3.1119 \(\int \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=126 \[ \frac {4 a^3 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3556, 3592, 3537, 63, 208} \[ \frac {4 a^3 (-4 d+i c) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 \frac {(a+i a \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {(2 a) \int \frac {(a+i a \tan (e+f x)) (a (i c+2 d)+a (c+4 i d) \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {(2 a) \int \frac {6 a^2 d+6 i a^2 d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\left (24 i a^5 d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {i x}{6 a^2}} \left (-36 a^4 d^2+6 a^2 d x\right )} \, dx,x,6 i a^2 d \tan (e+f x)\right )}{f}\\ &=\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\left (288 a^7 d\right ) \operatorname {Subst}\left (\int \frac {1}{-36 i a^4 c d-36 a^4 d^2+36 i a^4 d x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{f}\\ &=-\frac {8 i a^3 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {4 a^3 (i c-4 d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{3 d f}\\ \end {align*}

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Mathematica [A]  time = 5.20, size = 168, normalized size = 1.33 \[ \frac {a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac {2 (\sin (3 e)+i \cos (3 e)) (2 c-d \tan (e+f x)+9 i d) \sqrt {c+d \tan (e+f x)}}{3 d^2}-\frac {8 i e^{-3 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I
)*(e + f*x)))]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*E^((3*I)*e)) + (2*(I*Cos[3*e] + Sin[3*e])*(2*c + (9*I)*d - d*Tan
[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [B]  time = 0.51, size = 427, normalized size = 3.39 \[ \frac {3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (i \, c + d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c + d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {-\frac {64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} {\left ({\left (-i \, c - d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c - d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) + {\left (16 i \, a^{3} c - 64 \, a^{3} d - 16 \, {\left (-i \, a^{3} c + 5 \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \]

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))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/
((I*c + d)*f^2))*((I*c + d)*f*e^(2*I*f*x + 2*I*e) + (I*c + d)*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)) + (8*a^3*c - 8*I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 3*(d^2*f
*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2
))*((-I*c - d)*f*e^(2*I*f*x + 2*I*e) + (-I*c - d)*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)) + (8*a^3*c - 8*I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) + (16*I*a^3*c - 64*a^3
*d - 16*(-I*a^3*c + 5*a^3*d)*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^(2*I*f*x + 2*I*e) + d^2*f)

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giac [B]  time = 0.94, size = 243, normalized size = 1.93 \[ \frac {32 i \, a^{3} \arctan \left (\frac {4 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-8 \, c + 8 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} d^{4} f^{2} - 6 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} c d^{4} f^{2} + 18 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} d^{5} f^{2}}{3 \, d^{6} f^{3}} \]

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))^(1/2),x, algorithm="giac")

[Out]

32*I*a^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*sqr
t(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)) - 1/3*(2*I*(d*tan(f*x + e) + c)^(3/2)*a^3*d^4*f^2
- 6*I*sqrt(d*tan(f*x + e) + c)*a^3*c*d^4*f^2 + 18*sqrt(d*tan(f*x + e) + c)*a^3*d^5*f^2)/(d^6*f^3)

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maple [B]  time = 0.30, size = 1748, normalized size = 13.87 \[ \text {result too large to display} \]

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))^(1/2),x)

[Out]

-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*I/f*a^3*d^2/((c^2+d^2)^(1/2)*c+c^2+d^2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a
rctan(((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))-6/f*a^3/d*(c+d*t
an(f*x+e))^(1/2)+4*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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*I/f*a^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arc
tan((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/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))-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))-2/3*I/f*a^3/d^2*(c+d*tan(f*x+e))^(3/2)+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))+4*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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/(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*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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+2*I/f*a^3/d^2*c*(c+d*tan(f*x+e))^(1/2)-
2/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+
d^2)^(1/2)*c+c^2+d^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^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2
)/((c^2+d^2)^(1/2)*c+c^2+d^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/((c^2+d^2)^(1/2)*c+c^2+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-4/f*a^3*d/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^
2+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^2-4/f*a^3*d^3/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+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))

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

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))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.55, size = 119, normalized size = 0.94 \[ -\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {-c+d\,1{}\mathrm {i}}}\right )\,8{}\mathrm {i}}{f\,\sqrt {-c+d\,1{}\mathrm {i}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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))**(1/2),x)

[Out]

-I*a**3*(Integral(I/sqrt(c + d*tan(e + f*x)), x) + Integral(-3*tan(e + f*x)/sqrt(c + d*tan(e + f*x)), x) + Int
egral(tan(e + f*x)**3/sqrt(c + d*tan(e + f*x)), x) + Integral(-3*I*tan(e + f*x)**2/sqrt(c + d*tan(e + f*x)), x
))

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