3.8.34 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx\) [734]

Optimal. Leaf size=153 \[ \frac {(2 A-i B) x}{8 a^3 c}+\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))} \]

[Out]

1/8*(2*A-I*B)*x/a^3/c+1/12*(A+I*B)/a^3/c/f/(I-tan(f*x+e))^3-1/8*I*A/a^3/c/f/(I-tan(f*x+e))^2+1/16*(-3*A+I*B)/a
^3/c/f/(I-tan(f*x+e))+1/16*(A-I*B)/a^3/c/f/(I+tan(f*x+e))

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Rubi [A]
time = 0.15, antiderivative size = 153, 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 {3 A-i B}{16 a^3 c f (-\tan (e+f x)+i)}+\frac {A-i B}{16 a^3 c f (\tan (e+f x)+i)}+\frac {A+i B}{12 a^3 c f (-\tan (e+f x)+i)^3}+\frac {x (2 A-i B)}{8 a^3 c}-\frac {i A}{8 a^3 c f (-\tan (e+f x)+i)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A - I*B)*x)/(8*a^3*c) + (A + I*B)/(12*a^3*c*f*(I - Tan[e + f*x])^3) - ((I/8)*A)/(a^3*c*f*(I - Tan[e + f*x]
)^2) - (3*A - I*B)/(16*a^3*c*f*(I - Tan[e + f*x])) + (A - I*B)/(16*a^3*c*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))^3 (c-i c \tan (e+f x))} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {A+i B}{4 a^4 c^2 (-i+x)^4}+\frac {i A}{4 a^4 c^2 (-i+x)^3}+\frac {-3 A+i B}{16 a^4 c^2 (-i+x)^2}+\frac {-A+i B}{16 a^4 c^2 (i+x)^2}+\frac {2 A-i B}{8 a^4 c^2 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))}+\frac {(2 A-i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 c f}\\ &=\frac {(2 A-i B) x}{8 a^3 c}+\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.98, size = 164, normalized size = 1.07 \begin {gather*} -\frac {\sec ^2(e+f x) (18 i A+3 (B (-1-4 i f x)+A (2 i+8 f x)) \cos (2 (e+f x))+(-2 i A-4 B) \cos (4 (e+f x))+6 A \sin (2 (e+f x))+3 i B \sin (2 (e+f x))+24 i A f x \sin (2 (e+f x))+12 B f x \sin (2 (e+f x))+4 A \sin (4 (e+f x))-2 i B \sin (4 (e+f x)))}{96 a^3 c f (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(Sec[e + f*x]^2*((18*I)*A + 3*(B*(-1 - (4*I)*f*x) + A*(2*I + 8*f*x))*Cos[2*(e + f*x)] + ((-2*I)*A - 4*B)
*Cos[4*(e + f*x)] + 6*A*Sin[2*(e + f*x)] + (3*I)*B*Sin[2*(e + f*x)] + (24*I)*A*f*x*Sin[2*(e + f*x)] + 12*B*f*x
*Sin[2*(e + f*x)] + 4*A*Sin[4*(e + f*x)] - (2*I)*B*Sin[4*(e + f*x)]))/(a^3*c*f*(-I + Tan[e + f*x])^2)

