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

Optimal. Leaf size=152 \[ \frac {8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {8 a^3 d \sqrt {d \tan (e+f x)}}{f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{5/2}}{7 d f} \]

[Out]

8*(-1)^(1/4)*a^3*d^(3/2)*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f+8*a^3*d*(d*tan(f*x+e))^(1/2)/f+8/3*
I*a^3*(d*tan(f*x+e))^(3/2)/f-32/35*a^3*(d*tan(f*x+e))^(5/2)/d/f-2/7*(d*tan(f*x+e))^(5/2)*(a^3+I*a^3*tan(f*x+e)
)/d/f

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Rubi [A]  time = 0.27, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3556, 3592, 3528, 3533, 205} \[ \frac {8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {8 a^3 d \sqrt {d \tan (e+f x)}}{f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{5/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + (8*a^3*d*Sqrt[d*Tan[e + f*x]]
)/f + (((8*I)/3)*a^3*(d*Tan[e + f*x])^(3/2))/f - (32*a^3*(d*Tan[e + f*x])^(5/2))/(35*d*f) - (2*(d*Tan[e + f*x]
)^(5/2)*(a^3 + I*a^3*Tan[e + f*x]))/(7*d*f)

Rule 205

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

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 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

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]

Rubi steps

\begin {align*} \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3 \, dx &=-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac {(2 a) \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x)) (6 a d+8 i a d \tan (e+f x)) \, dx}{7 d}\\ &=-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac {(2 a) \int (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 (d \tan (e+f x))^{3/2}}{3 f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac {(2 a) \int \sqrt {d \tan (e+f x)} \left (-14 i a^2 d^2+14 a^2 d^2 \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac {8 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac {(2 a) \int \frac {-14 a^2 d^3-14 i a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{7 d}\\ &=\frac {8 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac {\left (112 a^5 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{-14 a^2 d^4+14 i a^2 d^3 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {8 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}\\ \end {align*}

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Mathematica [A]  time = 3.18, size = 138, normalized size = 0.91 \[ \frac {a^3 d \sqrt {d \tan (e+f x)} \left (\sqrt {i \tan (e+f x)} \sec ^3(e+f x) (95 i \sin (e+f x)+155 i \sin (3 (e+f x))+1197 \cos (e+f x)+483 \cos (3 (e+f x)))-1680 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{210 f \sqrt {i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*d*(-1680*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + Sec[e + f*x]^3*(1197*Cos[e
+ f*x] + 483*Cos[3*(e + f*x)] + (95*I)*Sin[e + f*x] + (155*I)*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*
Tan[e + f*x]])/(210*f*Sqrt[I*Tan[e + f*x]])

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fricas [B]  time = 0.60, size = 423, normalized size = 2.78 \[ -\frac {105 \, \sqrt {-\frac {64 i \, a^{6} d^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {-\frac {64 i \, a^{6} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d}\right ) - 105 \, \sqrt {-\frac {64 i \, a^{6} d^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt {-\frac {64 i \, a^{6} d^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d}\right ) - 16 \, {\left (319 \, a^{3} d e^{\left (6 i \, f x + 6 i \, e\right )} + 646 \, a^{3} d e^{\left (4 i \, f x + 4 i \, e\right )} + 551 \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + 164 \, a^{3} d\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{420 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(105*sqrt(-64*I*a^6*d^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e)
+ f)*log(1/4*(-8*I*a^3*d^2*e^(2*I*f*x + 2*I*e) + sqrt(-64*I*a^6*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I
*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*d)) - 105*sqrt(-64*I*a^6*d
^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d^2*
e^(2*I*f*x + 2*I*e) - sqrt(-64*I*a^6*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d
)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*d)) - 16*(319*a^3*d*e^(6*I*f*x + 6*I*e) + 646*a^3*d*e^
(4*I*f*x + 4*I*e) + 551*a^3*d*e^(2*I*f*x + 2*I*e) + 164*a^3*d)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f
*x + 2*I*e) + 1)))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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giac [A]  time = 1.93, size = 194, normalized size = 1.28 \[ -\frac {1}{105} \, {\left (\frac {840 i \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {30 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{3} + 126 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{2} - 280 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right ) - 840 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6}}{d^{21} f^{7}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(840*I*sqrt(2)*a^3*sqrt(d)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sq
rt(d^2)*sqrt(d)))/(f*(I*d/sqrt(d^2) + 1)) + (30*I*sqrt(d*tan(f*x + e))*a^3*d^21*f^6*tan(f*x + e)^3 + 126*sqrt(
d*tan(f*x + e))*a^3*d^21*f^6*tan(f*x + e)^2 - 280*I*sqrt(d*tan(f*x + e))*a^3*d^21*f^6*tan(f*x + e) - 840*sqrt(
d*tan(f*x + e))*a^3*d^21*f^6)/(d^21*f^7))*d

