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

Optimal. Leaf size=103 \[ \frac {2 a^2 (3 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 a^2 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3588, 77} \[ \frac {2 a^2 (3 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 a^2 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 11.92, size = 118, normalized size = 1.15 \[ \frac {a^2 \cos (e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (3 e+5 f x)+i \sin (3 e+5 f x)) (5 (A+3 i B) \sin (2 (e+f x))+(-21 B-i A) \cos (2 (e+f x))-i A+9 B)}{15 c^3 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[e + f*x]*((-I)*A + 9*B + ((-I)*A - 21*B)*Cos[2*(e + f*x)] + 5*(A + (3*I)*B)*Sin[2*(e + f*x)])*(Cos[3*
e + 5*f*x] + I*Sin[3*e + 5*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^2)

________________________________________________________________________________________

fricas [A]  time = 1.57, size = 100, normalized size = 0.97 \[ \frac {\sqrt {2} {\left ({\left (-3 i \, A - 3 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-4 i \, A + 6 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A - 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A - 18 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*((-3*I*A - 3*B)*a^2*e^(6*I*f*x + 6*I*e) + (-4*I*A + 6*B)*a^2*e^(4*I*f*x + 4*I*e) + (I*A - 9*B)*a^
2*e^(2*I*f*x + 2*I*e) + (2*I*A - 18*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maple [A]  time = 0.26, size = 80, normalized size = 0.78 \[ -\frac {2 i a^{2} \left (-\frac {c \left (-3 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 c^{2} \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.72, size = 76, normalized size = 0.74 \[ \frac {2 i \, {\left (15 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} B a^{2} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} c - 6 \, {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*I*(15*I*(-I*c*tan(f*x + e) + c)^2*B*a^2 + 5*(-I*c*tan(f*x + e) + c)*(A - 3*I*B)*a^2*c - 6*(A - I*B)*a^2*c
^2)/((-I*c*tan(f*x + e) + c)^(5/2)*c^2*f)

________________________________________________________________________________________

mupad [B]  time = 11.36, size = 208, normalized size = 2.02 \[ -\frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (18\,B-A\,2{}\mathrm {i}-A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,4{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+9\,B\,\cos \left (2\,e+2\,f\,x\right )-6\,B\,\cos \left (4\,e+4\,f\,x\right )+3\,B\,\cos \left (6\,e+6\,f\,x\right )+A\,\sin \left (2\,e+2\,f\,x\right )-4\,A\,\sin \left (4\,e+4\,f\,x\right )-3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,9{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,6{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^3\,f} \]

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)^2)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(a^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(18*B - A*2i - A*cos(2*e
+ 2*f*x)*1i + A*cos(4*e + 4*f*x)*4i + A*cos(6*e + 6*f*x)*3i + 9*B*cos(2*e + 2*f*x) - 6*B*cos(4*e + 4*f*x) + 3
*B*cos(6*e + 6*f*x) + A*sin(2*e + 2*f*x) - 4*A*sin(4*e + 4*f*x) - 3*A*sin(6*e + 6*f*x) + B*sin(2*e + 2*f*x)*9i
- B*sin(4*e + 4*f*x)*6i + B*sin(6*e + 6*f*x)*3i))/(30*c^3*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(-A/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*t
an(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(A*tan(e + f*x)**2/(-c**2*sqrt(-I*c*tan(e + f*x)
+ c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)),
x) + Integral(-B*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e +
f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)**3/(-c**2*sqrt(-I*c*
tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e +
f*x) + c)), x) + Integral(-2*I*A*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*s
qrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*B*tan(e + f*x)
**2/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c
**2*sqrt(-I*c*tan(e + f*x) + c)), x))

________________________________________________________________________________________