3.1107 \(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=181 \[ \frac{4 a^3 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 a^3 (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac{8 i a^3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

((-8*I)*a^3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (8*a^3*(I*c + d)*Sqrt[c + d*T
an[e + f*x]])/f + (((8*I)/3)*a^3*(c + d*Tan[e + f*x])^(3/2))/f + (4*a^3*(I*c - 8*d)*(c + d*Tan[e + f*x])^(5/2)
)/(35*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2))/(7*d*f)

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Rubi [A]  time = 0.516502, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{4 a^3 (-8 d+i c) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{8 a^3 (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac{8 i a^3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (8*a^3*(I*c + d)*Sqrt[c + d*T
an[e + f*x]])/f + (((8*I)/3)*a^3*(c + d*Tan[e + f*x])^(3/2))/f + (4*a^3*(I*c - 8*d)*(c + d*Tan[e + f*x])^(5/2)
)/(35*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2))/(7*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))^{3/2} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac{(2 a) \int (a+i a \tan (e+f x)) (a (i c+6 d)+a (c+8 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx}{7 d}\\ &=\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac{(2 a) \int (c+d \tan (e+f x))^{3/2} \left (14 a^2 d+14 i a^2 d \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac{(2 a) \int \sqrt{c+d \tan (e+f x)} \left (14 a^2 (c-i d) d+14 a^2 d (i c+d) \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac{8 a^3 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac{(2 a) \int \frac{14 a^2 (c-i d)^2 d+14 i a^2 (c-i d)^2 d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{7 d}\\ &=\frac{8 a^3 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}+\frac{\left (56 i a^5 (c-i d)^4 d\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{i x}{14 a^2 (c-i d)^2}} \left (-196 a^4 (c-i d)^4 d^2+14 a^2 (c-i d)^2 d x\right )} \, dx,x,14 i a^2 (c-i d)^2 d \tan (e+f x)\right )}{f}\\ &=\frac{8 a^3 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac{\left (1568 a^7 (c-i d)^6 d\right ) \operatorname{Subst}\left (\int \frac{1}{-196 i a^4 c (c-i d)^4 d-196 a^4 (c-i d)^4 d^2+196 i a^4 (c-i d)^4 d x^2} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 i a^3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{8 a^3 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{8 i a^3 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (i c-8 d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{7 d f}\\ \end{align*}

Mathematica [A]  time = 7.96996, size = 271, normalized size = 1.5 \[ \frac{a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac{(\sin (3 e)+i \cos (3 e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (d \left (-3 c^2+126 i c d+125 d^2\right ) \tan (e+f x)+\cos (2 (e+f x)) \left (d \left (-3 c^2+126 i c d+155 d^2\right ) \tan (e+f x)+63 i c^2 d+6 c^3+584 c d^2-483 i d^3\right )+63 i c^2 d+6 c^3+536 c d^2-357 i d^3\right )}{105 d^2}-8 i e^{-3 i e} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)
)))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^((3*I)*e) + (Sec[e + f*x]^2*(I*Cos[3*e] + Sin[3*e])*Sqrt[c +
d*Tan[e + f*x]]*(6*c^3 + (63*I)*c^2*d + 536*c*d^2 - (357*I)*d^3 + d*(-3*c^2 + (126*I)*c*d + 125*d^2)*Tan[e + f
*x] + Cos[2*(e + f*x)]*(6*c^3 + (63*I)*c^2*d + 584*c*d^2 - (483*I)*d^3 + d*(-3*c^2 + (126*I)*c*d + 155*d^2)*Ta
n[e + f*x])))/(105*d^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [B]  time = 0.048, size = 2535, normalized size = 14. \begin{align*} \text{result 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))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 4.16011, size = 2018, normalized size = 11.15 \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))^(3/2),x, algorithm="fricas")

[Out]

