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

Optimal. Leaf size=117 \[ \frac {A-i B}{8 a^2 c f (\tan (e+f x)+i)}-\frac {-B+i A}{8 a^2 c f (-\tan (e+f x)+i)^2}+\frac {x (3 A-i B)}{8 a^2 c}-\frac {A}{4 a^2 c f (-\tan (e+f x)+i)} \]

[Out]

1/8*(3*A-I*B)*x/a^2/c+1/8*(-I*A+B)/a^2/c/f/(-tan(f*x+e)+I)^2-1/4*A/a^2/c/f/(-tan(f*x+e)+I)+1/8*(A-I*B)/a^2/c/f
/(tan(f*x+e)+I)

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Rubi [A]  time = 0.19, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3588, 77, 203} \[ \frac {A-i B}{8 a^2 c f (\tan (e+f x)+i)}-\frac {-B+i A}{8 a^2 c f (-\tan (e+f x)+i)^2}+\frac {x (3 A-i B)}{8 a^2 c}-\frac {A}{4 a^2 c f (-\tan (e+f x)+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A - I*B)*x)/(8*a^2*c) - (I*A - B)/(8*a^2*c*f*(I - Tan[e + f*x])^2) - A/(4*a^2*c*f*(I - Tan[e + f*x])) + (A
- I*B)/(8*a^2*c*f*(I + 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 (c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {i (A+i B)}{4 a^3 c^2 (-i+x)^3}-\frac {A}{4 a^3 c^2 (-i+x)^2}+\frac {-A+i B}{8 a^3 c^2 (i+x)^2}+\frac {3 A-i B}{8 a^3 c^2 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A-B}{8 a^2 c f (i-\tan (e+f x))^2}-\frac {A}{4 a^2 c f (i-\tan (e+f x))}+\frac {A-i B}{8 a^2 c f (i+\tan (e+f x))}+\frac {(3 A-i B) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^2 c f}\\ &=\frac {(3 A-i B) x}{8 a^2 c}-\frac {i A-B}{8 a^2 c f (i-\tan (e+f x))^2}-\frac {A}{4 a^2 c f (i-\tan (e+f x))}+\frac {A-i B}{8 a^2 c f (i+\tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 2.31, size = 129, normalized size = 1.10 \[ -\frac {2 (A-3 i B) \cos (2 (e+f x))+(B+3 i A) \sin (3 (e+f x)) \sec (e+f x)-12 A f x \tan (e+f x)+6 i A \tan (e+f x)+12 i A f x-7 A-2 B \tan (e+f x)+4 i B f x \tan (e+f x)+4 B f x+i B}{32 a^2 c f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(-7*A + I*B + (12*I)*A*f*x + 4*B*f*x + 2*(A - (3*I)*B)*Cos[2*(e + f*x)] + ((3*I)*A + B)*Sec[e + f*x]*Sin
[3*(e + f*x)] + (6*I)*A*Tan[e + f*x] - 2*B*Tan[e + f*x] - 12*A*f*x*Tan[e + f*x] + (4*I)*B*f*x*Tan[e + f*x])/(a
^2*c*f*(-I + Tan[e + f*x]))

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fricas [A]  time = 0.66, size = 81, normalized size = 0.69 \[ \frac {{\left (4 \, {\left (3 \, A - i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-2 i \, A - 2 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (6 i \, A - 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*(3*A - I*B)*f*x*e^(4*I*f*x + 4*I*e) + (-2*I*A - 2*B)*e^(6*I*f*x + 6*I*e) + (6*I*A - 2*B)*e^(2*I*f*x +
2*I*e) + I*A - B)*e^(-4*I*f*x - 4*I*e)/(a^2*c*f)

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giac [A]  time = 4.75, size = 169, normalized size = 1.44 \[ \frac {\frac {2 \, {\left (3 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c} + \frac {2 \, {\left (-3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c} - \frac {2 \, {\left (3 \, A \tan \left (f x + e\right ) - i \, B \tan \left (f x + e\right ) + 5 i \, A + 3 \, B\right )}}{a^{2} c {\left (-i \, \tan \left (f x + e\right ) + 1\right )}} + \frac {9 i \, A \tan \left (f x + e\right )^{2} + 3 \, B \tan \left (f x + e\right )^{2} + 26 \, A \tan \left (f x + e\right ) - 6 i \, B \tan \left (f x + e\right ) - 21 i \, A + B}{a^{2} c {\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{32 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*(3*I*A + B)*log(tan(f*x + e) + I)/(a^2*c) + 2*(-3*I*A - B)*log(tan(f*x + e) - I)/(a^2*c) - 2*(3*A*tan(
f*x + e) - I*B*tan(f*x + e) + 5*I*A + 3*B)/(a^2*c*(-I*tan(f*x + e) + 1)) + (9*I*A*tan(f*x + e)^2 + 3*B*tan(f*x
+ e)^2 + 26*A*tan(f*x + e) - 6*I*B*tan(f*x + e) - 21*I*A + B)/(a^2*c*(tan(f*x + e) - I)^2))/f

