d^5)/f^2)*log((8*a^3*c^3 - 16*I*a^3*c^2*d - 8*a^3*c*d^2 - (-I*f*e^(2*I*f*x + 2*I*e) - I*f)*sqrt(((c -
I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(64*a^6*c^5 - 320*I*a^6*c^4*d - 640*a^6*
c^3*d^2 + 640*I*a^6*c^2*d^3 + 320*a^6*c*d^4 - 64*I*a^6*d^5)/f^2) + (8*a^3*c^3 - 24*I*a^3*c^2*d - 24*a^3*c*d^2
+ 8*I*a^3*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(4*a^3*c^2 - 8*I*a^3*c*d - 4*a^3*d^2)) - (80*I*a^3*c^
4 - 1040*a^3*c^3*d + 12816*I*a^3*c^2*d^2 + 18608*a^3*c*d^3 - 7936*I*a^3*d^4 + (80*I*a^3*c^4 - 1120*a^3*c^3*d +
19296*I*a^3*c^2*d^2 + 34912*a^3*c*d^3 - 16816*I*a^3*d^4)*e^(8*I*f*x + 8*I*e) + (320*I*a^3*c^4 - 4400*a^3*c^3*
d + 68304*I*a^3*c^2*d^2 + 107344*a^3*c*d^3 - 43760*I*a^3*d^4)*e^(6*I*f*x + 6*I*e) + (480*I*a^3*c^4 - 6480*a^3*
c^3*d + 91536*I*a^3*c^2*d^2 + 134640*a^3*c*d^3 - 58128*I*a^3*d^4)*e^(4*I*f*x + 4*I*e) + (320*I*a^3*c^4 - 4240*
a^3*c^3*d + 55344*I*a^3*c^2*d^2 + 80816*a^3*c*d^3 - 34640*I*a^3*d^4)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2
*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f*e^(6*I*f*x + 6*I*e
) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.74856, size = 572, normalized size = 2.65 \begin{align*} \frac{2 \,{\left (16 i \, a^{3} c^{3} + 48 \, a^{3} c^{2} d - 48 i \, a^{3} c d^{2} - 16 \, a^{3} d^{3}\right )} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{\sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} f{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{70 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}} a^{3} d^{16} f^{8} - 90 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{3} c d^{16} f^{8} + 270 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{3} d^{17} f^{8} - 504 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} d^{18} f^{8} - 840 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} c d^{18} f^{8} - 2520 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} c^{2} d^{18} f^{8} - 840 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} d^{19} f^{8} - 5040 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} c d^{19} f^{8} + 2520 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} d^{20} f^{8}}{315 \, d^{18} f^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2*(16*I*a^3*c^3 + 48*a^3*c^2*d - 48*I*a^3*c*d^2 - 16*a^3*d^3)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2
+ d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt
(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1
)) - 1/315*(70*I*(d*tan(f*x + e) + c)^(9/2)*a^3*d^16*f^8 - 90*I*(d*tan(f*x + e) + c)^(7/2)*a^3*c*d^16*f^8 + 27
0*(d*tan(f*x + e) + c)^(7/2)*a^3*d^17*f^8 - 504*I*(d*tan(f*x + e) + c)^(5/2)*a^3*d^18*f^8 - 840*I*(d*tan(f*x +
e) + c)^(3/2)*a^3*c*d^18*f^8 - 2520*I*sqrt(d*tan(f*x + e) + c)*a^3*c^2*d^18*f^8 - 840*(d*tan(f*x + e) + c)^(3
/2)*a^3*d^19*f^8 - 5040*sqrt(d*tan(f*x + e) + c)*a^3*c*d^19*f^8 + 2520*I*sqrt(d*tan(f*x + e) + c)*a^3*d^20*f^8
)/(d^18*f^9)