3.755 \(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=105 \[ \frac {2 a^2 (3 B+i A)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {4 a^2 (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 a^2 B}{3 c^2 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 105, 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} \[ \frac {2 a^2 (3 B+i A)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {4 a^2 (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 a^2 B}{3 c^2 f (c-i c \tan (e+f x))^{3/2}} \]

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])^(7/2),x]

[Out]

(-4*a^2*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (2*a^2*(I*A + 3*B))/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2
)) - (2*a^2*B)/(3*c^2*f*(c - I*c*Tan[e + f*x])^(3/2))

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

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Mathematica [A]  time = 13.08, size = 122, normalized size = 1.16 \[ \frac {a^2 \cos ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (4 e+6 f x)+i \sin (4 e+6 f x)) (7 (3 A+i B) \sin (2 (e+f x))+(-37 B-9 i A) \cos (2 (e+f x))-9 i A+33 B)}{105 c^4 f (\cos (f x)+i \sin (f x))^2} \]

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])^(7/2),x]

[Out]

(a^2*Cos[e + f*x]^2*((-9*I)*A + 33*B + ((-9*I)*A - 37*B)*Cos[2*(e + f*x)] + 7*(3*A + I*B)*Sin[2*(e + f*x)])*(C
os[4*e + 6*f*x] + I*Sin[4*e + 6*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(105*c^4*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [A]  time = 0.61, size = 120, normalized size = 1.14 \[ \frac {\sqrt {2} {\left ({\left (-15 i \, A - 15 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-39 i \, A + 3 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-27 i \, A + 29 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (3 i \, A - 11 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A - 22 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{420 \, c^{4} f} \]

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))^(7/2),x, algorithm="fricas")

[Out]

1/420*sqrt(2)*((-15*I*A - 15*B)*a^2*e^(8*I*f*x + 8*I*e) + (-39*I*A + 3*B)*a^2*e^(6*I*f*x + 6*I*e) + (-27*I*A +
29*B)*a^2*e^(4*I*f*x + 4*I*e) + (3*I*A - 11*B)*a^2*e^(2*I*f*x + 2*I*e) + (6*I*A - 22*B)*a^2)*sqrt(c/(e^(2*I*f
*x + 2*I*e) + 1))/(c^4*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

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))^(7/2),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)^(7/2), x)

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maple [A]  time = 0.30, size = 80, normalized size = 0.76 \[ -\frac {2 i a^{2} \left (\frac {2 c^{2} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}-\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {c \left (-3 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,c^{2}} \]

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))^(7/2),x)

[Out]

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

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maxima [A]  time = 0.57, size = 76, normalized size = 0.72 \[ \frac {2 i \, {\left (35 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} B a^{2} + 21 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} c - 30 \, {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} c^{2} f} \]

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))^(7/2),x, algorithm="maxima")

[Out]

2/105*I*(35*I*(-I*c*tan(f*x + e) + c)^2*B*a^2 + 21*(-I*c*tan(f*x + e) + c)*(A - 3*I*B)*a^2*c - 30*(A - I*B)*a^
2*c^2)/((-I*c*tan(f*x + e) + c)^(7/2)*c^2*f)

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mupad [B]  time = 11.81, size = 167, normalized size = 1.59 \[ -\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (-\frac {a^2\,\left (3\,A+B\,11{}\mathrm {i}\right )\,1{}\mathrm {i}}{210\,c^4\,f}-\frac {a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (3\,A+B\,11{}\mathrm {i}\right )\,1{}\mathrm {i}}{420\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (13\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{140\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (27\,A+B\,29{}\mathrm {i}\right )\,1{}\mathrm {i}}{420\,c^4\,f}+\frac {a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{28\,c^4\,f}\right ) \]

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)^(7/2),x)

[Out]

-(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(13*A + B*1i)*1i)/(140*c^4*f) - (a^2*ex
p(e*2i + f*x*2i)*(3*A + B*11i)*1i)/(420*c^4*f) - (a^2*(3*A + B*11i)*1i)/(210*c^4*f) + (a^2*exp(e*4i + f*x*4i)*
(27*A + B*29i)*1i)/(420*c^4*f) + (a^2*exp(e*8i + f*x*8i)*(A - B*1i)*1i)/(28*c^4*f))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {A}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \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 )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \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))**(7/2),x)

[Out]

-a**2*(Integral(-A/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*ta
n(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + In
tegral(A*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) +
c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x
) + Integral(-B*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*
x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)
), x) + Integral(B*tan(e + f*x)**3/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(
e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x
) + c)), x) + Integral(-2*I*A*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-
I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan
(e + f*x) + c)), x) + Integral(-2*I*B*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*
c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sq
rt(-I*c*tan(e + f*x) + c)), x))

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