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maple [B]  time = 0.43, size = 209, normalized size = 1.79 \[ \frac {A}{8 f \,a^{2} c \left (\tan \left (f x +e \right )+i\right )}-\frac {i B}{8 f \,a^{2} c \left (\tan \left (f x +e \right )+i\right )}+\frac {3 i \ln \left (\tan \left (f x +e \right )+i\right ) A}{16 f \,a^{2} c}+\frac {\ln \left (\tan \left (f x +e \right )+i\right ) B}{16 f \,a^{2} c}+\frac {B}{8 f \,a^{2} c \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i A}{8 f \,a^{2} c \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {A}{4 f \,a^{2} c \left (\tan \left (f x +e \right )-i\right )}-\frac {3 i \ln \left (\tan \left (f x +e \right )-i\right ) A}{16 f \,a^{2} c}-\frac {\ln \left (\tan \left (f x +e \right )-i\right ) B}{16 f \,a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f/a^2/c/(tan(f*x+e)+I)*A-1/8*I/f/a^2/c/(tan(f*x+e)+I)*B+3/16*I/f/a^2/c*ln(tan(f*x+e)+I)*A+1/16/f/a^2/c*ln(
tan(f*x+e)+I)*B+1/8/f/a^2/c/(tan(f*x+e)-I)^2*B-1/8*I/f/a^2/c/(tan(f*x+e)-I)^2*A+1/4/f/a^2/c*A/(tan(f*x+e)-I)-3
/16*I/f/a^2/c*ln(tan(f*x+e)-I)*A-1/16/f/a^2/c*ln(tan(f*x+e)-I)*B

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 8.77, size = 129, normalized size = 1.10 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,A}{8\,a^2\,c}-\frac {B\,1{}\mathrm {i}}{8\,a^2\,c}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {B}{8\,a^2\,c}+\frac {A\,3{}\mathrm {i}}{8\,a^2\,c}\right )-\frac {B}{4\,a^2\,c}+\frac {A\,1{}\mathrm {i}}{4\,a^2\,c}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,c} \]

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)),x)

[Out]

(tan(e + f*x)*((3*A)/(8*a^2*c) - (B*1i)/(8*a^2*c)) + tan(e + f*x)^2*((A*3i)/(8*a^2*c) + B/(8*a^2*c)) + (A*1i)/
(4*a^2*c) - B/(4*a^2*c))/(f*(tan(e + f*x)*1i + tan(e + f*x)^2 + tan(e + f*x)^3*1i + 1)) - (x*(A*3i + B)*1i)/(8
*a^2*c)

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sympy [A]  time = 0.52, size = 298, normalized size = 2.55 \[ \begin {cases} - \frac {\left (\left (- 256 i A a^{4} c^{2} f^{2} e^{2 i e} + 256 B a^{4} c^{2} f^{2} e^{2 i e}\right ) e^{- 4 i f x} + \left (- 1536 i A a^{4} c^{2} f^{2} e^{4 i e} + 512 B a^{4} c^{2} f^{2} e^{4 i e}\right ) e^{- 2 i f x} + \left (512 i A a^{4} c^{2} f^{2} e^{8 i e} + 512 B a^{4} c^{2} f^{2} e^{8 i e}\right ) e^{2 i f x}\right ) e^{- 6 i e}}{8192 a^{6} c^{3} f^{3}} & \text {for}\: 8192 a^{6} c^{3} f^{3} e^{6 i e} \neq 0 \\x \left (- \frac {3 A - i B}{8 a^{2} c} + \frac {\left (A e^{6 i e} + 3 A e^{4 i e} + 3 A e^{2 i e} + A - i B e^{6 i e} - i B e^{4 i e} + i B e^{2 i e} + i B\right ) e^{- 4 i e}}{8 a^{2} c}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 3 A + i B\right )}{8 a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-((-256*I*A*a**4*c**2*f**2*exp(2*I*e) + 256*B*a**4*c**2*f**2*exp(2*I*e))*exp(-4*I*f*x) + (-1536*I*A
*a**4*c**2*f**2*exp(4*I*e) + 512*B*a**4*c**2*f**2*exp(4*I*e))*exp(-2*I*f*x) + (512*I*A*a**4*c**2*f**2*exp(8*I*
e) + 512*B*a**4*c**2*f**2*exp(8*I*e))*exp(2*I*f*x))*exp(-6*I*e)/(8192*a**6*c**3*f**3), Ne(8192*a**6*c**3*f**3*
exp(6*I*e), 0)), (x*(-(3*A - I*B)/(8*a**2*c) + (A*exp(6*I*e) + 3*A*exp(4*I*e) + 3*A*exp(2*I*e) + A - I*B*exp(6
*I*e) - I*B*exp(4*I*e) + I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(8*a**2*c)), True)) - x*(-3*A + I*B)/(8*a**2*c)

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