3.1000 \(\int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=113 \[ \frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i a^{3/2} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]

[Out]

-I*a^(3/2)*c^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f+1/2*a*c*(a+I*a*
tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.14, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3523, 38, 63, 217, 203} \[ \frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i a^{3/2} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a^(3/2)*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (
a*c*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*f)

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3523

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

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \sqrt {a+i a x} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {\left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {\left (i a c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f}\\ &=\frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {\left (i a c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f}\\ &=-\frac {i a^{3/2} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}\\ \end {align*}

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Mathematica [A]  time = 3.87, size = 133, normalized size = 1.18 \[ \frac {a c^2 e^{-2 i e} \left (\cos \left (\frac {3 e}{2}\right )+i \sin \left (\frac {3 e}{2}\right )\right ) \sqrt {a+i a \tan (e+f x)} \left (\cos \left (\frac {e}{2}+f x\right )-i \sin \left (\frac {e}{2}+f x\right )\right ) \left (\tan (e+f x) \sec (e+f x)-2 i \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{2 \sqrt {2} f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2*(Cos[(3*e)/2] + I*Sin[(3*e)/2])*(Cos[e/2 + f*x] - I*Sin[e/2 + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*((-2*I)*
ArcTan[E^(I*(e + f*x))] + Sec[e + f*x]*Tan[e + f*x]))/(2*Sqrt[2]*E^((2*I)*e)*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]
*f)

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fricas [B]  time = 0.47, size = 356, normalized size = 3.15 \[ -\frac {\sqrt {\frac {a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {8 \, {\left (a c e^{\left (3 i \, f x + 3 i \, e\right )} + a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{3} c^{3}}{f^{2}}} {\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{a c e^{\left (2 i \, f x + 2 i \, e\right )} + a c}\right ) - \sqrt {\frac {a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {8 \, {\left (a c e^{\left (3 i \, f x + 3 i \, e\right )} + a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{3} c^{3}}{f^{2}}} {\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{a c e^{\left (2 i \, f x + 2 i \, e\right )} + a c}\right ) - 2 \, {\left (-2 i \, a c e^{\left (3 i \, f x + 3 i \, e\right )} + 2 i \, a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(a^3*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(a*c*e^(3*I*f*x + 3*I*e) + a*c*e^(I*f*x + I*e))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(a^3*c^3/f^2)*(4*I*f*e^(2*I*f*x + 2*I*e
) - 4*I*f))/(a*c*e^(2*I*f*x + 2*I*e) + a*c)) - sqrt(a^3*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(a*c*e^(3*
I*f*x + 3*I*e) + a*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sq
rt(a^3*c^3/f^2)*(-4*I*f*e^(2*I*f*x + 2*I*e) + 4*I*f))/(a*c*e^(2*I*f*x + 2*I*e) + a*c)) - 2*(-2*I*a*c*e^(3*I*f*
x + 3*I*e) + 2*I*a*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(f*
e^(2*I*f*x + 2*I*e) + f)

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giac [B]  time = 40.95, size = 307, normalized size = 2.72 \[ -\frac {7 \, {\left ({\left (a^{2} c - a c\right )} \sqrt {-a c} e^{\left (9 i \, f x + 9 i \, e\right )} + 6 \, {\left (a^{2} c - a c\right )} \sqrt {-a c} e^{\left (7 i \, f x + 7 i \, e\right )} + 12 \, {\left (a^{2} c - a c\right )} \sqrt {-a c} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 \, {\left (a^{2} c - a c\right )} \sqrt {-a c} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (a^{2} c - a c\right )} \sqrt {-a c} e^{\left (i \, f x + i \, e\right )}\right )}}{4 \, {\left ({\left (a - 1\right )} f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a - 1\right )} f\right )}} - \frac {i \, {\left (4 \, a^{\frac {3}{2}} c^{\frac {3}{2}} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {3 \, a^{\frac {3}{2}} c^{\frac {3}{2}} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{\frac {3}{2}} c^{\frac {3}{2}} e^{\left (i \, f x + i \, e\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}\right )}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/4*((a^2*c - a*c)*sqrt(-a*c)*e^(9*I*f*x + 9*I*e) + 6*(a^2*c - a*c)*sqrt(-a*c)*e^(7*I*f*x + 7*I*e) + 12*(a^2*
c - a*c)*sqrt(-a*c)*e^(5*I*f*x + 5*I*e) + 10*(a^2*c - a*c)*sqrt(-a*c)*e^(3*I*f*x + 3*I*e) + 3*(a^2*c - a*c)*sq
rt(-a*c)*e^(I*f*x + I*e))/((a - 1)*f*e^(10*I*f*x + 10*I*e) + 5*(a - 1)*f*e^(8*I*f*x + 8*I*e) + 10*(a - 1)*f*e^
(6*I*f*x + 6*I*e) + 10*(a - 1)*f*e^(4*I*f*x + 4*I*e) + 5*(a - 1)*f*e^(2*I*f*x + 2*I*e) + (a - 1)*f) - 1/4*I*(4
*a^(3/2)*c^(3/2)*arctan(e^(I*f*x + I*e)) - (3*a^(3/2)*c^(3/2)*e^(3*I*f*x + 3*I*e) + a^(3/2)*c^(3/2)*e^(I*f*x +
I*e))/(e^(2*I*f*x + 2*I*e) + 1)^2)/f

