3.8.12 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx\) [712]

Optimal. Leaf size=113 \[ \frac {(3 A+i B) x}{8 a c^2}-\frac {A+i B}{8 a c^2 f (i-\tan (e+f x))}+\frac {i A+B}{8 a c^2 f (i+\tan (e+f x))^2}+\frac {A}{4 a c^2 f (i+\tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 113, 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 = {3669, 78, 209} \begin {gather*} -\frac {A+i B}{8 a c^2 f (-\tan (e+f x)+i)}+\frac {B+i A}{8 a c^2 f (\tan (e+f x)+i)^2}+\frac {x (3 A+i B)}{8 a c^2}+\frac {A}{4 a c^2 f (\tan (e+f x)+i)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^2 (c-i c x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {-A-i B}{8 a^2 c^3 (-i+x)^2}-\frac {i (A-i B)}{4 a^2 c^3 (i+x)^3}-\frac {A}{4 a^2 c^3 (i+x)^2}+\frac {3 A+i B}{8 a^2 c^3 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {A+i B}{8 a c^2 f (i-\tan (e+f x))}+\frac {i A+B}{8 a c^2 f (i+\tan (e+f x))^2}+\frac {A}{4 a c^2 f (i+\tan (e+f x))}+\frac {(3 A+i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a c^2 f}\\ &=\frac {(3 A+i B) x}{8 a c^2}-\frac {A+i B}{8 a c^2 f (i-\tan (e+f x))}+\frac {i A+B}{8 a c^2 f (i+\tan (e+f x))^2}+\frac {A}{4 a c^2 f (i+\tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.24, size = 166, normalized size = 1.47 \begin {gather*} \frac {(2 (A (-3-6 i f x)+B (i+2 f x)) \cos (e+f x)+(A+3 i B) \cos (3 (e+f x))-(9 i A+B+12 A f x+4 i B f x+(6 i A-2 B) \cos (2 (e+f x))) \sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (A+B \tan (e+f x))}{32 a c^2 f (A \cos (e+f x)+B \sin (e+f x)) (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(A*(-3 - (6*I)*f*x) + B*(I + 2*f*x))*Cos[e + f*x] + (A + (3*I)*B)*Cos[3*(e + f*x)] - ((9*I)*A + B + 12*A*f
*x + (4*I)*B*f*x + ((6*I)*A - 2*B)*Cos[2*(e + f*x)])*Sin[e + f*x])*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(A
+ B*Tan[e + f*x]))/(32*a*c^2*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.24, size = 106, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {A}{4 i+4 \tan \left (f x +e \right )}+\left (\frac {3 i A}{16}-\frac {B}{16}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {-\frac {i A}{4}-\frac {B}{4}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}+\left (-\frac {3 i A}{16}+\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}-\frac {i B}{8}}{-i+\tan \left (f x +e \right )}}{f a \,c^{2}}\)	\(106\)
default	\(\frac {\frac {A}{4 i+4 \tan \left (f x +e \right )}+\left (\frac {3 i A}{16}-\frac {B}{16}\right ) \ln \left (i+\tan \left (f x +e \right )\right )-\frac {-\frac {i A}{4}-\frac {B}{4}}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}+\left (-\frac {3 i A}{16}+\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}-\frac {i B}{8}}{-i+\tan \left (f x +e \right )}}{f a \,c^{2}}\)	\(106\)
risch	\(\frac {i x B}{8 a \,c^{2}}+\frac {3 x A}{8 a \,c^{2}}-\frac {{\mathrm e}^{4 i \left (f x +e \right )} B}{32 a \,c^{2} f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} A}{32 a \,c^{2} f}-\frac {\cos \left (2 f x +2 e \right ) B}{8 a \,c^{2} f}-\frac {i \cos \left (2 f x +2 e \right ) A}{8 a \,c^{2} f}+\frac {A \sin \left (2 f x +2 e \right )}{4 a \,c^{2} f}\)	\(130\)
norman	\(\frac {\frac {\left (i B +3 A \right ) x}{8 a c}-\frac {i A +B}{4 a c f}+\frac {\left (i B +3 A \right ) \left (\tan ^{3}\left (f x +e \right )\right )}{8 a c f}+\frac {\left (i B +3 A \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{4 a c}+\frac {\left (i B +3 A \right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{8 a c}+\frac {\left (-i B +5 A \right ) \tan \left (f x +e \right )}{8 a c f}}{c \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}\)	\(154\)

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))/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))/(c-I*c*tan(f*x+e))^2,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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Fricas [A]
time = 4.97, size = 84, normalized size = 0.74 \begin {gather*} \frac {{\left (4 \, {\left (3 \, A + i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, {\left (3 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{32 \, a c^{2} f} \end {gather*}

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

[Out]

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

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Sympy [A]
time = 0.28, size = 284, normalized size = 2.51 \begin {gather*} \begin {cases} \frac {\left (\left (512 i A a^{2} c^{4} f^{2} - 512 B a^{2} c^{4} f^{2}\right ) e^{- 2 i f x} + \left (- 1536 i A a^{2} c^{4} f^{2} e^{4 i e} - 512 B a^{2} c^{4} f^{2} e^{4 i e}\right ) e^{2 i f x} + \left (- 256 i A a^{2} c^{4} f^{2} e^{6 i e} - 256 B a^{2} c^{4} f^{2} e^{6 i e}\right ) e^{4 i f x}\right ) e^{- 2 i e}}{8192 a^{3} c^{6} f^{3}} & \text {for}\: a^{3} c^{6} f^{3} e^{2 i e} \neq 0 \\x \left (- \frac {3 A + i B}{8 a c^{2}} + \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^{- 2 i e}}{8 a c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (3 A + i B\right )}{8 a c^{2}} \end {gather*}

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))/(c-I*c*tan(f*x+e))**2,x)

[Out]

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

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Giac [A]
time = 0.69, size = 169, normalized size = 1.50 \begin {gather*} \frac {\frac {2 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{2}} + \frac {2 \, {\left (-3 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{2}} - \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 c^{2} {\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 c^{2} {\left (\tan \left (f x + e\right ) + i\right )}^{2}}}{32 \, f} \end {gather*}

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

[Out]

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

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Mupad [B]
time = 9.06, size = 129, normalized size = 1.14 \begin {gather*} \frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {B}{8\,a\,c^2}+\frac {A\,3{}\mathrm {i}}{8\,a\,c^2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {3\,A}{8\,a\,c^2}+\frac {B\,1{}\mathrm {i}}{8\,a\,c^2}\right )+\frac {A}{4\,a\,c^2}-\frac {B\,1{}\mathrm {i}}{4\,a\,c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3+{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a\,c^2} \end {gather*}

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)^2),x)

[Out]

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

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