\(\int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [801]

Optimal result

Integrand size = 45, antiderivative size = 102 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(i A-4 B) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(5/2)-1/15*(I*A-4*B)*(a+I*a*tan(f*x+e))^(3/2)/c/f/(
c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3669, 79, 37} \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {(-4 B+i A) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[In]

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

[Out]

-1/5*((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) - ((I*A - 4*B)*(a + I*a*Tan[e +
f*x])^(3/2))/(15*c*f*(c - I*c*Tan[e + f*x])^(3/2))

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

Mathematica [A] (verified)

Time = 7.02 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a^2 \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (4 i A-B+(A+4 i B) \tan (e+f x))}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(a^2*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((4*I)*A - B + (A + (4*I)*B)*Tan[e + f*x]))/(15*c^
2*f*(I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-4 A +i A \tan \left (f x +e \right )-i B -4 B \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(90\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-4 A +i A \tan \left (f x +e \right )-i B -4 B \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(90\)
risch	\(-\frac {a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+3 B \,{\mathrm e}^{4 i \left (f x +e \right )}+5 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-5 B \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{30 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(104\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 \tan \left (f x +e \right )+i\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(147\)

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c^3*(1+tan(f*x+e)^2)*(-4*A+I*A*tan(f*x+e)-I*
B-4*B*tan(f*x+e))/(I+tan(f*x+e))^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} a e^{\left (7 i \, f x + 7 i \, e\right )} + 2 \, {\left (4 i \, A - B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/30*(3*(I*A + B)*a*e^(7*I*f*x + 7*I*e) + 2*(4*I*A - B)*a*e^(5*I*f*x + 5*I*e) + 5*(I*A - B)*a*e^(3*I*f*x + 3*
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^3*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+I*a*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(3/2)*(A + B*tan(e + f*x))/(-I*c*(tan(e + f*x) + I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {30 \, {\left (3 \, {\left (A - i \, B\right )} a \cos \left (7 \, f x + 7 \, e\right ) + 2 \, {\left (4 \, A + i \, B\right )} a \cos \left (5 \, f x + 5 \, e\right ) + 5 \, {\left (A + i \, B\right )} a \cos \left (3 \, f x + 3 \, e\right ) - 3 \, {\left (-i \, A - B\right )} a \sin \left (7 \, f x + 7 \, e\right ) - 2 \, {\left (-4 i \, A + B\right )} a \sin \left (5 \, f x + 5 \, e\right ) - 5 \, {\left (-i \, A + B\right )} a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {a} \sqrt {c}}{-900 \, {\left (i \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) - c^{3} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{3}\right )} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-30*(3*(A - I*B)*a*cos(7*f*x + 7*e) + 2*(4*A + I*B)*a*cos(5*f*x + 5*e) + 5*(A + I*B)*a*cos(3*f*x + 3*e) - 3*(-
I*A - B)*a*sin(7*f*x + 7*e) - 2*(-4*I*A + B)*a*sin(5*f*x + 5*e) - 5*(-I*A + B)*a*sin(3*f*x + 3*e))*sqrt(a)*sqr
t(c)/((-900*I*c^3*cos(2*f*x + 2*e) + 900*c^3*sin(2*f*x + 2*e) - 900*I*c^3)*f)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(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)*(I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.86 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-5\,B\,\cos \left (2\,e+2\,f\,x\right )+3\,B\,\cos \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2))/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(a*((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(2*e + 2*f*x)*5i + A
*cos(4*e + 4*f*x)*3i - 5*B*cos(2*e + 2*f*x) + 3*B*cos(4*e + 4*f*x) - 5*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*f*
x) - B*sin(2*e + 2*f*x)*5i + B*sin(4*e + 4*f*x)*3i))/(30*c^2*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))