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maple [A]  time = 0.23, size = 128, normalized size = 1.13 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c a \left (a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )+\tan \left (f x +e \right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c*a*(a*c*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^
2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))+tan(f*x+e)*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2)
)^(1/2)/(c*a)^(1/2)

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maxima [B]  time = 0.94, size = 643, normalized size = 5.69 \[ -\frac {{\left (4 \, a c \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 4 \, a c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4 i \, a c \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 4 i \, a c \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (2 \, a c \cos \left (4 \, f x + 4 \, e\right ) + 4 \, a c \cos \left (2 \, f x + 2 \, e\right ) + 2 i \, a c \sin \left (4 \, f x + 4 \, e\right ) + 4 i \, a c \sin \left (2 \, f x + 2 \, e\right ) + 2 \, a c\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + {\left (2 \, a c \cos \left (4 \, f x + 4 \, e\right ) + 4 \, a c \cos \left (2 \, f x + 2 \, e\right ) + 2 i \, a c \sin \left (4 \, f x + 4 \, e\right ) + 4 i \, a c \sin \left (2 \, f x + 2 \, e\right ) + 2 \, a c\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - {\left (-i \, a c \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, a c \cos \left (2 \, f x + 2 \, e\right ) + a c \sin \left (4 \, f x + 4 \, e\right ) + 2 \, a c \sin \left (2 \, f x + 2 \, e\right ) - i \, a c\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - {\left (i \, a c \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, a c \cos \left (2 \, f x + 2 \, e\right ) - a c \sin \left (4 \, f x + 4 \, e\right ) - 2 \, a c \sin \left (2 \, f x + 2 \, e\right ) + i \, a c\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right )\right )} \sqrt {a} \sqrt {c}}{f {\left (-4 i \, \cos \left (4 \, f x + 4 \, e\right ) - 8 i \, \cos \left (2 \, f x + 2 \, e\right ) + 4 \, \sin \left (4 \, f x + 4 \, e\right ) + 8 \, \sin \left (2 \, f x + 2 \, e\right ) - 4 i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*a*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 4*I*a*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*I*a*c*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) + (2*a*c*cos(4*f*x + 4*e) + 4*a*c*cos(2*f*x + 2*e) + 2*I*a*c*sin(4*f*x + 4*e) + 4*
I*a*c*sin(2*f*x + 2*e) + 2*a*c)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (2*a*c*cos(4*f*x + 4*e) + 4*a*c*cos(2*f*x + 2*e) + 2*I*a*c*sin(4*f
*x + 4*e) + 4*I*a*c*sin(2*f*x + 2*e) + 2*a*c)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -s
in(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (-I*a*c*cos(4*f*x + 4*e) - 2*I*a*c*cos(2*f*x + 2*e)
+ a*c*sin(4*f*x + 4*e) + 2*a*c*sin(2*f*x + 2*e) - I*a*c)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 1) - (I*a*c*cos(4*f*x + 4*e) + 2*I*a*c*cos(2*f*x + 2*e) - a*c*sin(4*f*x + 4*e) - 2*a*c*sin(2*f*x
+ 2*e) + I*a*c)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-4*I*
cos(4*f*x + 4*e) - 8*I*cos(2*f*x + 2*e) + 4*sin(4*f*x + 4*e) + 8*sin(2*f*x + 2*e) - 4*I))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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