3.2.58 \(\int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx\) [158]

Optimal. Leaf size=129 \[ \frac {8 (-1)^{3/4} a^3 \sqrt {d} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {8 i a^3 \sqrt {d \tan (e+f x)}}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3637, 3673, 3609, 3614, 211} \begin {gather*} \frac {8 (-1)^{3/4} a^3 \sqrt {d} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac {8 i a^3 \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))^{3/2}}{5 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3609

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 3614

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 3637

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 3673

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 \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3 \, dx &=-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac {(2 a) \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x)) (4 a d+6 i a d \tan (e+f x)) \, dx}{5 d}\\ &=-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac {(2 a) \int \sqrt {d \tan (e+f x)} \left (10 a^2 d+10 i a^2 d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac {8 i a^3 \sqrt {d \tan (e+f x)}}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}+\frac {(2 a) \int \frac {-10 i a^2 d^2+10 a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac {8 i a^3 \sqrt {d \tan (e+f x)}}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}-\frac {\left (80 a^5 d^3\right ) \text {Subst}\left (\int \frac {1}{-10 i a^2 d^3-10 a^2 d^2 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {8 (-1)^{3/4} a^3 \sqrt {d} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {8 i a^3 \sqrt {d \tan (e+f x)}}{f}-\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f}\\ \end {align*}

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Mathematica [A]
time = 3.32, size = 122, normalized size = 0.95 \begin {gather*} \frac {i a^3 \left (-40 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+\sec ^2(e+f x) (19+21 \cos (2 (e+f x))+5 i \sin (2 (e+f x))) \sqrt {i \tan (e+f x)}\right ) \sqrt {d \tan (e+f x)}}{5 f \sqrt {i \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*a^3*(-40*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + Sec[e + f*x]^2*(19 + 21*
Cos[2*(e + f*x)] + (5*I)*Sin[2*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[I*Tan[e + f*x]]
)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (107 ) = 214\).
time = 0.16, size = 327, normalized size = 2.53

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 i d^{2} \sqrt {d \tan \left (f x +e \right )}-4 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(327\)
default	\(\frac {2 a^{3} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 i d^{2} \sqrt {d \tan \left (f x +e \right )}-4 d^{3} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^3/d^2*(-1/5*I*(d*tan(f*x+e))^(5/2)-d*(d*tan(f*x+e))^(3/2)+4*I*d^2*(d*tan(f*x+e))^(1/2)-4*d^3*(1/8*I/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*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))-1/8/(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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (110) = 220\).
time = 0.58, size = 229, normalized size = 1.78 \begin {gather*} \frac {5 \, a^{3} d^{2} {\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 (-i \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} - 5 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + 20 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}\right )}}{d}}{5 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (110) = 220\).
time = 0.38, size = 376, normalized size = 2.91 \begin {gather*} \frac {5 \, \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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}}\right ) - 5 \, \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {64 i \, a^{6} d}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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}}\right ) - 16 \, {\left (-13 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 19 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\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}}}{20 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5*sqrt(64*I*a^6*d/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d*e^(2*I*
f*x + 2*I*e) + sqrt(64*I*a^6*d/f^2)*(I*f*e^(2*I*f*x + 2*I*e) + I*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) - 5*sqrt(64*I*a^6*d/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I
*f*x + 2*I*e) + f)*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + sqrt(64*I*a^6*d/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) -
I*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) - 16*(-13*I*a
^3*e^(4*I*f*x + 4*I*e) - 19*I*a^3*e^(2*I*f*x + 2*I*e) - 8*I*a^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I
*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a**3*(Integral(I*sqrt(d*tan(e + f*x)), x) + Integral(-3*sqrt(d*tan(e + f*x))*tan(e + f*x), x) + Integral(sq
rt(d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-3*I*sqrt(d*tan(e + f*x))*tan(e + f*x)**2, x))

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Giac [A]
time = 0.66, size = 161, normalized size = 1.25 \begin {gather*} \frac {8 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right ) - 20 i \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{10} f^{4}\right )}}{5 \, d^{10} f^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*sqrt(2)*a^3*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(
d)))/(f*(I*d/sqrt(d^2) + 1)) - 2/5*(I*sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x + e)^2 + 5*sqrt(d*tan(f*x + e)
)*a^3*d^10*f^4*tan(f*x + e) - 20*I*sqrt(d*tan(f*x + e))*a^3*d^10*f^4)/(d^10*f^5)

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Mupad [B]
time = 4.44, size = 96, normalized size = 0.74 \begin {gather*} \frac {a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,8{}\mathrm {i}}{f}-\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{d\,f}-\frac {a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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