3.1009 \(\int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=154 \[ -\frac {3 i \sqrt {a} c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3523, 50, 63, 217, 203} \[ -\frac {3 i \sqrt {a} c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 3.07, size = 155, normalized size = 1.01 \[ -\frac {i c e^{-i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (e^{i (e+f x)} \left (5+3 e^{2 i (e+f x)}\right )+3 \left (1+e^{2 i (e+f x)}\right )^2 \tan ^{-1}\left (e^{i (e+f x)}\right )\right ) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 357, normalized size = 2.32 \[ -\frac {3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {8 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a c^{5}}{f^{2}}} {\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {8 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a c^{5}}{f^{2}}} {\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 2 \, {\left (-6 i \, c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} - 10 i \, c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*sqrt(a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(c^2*e^(3*I*f*x + 3*I*e) + c^2*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*c^5/f^2)*(4*I*f*e^(2*I*f*x + 2*I*e)
- 4*I*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - 3*sqrt(a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(c^2*e^(3*I*
f*x + 3*I*e) + c^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)) + sqrt
(a*c^5/f^2)*(-4*I*f*e^(2*I*f*x + 2*I*e) + 4*I*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - 2*(-6*I*c^2*e^(3*I*f*x +
3*I*e) - 10*I*c^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)))/(f*e^(
2*I*f*x + 2*I*e) + f)

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giac [B]  time = 15.91, size = 313, normalized size = 2.03 \[ -\frac {11 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (9 i \, f x + 9 i \, e\right )} + 50 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (7 i \, f x + 7 i \, e\right )} + 84 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (5 i \, f x + 5 i \, e\right )} + 62 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (3 i \, f x + 3 i \, e\right )} + 17 \, {\left (a c^{2} - c^{2}\right )} \sqrt {-a c} e^{\left (i \, f x + i \, e\right )}}{4 \, {\left ({\left (a - 1\right )} f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (a - 1\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (a - 1\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a - 1\right )} f\right )}} - \frac {i \, {\left (32 \, \sqrt {a} c^{\frac {5}{2}} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) + \frac {11 \, \sqrt {a} c^{\frac {5}{2}} e^{\left (3 i \, f x + 3 i \, e\right )} + 9 \, \sqrt {a} c^{\frac {5}{2}} e^{\left (i \, f x + i \, e\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}\right )}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(11*(a*c^2 - c^2)*sqrt(-a*c)*e^(9*I*f*x + 9*I*e) + 50*(a*c^2 - c^2)*sqrt(-a*c)*e^(7*I*f*x + 7*I*e) + 84*(
a*c^2 - c^2)*sqrt(-a*c)*e^(5*I*f*x + 5*I*e) + 62*(a*c^2 - c^2)*sqrt(-a*c)*e^(3*I*f*x + 3*I*e) + 17*(a*c^2 - c^
2)*sqrt(-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*(32*sqrt(a)*c^(5/2)*arctan(e^(I*f*x + I*e)) + (11*sqrt(a)*c^(5/2)*e^(3*I*f*x + 3*I*e) + 9*sqrt(a)*c^(5/2)*e
^(I*f*x + I*e))/(e^(2*I*f*x + 2*I*e) + 1)^2)/f

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maple [A]  time = 0.30, size = 154, normalized size = 1.00 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )-4 i \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-\tan \left (f x +e \right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^2*(3*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))-4*I*(c*a*(1+tan(f*x+e)^2))^(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 = 1.09, size = 450, normalized size = 2.92 \[ -\frac {{\left (12 \, c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 20 \, c^{2} \cos \left (f x + e\right ) + 12 i \, c^{2} \sin \left (3 \, f x + 3 \, e\right ) + 20 i \, c^{2} \sin \left (f x + e\right ) + {\left (6 \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 12 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 6 i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 12 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + {\left (6 \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 12 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 6 i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 12 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - {\left (-3 i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) - 6 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 6 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) - 3 i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - {\left (3 i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 6 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) - 3 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) - 6 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )} \sqrt {a} \sqrt {c}}{f {\left (-4 i \, \cos \left (4 \, f x + 4 \, e\right ) - 8 i \, \cos \left (2 \, f x + 2 \, e\right ) + 4 \, \sin \left (4 \, f x + 4 \, e\right ) + 8 \, \sin \left (2 \, f x + 2 \, e\right ) - 4 i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(12*c^2*cos(3*f*x + 3*e) + 20*c^2*cos(f*x + e) + 12*I*c^2*sin(3*f*x + 3*e) + 20*I*c^2*sin(f*x + e) + (6*c^2*c
os(4*f*x + 4*e) + 12*c^2*cos(2*f*x + 2*e) + 6*I*c^2*sin(4*f*x + 4*e) + 12*I*c^2*sin(2*f*x + 2*e) + 6*c^2)*arct
an2(cos(f*x + e), sin(f*x + e) + 1) + (6*c^2*cos(4*f*x + 4*e) + 12*c^2*cos(2*f*x + 2*e) + 6*I*c^2*sin(4*f*x +
4*e) + 12*I*c^2*sin(2*f*x + 2*e) + 6*c^2)*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - (-3*I*c^2*cos(4*f*x + 4*e
) - 6*I*c^2*cos(2*f*x + 2*e) + 3*c^2*sin(4*f*x + 4*e) + 6*c^2*sin(2*f*x + 2*e) - 3*I*c^2)*log(cos(f*x + e)^2 +
sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - (3*I*c^2*cos(4*f*x + 4*e) + 6*I*c^2*cos(2*f*x + 2*e) - 3*c^2*sin(4*f*x
+ 4*e) - 6*c^2*sin(2*f*x + 2*e) + 3*I*c^2)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + 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 \[ \int \sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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