3.1092 \(\int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=202 \[ \frac{x \left (3 i c^2 d+c^3+3 c d^2-3 i d^3\right )}{2 a (c-i d)^2 (c+i d)^3}+\frac{d^2 (3 c-i d) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (-d+i c)^3 (c-i d)^2}+\frac{d (c-3 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

[Out]

((c^3 + (3*I)*c^2*d + 3*c*d^2 - (3*I)*d^3)*x)/(2*a*(c - I*d)^2*(c + I*d)^3) + ((3*c - I*d)*d^2*Log[c*Cos[e + f
*x] + d*Sin[e + f*x]])/(a*(I*c - d)^3*(c - I*d)^2*f) + ((c - (3*I)*d)*d)/(2*a*(c - I*d)*(c + I*d)^2*f*(c + d*T
an[e + f*x])) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x]))

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Rubi [A]  time = 0.32224, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3552, 3529, 3531, 3530} \[ \frac{x \left (3 i c^2 d+c^3+3 c d^2-3 i d^3\right )}{2 a (c-i d)^2 (c+i d)^3}+\frac{d^2 (3 c-i d) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (-d+i c)^3 (c-i d)^2}+\frac{d (c-3 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^2),x]

[Out]

((c^3 + (3*I)*c^2*d + 3*c*d^2 - (3*I)*d^3)*x)/(2*a*(c - I*d)^2*(c + I*d)^3) + ((3*c - I*d)*d^2*Log[c*Cos[e + f
*x] + d*Sin[e + f*x]])/(a*(I*c - d)^3*(c - I*d)^2*f) + ((c - (3*I)*d)*d)/(2*a*(c - I*d)*(c + I*d)^2*f*(c + d*T
an[e + f*x])) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x]))

Rule 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx &=-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}+\frac{\int \frac{a (i c-3 d)+2 i a d \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx}{2 a^2 (i c-d)}\\ &=\frac{(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}+\frac{\int \frac{-a \left (3 c d-i \left (c^2+2 d^2\right )\right )+a d (i c+3 d) \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{2 a^2 (i c-d) \left (c^2+d^2\right )}\\ &=\frac{\left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) x}{2 a (c-i d)^2 (c+i d)^3}+\frac{(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}-\frac{\left ((3 c-i d) d^2\right ) \int \frac{d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{a (i c-d) \left (c^2+d^2\right )^2}\\ &=\frac{\left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) x}{2 a (c-i d)^2 (c+i d)^3}+\frac{(3 c-i d) d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{a (i c-d)^3 (c-i d)^2 f}+\frac{(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.58913, size = 385, normalized size = 1.91 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac{2 x \left (3 i c^2 d+c^3+3 c d^2-3 i d^3\right ) (\cos (e)+i \sin (e))}{(c-i d)^2}+\frac{4 i d^3 (c+i d) (\cos (e)+i \sin (e)) \sin (f x)}{f (c-i d) (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{2 d^2 (d+3 i c) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right )^2 \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{f (c-i d)^2}-\frac{4 d^2 (3 c-i d) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right )^2 \tan ^{-1}\left (\frac{c \sin (f x)+d \cos (f x)}{d \sin (f x)-c \cos (f x)}\right )}{f (c-i d)^2}-\frac{4 d^2 x (3 c-i d) (\cos (e)+i \sin (e))}{(c-i d)^2}+\frac{(c+i d) (\sin (e)+i \cos (e)) \cos (2 f x)}{f}+\frac{(c+i d) (\cos (e)-i \sin (e)) \sin (2 f x)}{f}\right )}{4 (c+i d)^3 (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^2),x]

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((-4*(3*c - I*d)*d^2*ArcTan[(d*Cos[f*x] + c*Sin[f*x])/(-(c*Cos[f*x]) + d
*Sin[f*x])]*(Cos[e/2] + I*Sin[e/2])^2)/((c - I*d)^2*f) + (2*d^2*((3*I)*c + d)*Log[(c*Cos[e + f*x] + d*Sin[e +
f*x])^2]*(Cos[e/2] + I*Sin[e/2])^2)/((c - I*d)^2*f) - (4*(3*c - I*d)*d^2*x*(Cos[e] + I*Sin[e]))/(c - I*d)^2 +
(2*(c^3 + (3*I)*c^2*d + 3*c*d^2 - (3*I)*d^3)*x*(Cos[e] + I*Sin[e]))/(c - I*d)^2 + ((c + I*d)*Cos[2*f*x]*(I*Cos
[e] + Sin[e]))/f + ((c + I*d)*(Cos[e] - I*Sin[e])*Sin[2*f*x])/f + ((4*I)*(c + I*d)*d^3*(Cos[e] + I*Sin[e])*Sin
[f*x])/((c - I*d)*f*(c*Cos[e] + d*Sin[e])*(c*Cos[e + f*x] + d*Sin[e + f*x]))))/(4*(c + I*d)^3*(a + I*a*Tan[e +
f*x]))

