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

Optimal. Leaf size=217 \[ -\frac{\sqrt{a} c^{5/2} (-2 B+3 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}-\frac{c^2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{c (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 f}+\frac{B \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.295699, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3588, 80, 50, 63, 217, 203} \[ -\frac{\sqrt{a} c^{5/2} (-2 B+3 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}-\frac{c^2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{c (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 f}+\frac{B \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 80

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 6.64861, size = 226, normalized size = 1.04 \[ \frac{\sqrt{a+i a \tan (e+f x)} (A+B \tan (e+f x)) \left (\frac{c^3 (2 B-3 i A) e^{-i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}+\frac{1}{12} c^2 \sec ^{\frac{5}{2}}(e+f x) \sqrt{c-i c \tan (e+f x)} (-3 (A+2 i B) \sin (2 (e+f x))+12 (B-i A) \cos (2 (e+f x))-12 i A+8 B)\right )}{f \sec ^{\frac{3}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x])*((((-3*I)*A + 2*B)*c^3*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e
+ f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^(I*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + (c^2*Sec[e + f*x]^(5/
2)*((-12*I)*A + 8*B + 12*((-I)*A + B)*Cos[2*(e + f*x)] - 3*(A + (2*I)*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e
+ f*x]])/12))/(f*Sec[e + f*x]^(3/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]  time = 0.154, size = 285, normalized size = 1.3 \begin{align*} -{\frac{{c}^{2}}{6\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( -6\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac+6\,iB\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +2\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+12\,iA\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }-9\,A\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) ac+3\,A\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -10\,B\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^2*(-6*I*B*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+
e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c+6*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+2*B*tan(f
*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+12*I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)-9*A*ln((a*c*t
an(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c+3*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1
/2)*tan(f*x+e)-10*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2)/(a*c*(1+tan(f*x+e)^2))^(1/2)

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Maxima [B]  time = 3.48267, size = 1455, normalized size = 6.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((216*A + 144*I*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (576*A + 384*I*B)*c^2*cos(3/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (360*A + 432*I*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 72*(3*I*A - 2*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 192*(3*I*A - 2*B)*c^2
*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 72*(5*I*A - 6*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) + ((108*A + 72*I*B)*c^2*cos(6*f*x + 6*e) + (324*A + 216*I*B)*c^2*cos(4*f*x + 4*e) + (324*A
+ 216*I*B)*c^2*cos(2*f*x + 2*e) + 36*(3*I*A - 2*B)*c^2*sin(6*f*x + 6*e) + 108*(3*I*A - 2*B)*c^2*sin(4*f*x + 4
*e) + 108*(3*I*A - 2*B)*c^2*sin(2*f*x + 2*e) + (108*A + 72*I*B)*c^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) + ((108*A + 72*I*B)*c^2*cos(6*f
*x + 6*e) + (324*A + 216*I*B)*c^2*cos(4*f*x + 4*e) + (324*A + 216*I*B)*c^2*cos(2*f*x + 2*e) + 36*(3*I*A - 2*B)
*c^2*sin(6*f*x + 6*e) + 108*(3*I*A - 2*B)*c^2*sin(4*f*x + 4*e) + 108*(3*I*A - 2*B)*c^2*sin(2*f*x + 2*e) + (108
*A + 72*I*B)*c^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) + (18*(3*I*A - 2*B)*c^2*cos(6*f*x + 6*e) + 54*(3*I*A - 2*B)*c^2*cos(4*f*x + 4*e)
+ 54*(3*I*A - 2*B)*c^2*cos(2*f*x + 2*e) - (54*A + 36*I*B)*c^2*sin(6*f*x + 6*e) - (162*A + 108*I*B)*c^2*sin(4*f
*x + 4*e) - (162*A + 108*I*B)*c^2*sin(2*f*x + 2*e) + 18*(3*I*A - 2*B)*c^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) + (18*(-3*I*A + 2*B)*c^2*cos(6*f*x + 6*e) + 54*(-3*I*A + 2*B)*c^2*cos(4*f*x +
4*e) + 54*(-3*I*A + 2*B)*c^2*cos(2*f*x + 2*e) + (54*A + 36*I*B)*c^2*sin(6*f*x + 6*e) + (162*A + 108*I*B)*c^2*
sin(4*f*x + 4*e) + (162*A + 108*I*B)*c^2*sin(2*f*x + 2*e) + 18*(-3*I*A + 2*B)*c^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(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-72*I*cos(6*f*x + 6*e) - 216*I*cos(4*f*x + 4*e)
- 216*I*cos(2*f*x + 2*e) + 72*sin(6*f*x + 6*e) + 216*sin(4*f*x + 4*e) + 216*sin(2*f*x + 2*e) - 72*I))

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Fricas [B]  time = 1.68716, size = 1413, normalized size = 6.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*((-18*I*A + 12*B)*c^2*e^(4*I*f*x + 4*I*e) + (-48*I*A + 32*B)*c^2*e^(2*I*f*x + 2*I*e) + (-30*I*A + 36*B
)*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 3*sqrt((9*A^2 + 1
2*I*A*B - 4*B^2)*a*c^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-12*I*A + 8*B)*c^2*
e^(2*I*f*x + 2*I*e) + (-12*I*A + 8*B)*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))
*e^(I*f*x + I*e) + 2*sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-3*I*A + 2*B)*c
^2*e^(2*I*f*x + 2*I*e) + (-3*I*A + 2*B)*c^2)) + 3*sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c^5/f^2)*(f*e^(4*I*f*x + 4
*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-12*I*A + 8*B)*c^2*e^(2*I*f*x + 2*I*e) + (-12*I*A + 8*B)*c^2)*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 2*sqrt((9*A^2 + 12*I*A*B -
4*B^2)*a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-3*I*A + 2*B)*c^2*e^(2*I*f*x + 2*I*e) + (-3*I*A + 2*B)*c^2))
)/(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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt{i \, a \tan \left (f x + e\right ) + a}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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