3.9.15 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\) [815]

Optimal. Leaf size=261 \[ -\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

-1/13*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(13/2)-1/143*(4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/
c/f/(c-I*c*tan(f*x+e))^(11/2)-1/429*(4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(9/2)-2/3003
*(4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c^3/f/(c-I*c*tan(f*x+e))^(7/2)-2/15015*(4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5
/2)/c^4/f/(c-I*c*tan(f*x+e))^(5/2)

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Rubi [A]
time = 0.22, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \begin {gather*} -\frac {2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

-1/13*((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*(c - I*c*Tan[e + f*x])^(13/2)) - (((4*I)*A - 9*B)*(a + I*a*T
an[e + f*x])^(5/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(
429*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3003*c^3*f*(c - I*
c*Tan[e + f*x])^(7/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(15015*c^4*f*(c - I*c*Tan[e + f*x])^
(5/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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

Rule 79

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

Rule 3669

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 \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}+\frac {(a (4 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{13 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}+\frac {(3 a (4 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{143 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}+\frac {(2 a (4 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{429 c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}+\frac {(2 a (4 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{3003 c^3 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 11.32, size = 183, normalized size = 0.70 \begin {gather*} -\frac {i a^2 \cos (e+f x) (5005 A+780 (9 A+i B) \cos (2 (e+f x))+231 (9 A+4 i B) \cos (4 (e+f x))-1560 i A \sin (2 (e+f x))+3510 B \sin (2 (e+f x))-924 i A \sin (4 (e+f x))+2079 B \sin (4 (e+f x))) (\cos (9 e+11 f x)+i \sin (9 e+11 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{120120 c^7 f (\cos (f x)+i \sin (f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

((-1/120120*I)*a^2*Cos[e + f*x]*(5005*A + 780*(9*A + I*B)*Cos[2*(e + f*x)] + 231*(9*A + (4*I)*B)*Cos[4*(e + f*
x)] - (1560*I)*A*Sin[2*(e + f*x)] + 3510*B*Sin[2*(e + f*x)] - (924*I)*A*Sin[4*(e + f*x)] + 2079*B*Sin[4*(e + f
*x)])*(Cos[9*e + 11*f*x] + I*Sin[9*e + 11*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c^7*f*
(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]
time = 0.47, size = 183, normalized size = 0.70

method	result	size
risch	\(-\frac {a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (1155 i A \,{\mathrm e}^{12 i \left (f x +e \right )}+1155 B \,{\mathrm e}^{12 i \left (f x +e \right )}+5460 i A \,{\mathrm e}^{10 i \left (f x +e \right )}+2730 B \,{\mathrm e}^{10 i \left (f x +e \right )}+10010 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+8580 i A \,{\mathrm e}^{6 i \left (f x +e \right )}-4290 B \,{\mathrm e}^{6 i \left (f x +e \right )}+3003 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-3003 B \,{\mathrm e}^{4 i \left (f x +e \right )}\right )}{240240 c^{6} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(169\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (18 i B \left (\tan ^{5}\left (f x +e \right )\right )+64 i A \left (\tan ^{4}\left (f x +e \right )\right )+8 A \left (\tan ^{5}\left (f x +e \right )\right )-531 i B \left (\tan ^{3}\left (f x +e \right )\right )-144 B \left (\tan ^{4}\left (f x +e \right )\right )-544 i A \left (\tan ^{2}\left (f x +e \right )\right )-236 A \left (\tan ^{3}\left (f x +e \right )\right )-1704 i B \tan \left (f x +e \right )+1224 B \left (\tan ^{2}\left (f x +e \right )\right )-1763 i A +911 A \tan \left (f x +e \right )+213 B \right )}{15015 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}\)	\(183\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (18 i B \left (\tan ^{5}\left (f x +e \right )\right )+64 i A \left (\tan ^{4}\left (f x +e \right )\right )+8 A \left (\tan ^{5}\left (f x +e \right )\right )-531 i B \left (\tan ^{3}\left (f x +e \right )\right )-144 B \left (\tan ^{4}\left (f x +e \right )\right )-544 i A \left (\tan ^{2}\left (f x +e \right )\right )-236 A \left (\tan ^{3}\left (f x +e \right )\right )-1704 i B \tan \left (f x +e \right )+1224 B \left (\tan ^{2}\left (f x +e \right )\right )-1763 i A +911 A \tan \left (f x +e \right )+213 B \right )}{15015 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x,method=_RETURNVERBOSE)