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Maple [A]
time = 0.31, size = 128, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {i A}{8 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {3 A}{16}+\frac {i B}{16}}{-i+\tan \left (f x +e \right )}-\frac {\frac {A}{4}+\frac {i B}{4}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\left (-\frac {B}{16}-\frac {i A}{8}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{16}+\frac {i B}{16}}{i+\tan \left (f x +e \right )}+\left (\frac {B}{16}+\frac {i A}{8}\right ) \ln \left (i+\tan \left (f x +e \right )\right )}{f \,a^{3} c}\)	\(128\)
default	\(\frac {-\frac {i A}{8 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {-\frac {3 A}{16}+\frac {i B}{16}}{-i+\tan \left (f x +e \right )}-\frac {\frac {A}{4}+\frac {i B}{4}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\left (-\frac {B}{16}-\frac {i A}{8}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{16}+\frac {i B}{16}}{i+\tan \left (f x +e \right )}+\left (\frac {B}{16}+\frac {i A}{8}\right ) \ln \left (i+\tan \left (f x +e \right )\right )}{f \,a^{3} c}\)	\(128\)
risch	\(-\frac {i x B}{8 a^{3} c}+\frac {x A}{4 a^{3} c}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} B}{32 a^{3} c f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} A}{16 a^{3} c f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{96 a^{3} c f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{96 a^{3} c f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {5 i \cos \left (2 f x +2 e \right ) A}{32 a^{3} c f}-\frac {i \sin \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {7 \sin \left (2 f x +2 e \right ) A}{32 a^{3} c f}\)	\(195\)
norman	\(\frac {\frac {\left (-i B +2 A \right ) x}{8 a c}-\frac {-4 i A +B}{12 a c f}+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{4 a c f}+\frac {\left (-i B +2 A \right ) \left (\tan ^{3}\left (f x +e \right )\right )}{3 a c f}+\frac {\left (-i B +2 A \right ) \left (\tan ^{5}\left (f x +e \right )\right )}{8 a c f}+\frac {3 \left (-i B +2 A \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{8 a c}+\frac {3 \left (-i B +2 A \right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{8 a c}+\frac {\left (-i B +2 A \right ) x \left (\tan ^{6}\left (f x +e \right )\right )}{8 a c}+\frac {\left (i B +6 A \right ) \tan \left (f x +e \right )}{8 a c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{3} a^{2}}\)	\(226\)

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))^3/(c-I*c*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f/a^3/c*(-1/8*I*A/(-I+tan(f*x+e))^2-(-3/16*A+1/16*I*B)/(-I+tan(f*x+e))-1/3*(1/4*A+1/4*I*B)/(-I+tan(f*x+e))^3
+(-1/16*B-1/8*I*A)*ln(-I+tan(f*x+e))-(-1/16*A+1/16*I*B)/(I+tan(f*x+e))+(1/16*B+1/8*I*A)*ln(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))^3/(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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Fricas [A]
time = 6.83, size = 96, normalized size = 0.63 \begin {gather*} \frac {{\left (12 \, {\left (2 \, A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 18 i \, A e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-2 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} c 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))^3/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/96*(12*(2*A - I*B)*f*x*e^(6*I*f*x + 6*I*e) - 3*(I*A + B)*e^(8*I*f*x + 8*I*e) + 18*I*A*e^(4*I*f*x + 4*I*e) -
3*(-2*I*A + B)*e^(2*I*f*x + 2*I*e) + I*A - B)*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)

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Sympy [A]
time = 0.36, size = 340, normalized size = 2.22 \begin {gather*} \begin {cases} \frac {\left (294912 i A a^{9} c^{3} f^{3} e^{10 i e} e^{- 2 i f x} + \left (16384 i A a^{9} c^{3} f^{3} e^{6 i e} - 16384 B a^{9} c^{3} f^{3} e^{6 i e}\right ) e^{- 6 i f x} + \left (98304 i A a^{9} c^{3} f^{3} e^{8 i e} - 49152 B a^{9} c^{3} f^{3} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 49152 i A a^{9} c^{3} f^{3} e^{14 i e} - 49152 B a^{9} c^{3} f^{3} e^{14 i e}\right ) e^{2 i f x}\right ) e^{- 12 i e}}{1572864 a^{12} c^{4} f^{4}} & \text {for}\: a^{12} c^{4} f^{4} e^{12 i e} \neq 0 \\x \left (- \frac {2 A - i B}{8 a^{3} c} + \frac {\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{16 a^{3} c}\right ) & \text {otherwise} \end {cases} + \frac {x \left (2 A - i B\right )}{8 a^{3} c} \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))**3/(c-I*c*tan(f*x+e)),x)