-1/420*(105*(d^2*f*e^(6*I*f*x + 6*I*e) + 3*d^2*f*e^(4*I*f*x + 4*I*e) + 3*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sq
rt(-(64*a^6*c^3 - 192*I*a^6*c^2*d - 192*a^6*c*d^2 + 64*I*a^6*d^3)/f^2)*log(-1/4*(8*I*a^3*c^2 + 8*a^3*c*d + (f*
e^(2*I*f*x + 2*I*e) + 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^3 - 192*I*a^6*c^2*d - 192*a^6*c*d^2 + 64*I*a^6*d^3)/f^2) + (8*I*a^3*c^2 + 16*a^3*c*d - 8*I*a^3*d^2)*e^(2*
I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a^3*c - a^3*d)) - 105*(d^2*f*e^(6*I*f*x + 6*I*e) + 3*d^2*f*e^(4*I*f*x
+ 4*I*e) + 3*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(64*a^6*c^3 - 192*I*a^6*c^2*d - 192*a^6*c*d^2 + 64*I*a^
6*d^3)/f^2)*log(-1/4*(8*I*a^3*c^2 + 8*a^3*c*d - (f*e^(2*I*f*x + 2*I*e) + 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^3 - 192*I*a^6*c^2*d - 192*a^6*c*d^2 + 64*I*a^6*d^3)/f^
2) + (8*I*a^3*c^2 + 16*a^3*c*d - 8*I*a^3*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a^3*c - a^3*d)) -
(48*I*a^3*c^3 - 480*a^3*c^2*d + 3664*I*a^3*c*d^2 + 2624*a^3*d^3 + (48*I*a^3*c^3 - 528*a^3*c^2*d + 5680*I*a^3*c
*d^2 + 5104*a^3*d^3)*e^(6*I*f*x + 6*I*e) + (144*I*a^3*c^3 - 1536*a^3*c^2*d + 14256*I*a^3*c*d^2 + 10336*a^3*d^3
)*e^(4*I*f*x + 4*I*e) + (144*I*a^3*c^3 - 1488*a^3*c^2*d + 12240*I*a^3*c*d^2 + 8816*a^3*d^3)*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^(6*I*f*x + 6*I*e) + 3*d
^2*f*e^(4*I*f*x + 4*I*e) + 3*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int c \sqrt{c + d \tan{\left (e + f x \right )}}\, dx + \int - 3 c \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int - 3 d \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int 3 i c \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int - i c \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int 3 i d \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int - i d \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx\right ) \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))**(3/2),x)

[Out]

a**3*(Integral(c*sqrt(c + d*tan(e + f*x)), x) + Integral(-3*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + I
ntegral(d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-3*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3,
x) + Integral(3*I*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-I*c*sqrt(c + d*tan(e + f*x))*tan(e +
f*x)**3, x) + Integral(3*I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-I*d*sqrt(c + d*tan(e +
f*x))*tan(e + f*x)**4, x))

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Giac [B]  time = 1.60025, size = 455, normalized size = 2.51 \begin{align*} \frac{2 \,{\left (16 i \, a^{3} c^{2} + 32 \, a^{3} c d - 16 i \, a^{3} d^{2}\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{30 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{3} d^{12} f^{6} - 42 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} c d^{12} f^{6} + 126 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} d^{13} f^{6} - 280 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} d^{14} f^{6} - 840 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} c d^{14} f^{6} - 840 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} d^{15} f^{6}}{105 \, d^{14} f^{7}} \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))^(3/2),x, algorithm="giac")

[Out]

2*(16*I*a^3*c^2 + 32*a^3*c*d - 16*I*a^3*d^2)*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)) - 1/105*(30*I*
(d*tan(f*x + e) + c)^(7/2)*a^3*d^12*f^6 - 42*I*(d*tan(f*x + e) + c)^(5/2)*a^3*c*d^12*f^6 + 126*(d*tan(f*x + e)
+ c)^(5/2)*a^3*d^13*f^6 - 280*I*(d*tan(f*x + e) + c)^(3/2)*a^3*d^14*f^6 - 840*I*sqrt(d*tan(f*x + e) + c)*a^3*
c*d^14*f^6 - 840*sqrt(d*tan(f*x + e) + c)*a^3*d^15*f^6)/(d^14*f^7)