[Out]

1/15015/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c^7*(1+tan(f*x+e)^2)*(18*I*B*tan(f*x+e)^5
+64*I*A*tan(f*x+e)^4+8*A*tan(f*x+e)^5-531*I*B*tan(f*x+e)^3-144*B*tan(f*x+e)^4-544*I*A*tan(f*x+e)^2-236*A*tan(f
*x+e)^3-1704*I*B*tan(f*x+e)+1224*B*tan(f*x+e)^2-1763*I*A+911*A*tan(f*x+e)+213*B)/(I+tan(f*x+e))^8

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Maxima [A]
time = 0.64, size = 352, normalized size = 1.35 \begin {gather*} \frac {{\left (1155 \, {\left (-i \, A - B\right )} a^{2} \cos \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 2730 \, {\left (-2 i \, A - B\right )} a^{2} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 10010 i \, A a^{2} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4290 \, {\left (-2 i \, A + B\right )} a^{2} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3003 \, {\left (-i \, A + B\right )} a^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1155 \, {\left (A - i \, B\right )} a^{2} \sin \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 2730 \, {\left (2 \, A - i \, B\right )} a^{2} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 10010 \, A a^{2} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4290 \, {\left (2 \, A + i \, B\right )} a^{2} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3003 \, {\left (A + i \, B\right )} a^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{240240 \, c^{\frac {13}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

1/240240*(1155*(-I*A - B)*a^2*cos(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2730*(-2*I*A - B)*a^2*co
s(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 10010*I*A*a^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 4290*(-2*I*A + B)*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3003*(-I*A + B)*a^2*
cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1155*(A - I*B)*a^2*sin(13/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) + 2730*(2*A - I*B)*a^2*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 10010*A*a^2*s
in(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4290*(2*A + I*B)*a^2*sin(7/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) + 3003*(A + I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(13/2)
*f)

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Fricas [A]
time = 3.10, size = 175, normalized size = 0.67 \begin {gather*} -\frac {{\left (1155 \, {\left (i \, A + B\right )} a^{2} e^{\left (15 i \, f x + 15 i \, e\right )} + 105 \, {\left (63 i \, A + 37 \, B\right )} a^{2} e^{\left (13 i \, f x + 13 i \, e\right )} + 910 \, {\left (17 i \, A + 3 \, B\right )} a^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + 1430 \, {\left (13 i \, A - 3 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + 429 \, {\left (27 i \, A - 17 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + 3003 \, {\left (i \, A - B\right )} a^{2} e^{\left (5 i \, f x + 5 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}}}{240240 \, c^{7} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

-1/240240*(1155*(I*A + B)*a^2*e^(15*I*f*x + 15*I*e) + 105*(63*I*A + 37*B)*a^2*e^(13*I*f*x + 13*I*e) + 910*(17*
I*A + 3*B)*a^2*e^(11*I*f*x + 11*I*e) + 1430*(13*I*A - 3*B)*a^2*e^(9*I*f*x + 9*I*e) + 429*(27*I*A - 17*B)*a^2*e
^(7*I*f*x + 7*I*e) + 3003*(I*A - B)*a^2*e^(5*I*f*x + 5*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^7*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(13/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/(-I*c*tan(f*x + e) + c)^(13/2), x)

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Mupad [B]
time = 13.73, size = 191, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (2\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{56\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{88\,c^6\,f}+\frac {A\,a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1{}\mathrm {i}}{24\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{80\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,12{}\mathrm {i}+f\,x\,12{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{208\,c^6\,f}\right )}{\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}} \end {gather*}

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)^(5/2))/(c - c*tan(e + f*x)*1i)^(13/2),x)

[Out]

-((a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(2*A + B*1i)*1i)/(56*c^6*f) + (a^2*exp
(e*10i + f*x*10i)*(2*A - B*1i)*1i)/(88*c^6*f) + (A*a^2*exp(e*8i + f*x*8i)*1i)/(24*c^6*f) + (a^2*exp(e*4i + f*x
*4i)*(A + B*1i)*1i)/(80*c^6*f) + (a^2*exp(e*12i + f*x*12i)*(A - B*1i)*1i)/(208*c^6*f)))/(c - (c*sin(e + f*x)*1
i)/cos(e + f*x))^(1/2)

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