\(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\) [813]

Optimal result

Integrand size = 45, antiderivative size = 155 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

-1/9*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(9/2)-1/63*(2*I*A-7*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f
/(c-I*c*tan(f*x+e))^(7/2)-1/315*(2*I*A-7*B)*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(9/2),x]

[Out]

-1/9*((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*(c - I*c*Tan[e + f*x])^(9/2)) - (((2*I)*A - 7*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(63*c*f*(c - I*c*Tan[e + f*x])^(7/2)) - (((2*I)*A - 7*B)*(a + I*a*Tan[e + f*x])^(5/2))/(315*
c^2*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*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac {(a (2 A+7 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 )}{9 f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (2 A+7 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 )}{63 c f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\frac {a^3 \sec ^5(e+f x) (45 A+7 (7 A+2 i B) \cos (2 (e+f x))+7 (-2 i A+7 B) \sin (2 (e+f x))) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))}{630 c^4 f (i+\tan (e+f x))^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(9/2),x]

[Out]

(a^3*Sec[e + f*x]^5*(45*A + 7*(7*A + (2*I)*B)*Cos[2*(e + f*x)] + 7*((-2*I)*A + 7*B)*Sin[2*(e + f*x)])*((-I)*Co
s[3*(e + f*x)] + Sin[3*(e + f*x)]))/(630*c^4*f*(I + Tan[e + f*x])^4*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Ta
n[e + f*x]])

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (35 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+35 B \,{\mathrm e}^{8 i \left (f x +e \right )}+90 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+63 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-63 B \,{\mathrm e}^{4 i \left (f x +e \right )}\right )}{1260 c^{4} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(119\)
derivativedivides	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-47 A +2 i A \tan \left (f x +e \right )^{3}-12 A \tan \left (f x +e \right )^{2}-33 i A \tan \left (f x +e \right )-7 i B -42 B \tan \left (f x +e \right )-42 i B \tan \left (f x +e \right )^{2}-7 B \tan \left (f x +e \right )^{3}\right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(138\)
default	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-47 A +2 i A \tan \left (f x +e \right )^{3}-12 A \tan \left (f x +e \right )^{2}-33 i A \tan \left (f x +e \right )-7 i B -42 B \tan \left (f x +e \right )-42 i B \tan \left (f x +e \right )^{2}-7 B \tan \left (f x +e \right )^{3}\right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(138\)
parts	\(-\frac {i A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-47-33 i \tan \left (f x +e \right )-12 \tan \left (f x +e \right )^{2}+2 i \tan \left (f x +e \right )^{3}\right )}{315 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (6 i \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right )^{3}+i+6 \tan \left (f x +e \right )\right )}{45 f \,c^{5} \left (i+\tan \left (f x +e \right )\right )^{6}}\)	\(194\)

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/1260*a^2/c^4*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2*I*(f*x+e))+1))^(1/2)/f*(35*I*A*exp(8
*I*(f*x+e))+35*B*exp(8*I*(f*x+e))+90*I*A*exp(6*I*(f*x+e))+63*I*A*exp(4*I*(f*x+e))-63*B*exp(4*I*(f*x+e)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {{\left (35 \, {\left (i \, A + B\right )} a^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + 5 \, {\left (25 i \, A + 7 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + 9 \, {\left (17 i \, A - 7 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + 63 \, {\left (i \, A - B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{1260 \, c^{5} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/1260*(35*(I*A + B)*a^2*e^(11*I*f*x + 11*I*e) + 5*(25*I*A + 7*B)*a^2*e^(9*I*f*x + 9*I*e) + 9*(17*I*A - 7*B)*
a^2*e^(7*I*f*x + 7*I*e) + 63*(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^5*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(9/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.28 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\frac {{\left (35 \, {\left (-i \, A - B\right )} a^{2} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 90 i \, A a^{2} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (-i \, A + B\right )} a^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 35 \, {\left (A - i \, B\right )} a^{2} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 90 \, A a^{2} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (A + i \, B\right )} a^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{1260 \, c^{\frac {9}{2}} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

1/1260*(35*(-I*A - B)*a^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 90*I*A*a^2*cos(7/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 63*(-I*A + B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
35*(A - I*B)*a^2*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 90*A*a^2*sin(7/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + 63*(A + I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(9
/2)*f)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),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)^(9/2), x)

Mupad [B] (verification not implemented)

Time = 10.56 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.40 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx=-\frac {a^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,90{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}-63\,B\,\cos \left (4\,e+4\,f\,x\right )+35\,B\,\cos \left (8\,e+8\,f\,x\right )-63\,A\,\sin \left (4\,e+4\,f\,x\right )-90\,A\,\sin \left (6\,e+6\,f\,x\right )-35\,A\,\sin \left (8\,e+8\,f\,x\right )-B\,\sin \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}\right )}{1260\,c^4\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(5/2))/(c - c*tan(e + f*x)*1i)^(9/2),x)

[Out]

-(a^2*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(4*e + 4*f*x)*63i
+ A*cos(6*e + 6*f*x)*90i + A*cos(8*e + 8*f*x)*35i - 63*B*cos(4*e + 4*f*x) + 35*B*cos(8*e + 8*f*x) - 63*A*sin(4
*e + 4*f*x) - 90*A*sin(6*e + 6*f*x) - 35*A*sin(8*e + 8*f*x) - B*sin(4*e + 4*f*x)*63i + B*sin(8*e + 8*f*x)*35i)
)/(1260*c^4*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))