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

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

[Out]

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

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Rubi [A]  time = 0.290031, 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} \[ -\frac{a^{5/2} \sqrt{c} (2 B+3 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a^2 (2 B+3 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{a (2 B+3 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 f}+\frac{B (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 8.35172, size = 253, normalized size = 1.17 \[ \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\frac{(\sin (2 e)+i \cos (2 e)) \sec ^{\frac{5}{2}}(e+f x) \sqrt{c-i c \tan (e+f x)} ((6 B+3 i A) \sin (2 (e+f x))+12 (A-i B) \cos (2 (e+f x))+12 A-8 i B)}{12 (\cos (f x)+i \sin (f x))^2}-\frac{i c (3 A-2 i B) e^{-3 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}\right )}{f \sec ^{\frac{7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x])*(((-I)*(3*A - (2*I)*B)*c*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)
*(e + f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((3*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + (Sec[e + f*x]
^(5/2)*(I*Cos[2*e] + Sin[2*e])*(12*A - (8*I)*B + 12*(A - I*B)*Cos[2*(e + f*x)] + ((3*I)*A + 6*B)*Sin[2*(e + f*
x)])*Sqrt[c - I*c*Tan[e + f*x]])/(12*(Cos[f*x] + I*Sin[f*x])^2)))/(f*Sec[e + f*x]^(7/2)*(A*Cos[e + f*x] + B*Si
n[e + f*x]))

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Maple [A]  time = 0.102, size = 285, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}}{6\,f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 3.48748, size = 1457, 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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

((360*A - 432*I*B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (576*A - 384*I*B)*a^2*cos(3/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (216*A - 144*I*B)*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 72*(5*I*A + 6*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 192*(3*I*A + 2*B)*a^2*
sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 72*(3*I*A + 2*B)*a^2*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) - ((108*A - 72*I*B)*a^2*cos(6*f*x + 6*e) + (324*A - 216*I*B)*a^2*cos(4*f*x + 4*e) + (324*A
- 216*I*B)*a^2*cos(2*f*x + 2*e) - 36*(-3*I*A - 2*B)*a^2*sin(6*f*x + 6*e) - 108*(-3*I*A - 2*B)*a^2*sin(4*f*x +
4*e) - 108*(-3*I*A - 2*B)*a^2*sin(2*f*x + 2*e) + (108*A - 72*I*B)*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) - ((108*A - 72*I*B)*a^2*cos(6
*f*x + 6*e) + (324*A - 216*I*B)*a^2*cos(4*f*x + 4*e) + (324*A - 216*I*B)*a^2*cos(2*f*x + 2*e) - 36*(-3*I*A - 2
*B)*a^2*sin(6*f*x + 6*e) - 108*(-3*I*A - 2*B)*a^2*sin(4*f*x + 4*e) - 108*(-3*I*A - 2*B)*a^2*sin(2*f*x + 2*e) +
(108*A - 72*I*B)*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) + (18*(-3*I*A - 2*B)*a^2*cos(6*f*x + 6*e) + 54*(-3*I*A - 2*B)*a^2*cos(4*f*x
+ 4*e) + 54*(-3*I*A - 2*B)*a^2*cos(2*f*x + 2*e) + (54*A - 36*I*B)*a^2*sin(6*f*x + 6*e) + (162*A - 108*I*B)*a^2
*sin(4*f*x + 4*e) + (162*A - 108*I*B)*a^2*sin(2*f*x + 2*e) + 18*(-3*I*A - 2*B)*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) + (18*(3*I*A + 2*B)*a^2*cos(6*f*x + 6*e) + 54*(3*I*A + 2*B)*a^2*cos(
4*f*x + 4*e) + 54*(3*I*A + 2*B)*a^2*cos(2*f*x + 2*e) - (54*A - 36*I*B)*a^2*sin(6*f*x + 6*e) - (162*A - 108*I*B
)*a^2*sin(4*f*x + 4*e) - (162*A - 108*I*B)*a^2*sin(2*f*x + 2*e) + 18*(3*I*A + 2*B)*a^2)*log(cos(1/2*arctan2(si
n(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*arct
an2(sin(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 = 1.59996, size = 1399, normalized size = 6.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*((30*I*A + 36*B)*a^2*e^(4*I*f*x + 4*I*e) + (48*I*A + 32*B)*a^2*e^(2*I*f*x + 2*I*e) + (18*I*A + 12*B)*a
^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 - 12*I
*A*B - 4*B^2)*a^5*c/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)*a^2*e^(2
*I*f*x + 2*I*e) + (12*I*A + 8*B)*a^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^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((3*I*A + 2*B)*a^2*e^(
2*I*f*x + 2*I*e) + (3*I*A + 2*B)*a^2)) - 3*sqrt((9*A^2 - 12*I*A*B - 4*B^2)*a^5*c/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)*a^2*e^(2*I*f*x + 2*I*e) + (12*I*A + 8*B)*a^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
^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((3*I*A + 2*B)*a^2*e^(2*I*f*x + 2*I*e) + (3*I*A + 2*B)*a^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((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e)),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 )}{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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