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]

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

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Rubi [A]  time = 0.30, antiderivative size = 217, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 7.06, 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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fricas [B]  time = 0.73, size = 543, normalized size = 2.50 \[ -\frac {3 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-12 i \, A + 8 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-12 i \, A + 8 \, B\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}} + 2 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-3 i \, A + 2 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A + 2 \, B\right )} c^{2}}\right ) - 3 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-12 i \, A + 8 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-12 i \, A + 8 \, B\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}} - 2 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-3 i \, A + 2 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A + 2 \, B\right )} c^{2}}\right ) - 2 \, {\left ({\left (-18 i \, A + 12 \, B\right )} c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-48 i \, A + 32 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-30 i \, A + 36 \, B\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}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/12*(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^(3*I*f*x + 3*I*e) + (-12*I*A + 8*B)*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)) + 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^(3*I*f*x + 3*I*e) +
(-12*I*A + 8*B)*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)) - 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)) - 2*((-18*I*A + 12*B)*c^2*e^(5*I*f*x + 5*I*e) + (-48*I*A + 32*B)*c^2*e^(3*I*f*x + 3*I
*e) + (-30*I*A + 36*B)*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^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.68, size = 285, normalized size = 1.31 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (-6 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +6 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+2 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+12 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-9 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +3 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-10 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{6 f \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}} \]

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

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maxima [B]  time = 1.38, size = 1086, normalized size = 5.00 \[ \text {result too large to display} \]

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]

-(72*(3*A + 2*I*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 192*(3*A + 2*I*B)*c^2*cos(3/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 72*(5*A + 6*I*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) - (-216*I*A + 144*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-576*I*A + 384*B)*c^
2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-360*I*A + 432*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) + (36*(3*A + 2*I*B)*c^2*cos(6*f*x + 6*e) + 108*(3*A + 2*I*B)*c^2*cos(4*f*x + 4*e) + 108
*(3*A + 2*I*B)*c^2*cos(2*f*x + 2*e) - (-108*I*A + 72*B)*c^2*sin(6*f*x + 6*e) - (-324*I*A + 216*B)*c^2*sin(4*f*
x + 4*e) - (-324*I*A + 216*B)*c^2*sin(2*f*x + 2*e) + 36*(3*A + 2*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) + (36*(3*A + 2*I*B)*c^2*c
os(6*f*x + 6*e) + 108*(3*A + 2*I*B)*c^2*cos(4*f*x + 4*e) + 108*(3*A + 2*I*B)*c^2*cos(2*f*x + 2*e) - (-108*I*A
+ 72*B)*c^2*sin(6*f*x + 6*e) - (-324*I*A + 216*B)*c^2*sin(4*f*x + 4*e) - (-324*I*A + 216*B)*c^2*sin(2*f*x + 2*
e) + 36*(3*A + 2*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) - ((-54*I*A + 36*B)*c^2*cos(6*f*x + 6*e) + (-162*I*A + 108*B)*c^2*cos(4*
f*x + 4*e) + (-162*I*A + 108*B)*c^2*cos(2*f*x + 2*e) + 18*(3*A + 2*I*B)*c^2*sin(6*f*x + 6*e) + 54*(3*A + 2*I*B
)*c^2*sin(4*f*x + 4*e) + 54*(3*A + 2*I*B)*c^2*sin(2*f*x + 2*e) + (-54*I*A + 36*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*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((54*I*A - 36*B)*c^2*cos(6*f*x + 6*e) + (162*I*A - 108*B)*c^2*c
os(4*f*x + 4*e) + (162*I*A - 108*B)*c^2*cos(2*f*x + 2*e) - 18*(3*A + 2*I*B)*c^2*sin(6*f*x + 6*e) - 54*(3*A + 2
*I*B)*c^2*sin(4*f*x + 4*e) - 54*(3*A + 2*I*B)*c^2*sin(2*f*x + 2*e) + (54*I*A - 36*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*ar
ctan2(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*tan(e + f*x))*(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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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