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maple [B]  time = 0.26, size = 426, normalized size = 2.80 \[ -\frac {2 i a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 f \,d^{2}}-\frac {6 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 d f}+\frac {8 i a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}+\frac {8 a^{3} d \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {a^{3} d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{f}-\frac {2 a^{3} d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}+\frac {2 a^{3} d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}-\frac {i a^{3} d^{2} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 i a^{3} d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {2 i a^{3} d^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*I/f*a^3/d^2*(d*tan(f*x+e))^(7/2)-6/5*a^3*(d*tan(f*x+e))^(5/2)/d/f+8/3*I*a^3*(d*tan(f*x+e))^(3/2)/f+8*a^3*
d*(d*tan(f*x+e))^(1/2)/f-1/f*a^3*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/f*a^3*d*(d^2)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+2/f*a^3*d*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1
/4)*(d*tan(f*x+e))^(1/2)+1)-I/f*a^3*d^2/(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*I/f*a^3*d^2/(d^2)^
(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+2*I/f*a^3*d^2/(d^2)^(1/4)*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 = 0.47, size = 238, normalized size = 1.57 \[ -\frac {105 \, a^{3} d^{3} {\left (\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {2 \, {\left (-15 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} - 63 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d + 140 i \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{2} + 420 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{3}\right )}}{d}}{105 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(105*a^3*d^3*((2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/
sqrt(d) + (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) -
(I - 1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + (I - 1)*sqrt(2)*log(d
*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 2*(-15*I*(d*tan(f*x + e))^(7/2)*a^3 - 63*
(d*tan(f*x + e))^(5/2)*a^3*d + 140*I*(d*tan(f*x + e))^(3/2)*a^3*d^2 + 420*sqrt(d*tan(f*x + e))*a^3*d^3)/d)/(d*
f)

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mupad [B]  time = 4.77, size = 119, normalized size = 0.78 \[ \frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,8{}\mathrm {i}}{3\,f}-\frac {6\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}-\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}\,2{}\mathrm {i}}{7\,d^2\,f}+\frac {8\,a^3\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {\sqrt {16{}\mathrm {i}}\,a^3\,{\left (-d\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*(d*tan(e + f*x))^(3/2)*8i)/(3*f) - (6*a^3*(d*tan(e + f*x))^(5/2))/(5*d*f) - (a^3*(d*tan(e + f*x))^(7/2)*2
i)/(7*d^2*f) + (8*a^3*d*(d*tan(e + f*x))^(1/2))/f - (16i^(1/2)*a^3*(-d)^(3/2)*atan((16i^(1/2)*(d*tan(e + f*x))
^(1/2))/(4*(-d)^(1/2)))*2i)/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx + \int \left (- 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a**3*(Integral(I*(d*tan(e + f*x))**(3/2), x) + Integral(-3*(d*tan(e + f*x))**(3/2)*tan(e + f*x), x) + Integ
ral((d*tan(e + f*x))**(3/2)*tan(e + f*x)**3, x) + Integral(-3*I*(d*tan(e + f*x))**(3/2)*tan(e + f*x)**2, x))

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