[Out]

Piecewise(((294912*I*A*a**9*c**3*f**3*exp(10*I*e)*exp(-2*I*f*x) + (16384*I*A*a**9*c**3*f**3*exp(6*I*e) - 16384
*B*a**9*c**3*f**3*exp(6*I*e))*exp(-6*I*f*x) + (98304*I*A*a**9*c**3*f**3*exp(8*I*e) - 49152*B*a**9*c**3*f**3*ex
p(8*I*e))*exp(-4*I*f*x) + (-49152*I*A*a**9*c**3*f**3*exp(14*I*e) - 49152*B*a**9*c**3*f**3*exp(14*I*e))*exp(2*I
*f*x))*exp(-12*I*e)/(1572864*a**12*c**4*f**4), Ne(a**12*c**4*f**4*exp(12*I*e), 0)), (x*(-(2*A - I*B)/(8*a**3*c
) + (A*exp(8*I*e) + 4*A*exp(6*I*e) + 6*A*exp(4*I*e) + 4*A*exp(2*I*e) + A - I*B*exp(8*I*e) - 2*I*B*exp(6*I*e) +
2*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(16*a**3*c)), True)) + x*(2*A - I*B)/(8*a**3*c)

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Giac [A]
time = 0.78, size = 192, normalized size = 1.25 \begin {gather*} -\frac {\frac {6 \, {\left (-2 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A \tan \left (f x + e\right ) + B \tan \left (f x + e\right ) - 3 \, A + 2 i \, B\right )}}{a^{3} c {\left (\tan \left (f x + e\right ) + i\right )}} + \frac {-22 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 84 \, A \tan \left (f x + e\right )^{2} + 39 i \, B \tan \left (f x + e\right )^{2} + 114 i \, A \tan \left (f x + e\right ) + 45 \, B \tan \left (f x + e\right ) + 60 \, A - 9 i \, B}{a^{3} c {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, 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))^3/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/96*(6*(-2*I*A - B)*log(tan(f*x + e) + I)/(a^3*c) + 6*(2*I*A + B)*log(tan(f*x + e) - I)/(a^3*c) + 6*(2*I*A*t
an(f*x + e) + B*tan(f*x + e) - 3*A + 2*I*B)/(a^3*c*(tan(f*x + e) + I)) + (-22*I*A*tan(f*x + e)^3 - 11*B*tan(f*
x + e)^3 - 84*A*tan(f*x + e)^2 + 39*I*B*tan(f*x + e)^2 + 114*I*A*tan(f*x + e) + 45*B*tan(f*x + e) + 60*A - 9*I
*B)/(a^3*c*(tan(f*x + e) - I)^3))/f

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Mupad [B]
time = 9.06, size = 161, normalized size = 1.05 \begin {gather*} \frac {\frac {A}{3\,a^3\,c}+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {A}{2\,a^3\,c}-\frac {B\,1{}\mathrm {i}}{4\,a^3\,c}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {B}{8\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{4\,a^3\,c}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {B}{24\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{12\,a^3\,c}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3\,c}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {x\,\left (B+A\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3\,c} \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)^3*(c - c*tan(e + f*x)*1i)),x)

[Out]

(tan(e + f*x)^2*(A/(2*a^3*c) - (B*1i)/(4*a^3*c)) - tan(e + f*x)*((A*1i)/(12*a^3*c) + B/(24*a^3*c)) + tan(e + f
*x)^3*((A*1i)/(4*a^3*c) + B/(8*a^3*c)) + A/(3*a^3*c) + (B*1i)/(12*a^3*c))/(f*(2*tan(e + f*x) + 2*tan(e + f*x)^
3 + tan(e + f*x)^4*1i - 1i)) - (x*(A*2i + B)*1i)/(8*a^3*c)

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