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

Optimal. Leaf size=62 \[ \frac {2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3588, 43} \[ \frac {2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 7.73, size = 100, normalized size = 1.61 \[ \frac {2 a \cos ^2(e+f x) (\cos (f x)-i \sin (f x)) \sqrt {c-i c \tan (e+f x)} (\cos (3 e+4 f x)+i \sin (3 e+4 f x)) ((2 B-3 i A) \cos (e+f x)-5 i B \sin (e+f x))}{15 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 90, normalized size = 1.45 \[ \frac {\sqrt {2} {\left ({\left (-3 i \, A - 3 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-9 i \, A + B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-9 i \, A + 11 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A + 7 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*sqrt(2)*((-3*I*A - 3*B)*a*e^(6*I*f*x + 6*I*e) + (-9*I*A + B)*a*e^(4*I*f*x + 4*I*e) + (-9*I*A + 11*B)*a*e^
(2*I*f*x + 2*I*e) + (-3*I*A + 7*B)*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.28, size = 53, normalized size = 0.85 \[ \frac {2 i a \left (-\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {c \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 46, normalized size = 0.74 \[ -\frac {2 i \, {\left (5 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a + 3 \, {\left (A - i \, B\right )} a c\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 10.95, size = 232, normalized size = 3.74 \[ \frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (7\,B+11\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+9\,A\,\sin \left (2\,e+2\,f\,x\right )+9\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )-3\,B\,\left (2\,{\cos \left (3\,e+3\,f\,x\right )}^2-1\right )-A\,3{}\mathrm {i}-A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,9{}\mathrm {i}+B\,\sin \left (2\,e+2\,f\,x\right )\,11{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-A\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )\,9{}\mathrm {i}-A\,\left (2\,{\cos \left (3\,e+3\,f\,x\right )}^2-1\right )\,3{}\mathrm {i}\right )}{60\,c^3\,f} \]

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))/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2)*(7*B - A*3i - A*(2*cos(e + f*x)^2
- 1)*9i + 11*B*(2*cos(e + f*x)^2 - 1) + 9*A*sin(2*e + 2*f*x) + 9*A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x) + B
*sin(2*e + 2*f*x)*11i + B*sin(4*e + 4*f*x)*1i - B*sin(6*e + 6*f*x)*3i - A*(2*cos(2*e + 2*f*x)^2 - 1)*9i - A*(2
*cos(3*e + 3*f*x)^2 - 1)*3i + B*(2*cos(2*e + 2*f*x)^2 - 1) - 3*B*(2*cos(3*e + 3*f*x)^2 - 1)))/(60*c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*(Integral(-I*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)/(-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)**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(-I*B*tan(e + f*x)/(-c**2*sqrt(-I*c*t
an(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))

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