3.1017 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=153 \[ \frac {6 i a^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac {6 i a^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(5/2)/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((6*I)*a^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[c]*f)
- ((2*I)*a*(a + I*a*Tan[e + f*x])^(3/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - ((3*I)*a^2*Sqrt[a + I*a*Tan[e + f*x]
]*Sqrt[c - I*c*Tan[e + f*x]])/(c*f)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [A]  time = 4.52, size = 155, normalized size = 1.01 \[ -\frac {2 i e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)} \left (3+2 e^{2 i (e+f x)}\right )-3 \left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )\right ) (a+i a \tan (e+f x))^{5/2}}{c f \sec ^{\frac {5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(5/2)/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((-2*I)*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))*(3
+ 2*E^((2*I)*(e + f*x))) - 3*(1 + E^((2*I)*(e + f*x)))*ArcTan[E^(I*(e + f*x))])*(a + I*a*Tan[e + f*x])^(5/2))/
(c*E^((3*I)*(e + f*x))*f*Sec[e + f*x]^(5/2))

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fricas [B]  time = 0.63, size = 339, normalized size = 2.22 \[ \frac {3 \, \sqrt {\frac {a^{5}}{c f^{2}}} c f \log \left (\frac {2 \, {\left (4 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (2 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, c f\right )} \sqrt {\frac {a^{5}}{c f^{2}}}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) - 3 \, \sqrt {\frac {a^{5}}{c f^{2}}} c f \log \left (\frac {2 \, {\left (4 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (-2 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c f\right )} \sqrt {\frac {a^{5}}{c f^{2}}}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) + {\left (-8 i \, a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} - 12 i \, a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(5/2)/sqrt(-I*c*tan(f*x + e) + c), x)

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maple [B]  time = 0.31, size = 299, normalized size = 1.95 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -6 i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-6 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )+5 \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{f c \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c*(3*I*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)^2*a*c-3*I*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))*a*c-6*I*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-6*ln((c*a*tan(f*x+e)+(c*a*(1+t
an(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c-(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+
e)^2+5*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^2/(c*a)^(1/2)

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maxima [B]  time = 0.57, size = 541, normalized size = 3.54 \[ \frac {{\left (6 \, {\left (a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{2}\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 ) + 6 \, {\left (a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{2}\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 ) - 4 \, {\left (2 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 2 i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 \, a^{2}\right )} \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 ) - {\left (-3 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) - 3 i \, a^{2}\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 (3 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) - 3 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 i \, a^{2}\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 (8 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) - 8 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 12 i \, a^{2}\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 )\right )} \sqrt {a} \sqrt {c}}{{\left (-2 i \, c \cos \left (2 \, f x + 2 \, e\right ) + 2 \, c \sin \left (2 \, f x + 2 \, e\right ) - 2 i \, c\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

(6*(a^2*cos(2*f*x + 2*e) + I*a^2*sin(2*f*x + 2*e) + a^2)*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) + 6*(a^2*cos(2*f*x + 2*e) + I*a^2*sin(2*f*x
+ 2*e) + a^2)*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) - 4*(2*a^2*cos(2*f*x + 2*e) + 2*I*a^2*sin(2*f*x + 2*e) + 3*a^2)*cos(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) - (-3*I*a^2*cos(2*f*x + 2*e) + 3*a^2*sin(2*f*x + 2*e) - 3*I*a^2)*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*si
n(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (3*I*a^2*cos(2*f*x + 2*e) - 3*a^2*sin(2*f*x + 2*e) +
3*I*a^2)*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) - (8*I*a^2*cos(2*f*x + 2*e) - 8*
a^2*sin(2*f*x + 2*e) + 12*I*a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-2*I*
c*cos(2*f*x + 2*e) + 2*c*sin(2*f*x + 2*e) - 2*I*c)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(5/2)/(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(5/2)/(c - c*tan(e + f*x)*1i)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(5/2)/sqrt(-I*c*(tan(e + f*x) + I)), x)

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