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Maple [A]  time = 0.064, size = 281, normalized size = 1.4 \begin{align*}{\frac{-{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) c}{af \left ( c+id \right ) ^{3}}}+{\frac{5\,\ln \left ( \tan \left ( fx+e \right ) -i \right ) d}{4\,af \left ( c+id \right ) ^{3}}}+{\frac{1}{2\,af \left ( c+id \right ) ^{2} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{af \left ( id-c \right ) ^{2}}}-{\frac{i{d}^{2}{c}^{2}}{af \left ( id-c \right ) ^{2} \left ( c+id \right ) ^{3} \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{i{d}^{4}}{af \left ( id-c \right ) ^{2} \left ( c+id \right ) ^{3} \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{3\,i{d}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) c}{af \left ( id-c \right ) ^{2} \left ( c+id \right ) ^{3}}}+{\frac{{d}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{af \left ( id-c \right ) ^{2} \left ( c+id \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^2,x)

[Out]

-1/4*I/f/a/(c+I*d)^3*ln(tan(f*x+e)-I)*c+5/4/f/a/(c+I*d)^3*ln(tan(f*x+e)-I)*d+1/2/f/a/(c+I*d)^2/(tan(f*x+e)-I)+
1/4*I/f/a/(I*d-c)^2*ln(tan(f*x+e)+I)-I/f/a*d^2/(I*d-c)^2/(c+I*d)^3/(c+d*tan(f*x+e))*c^2-I/f/a*d^4/(I*d-c)^2/(c
+I*d)^3/(c+d*tan(f*x+e))+3*I/f/a*d^2/(I*d-c)^2/(c+I*d)^3*ln(c+d*tan(f*x+e))*c+1/f/a*d^3/(I*d-c)^2/(c+I*d)^3*ln
(c+d*tan(f*x+e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.73717, size = 801, normalized size = 3.97 \begin{align*} \frac{i \, c^{4} + 2 i \, c^{2} d^{2} + i \, d^{4} +{\left (2 \, c^{4} + 4 i \, c^{3} d + 24 \, c^{2} d^{2} - 28 i \, c d^{3} - 10 \, d^{4}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (i \, c^{4} + 2 \, c^{3} d - 6 \, c d^{3} - 9 i \, d^{4} +{\left (2 \, c^{4} + 8 i \, c^{3} d + 12 \, c^{2} d^{2} + 8 i \, c d^{3} + 10 \, d^{4}\right )} f x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left ({\left (12 i \, c^{2} d^{2} + 16 \, c d^{3} - 4 i \, d^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (12 i \, c^{2} d^{2} - 8 \, c d^{3} + 4 i \, d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac{{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{4 \,{\left (a c^{6} + 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} + a d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (4 \, a c^{6} + 8 i \, a c^{5} d + 4 \, a c^{4} d^{2} + 16 i \, a c^{3} d^{3} - 4 \, a c^{2} d^{4} + 8 i \, a c d^{5} - 4 \, a d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

(I*c^4 + 2*I*c^2*d^2 + I*d^4 + (2*c^4 + 4*I*c^3*d + 24*c^2*d^2 - 28*I*c*d^3 - 10*d^4)*f*x*e^(4*I*f*x + 4*I*e)
+ (I*c^4 + 2*c^3*d - 6*c*d^3 - 9*I*d^4 + (2*c^4 + 8*I*c^3*d + 12*c^2*d^2 + 8*I*c*d^3 + 10*d^4)*f*x)*e^(2*I*f*x
+ 2*I*e) + ((12*I*c^2*d^2 + 16*c*d^3 - 4*I*d^4)*e^(4*I*f*x + 4*I*e) + (12*I*c^2*d^2 - 8*c*d^3 + 4*I*d^4)*e^(2
*I*f*x + 2*I*e))*log(((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c - d)/(I*c + d)))/(4*(a*c^6 + 3*a*c^4*d^2 + 3*a*c^2*d
^4 + a*d^6)*f*e^(4*I*f*x + 4*I*e) + (4*a*c^6 + 8*I*a*c^5*d + 4*a*c^4*d^2 + 16*I*a*c^3*d^3 - 4*a*c^2*d^4 + 8*I*
a*c*d^5 - 4*a*d^6)*f*e^(2*I*f*x + 2*I*e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.44859, size = 463, normalized size = 2.29 \begin{align*} \frac{\frac{16 \,{\left (3 \, c d^{3} - i \, d^{4}\right )} \log \left (i \, d \tan \left (f x + e\right ) + i \, c\right )}{-2 i \, a c^{5} d + 2 \, a c^{4} d^{2} - 4 i \, a c^{3} d^{3} + 4 \, a c^{2} d^{4} - 2 i \, a c d^{5} + 2 \, a d^{6}} - \frac{16 \,{\left (i \, c - 5 \, d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{8 \, a c^{3} + 24 i \, a c^{2} d - 24 \, a c d^{2} - 8 i \, a d^{3}} - \frac{16 \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{8 i \, a c^{2} + 16 \, a c d - 8 i \, a d^{2}} + \frac{i \, c^{2} d \tan \left (f x + e\right )^{2} - 2 \, c d^{2} \tan \left (f x + e\right )^{2} - i \, d^{3} \tan \left (f x + e\right )^{2} + i \, c^{3} \tan \left (f x + e\right ) + 3 \, c^{2} d \tan \left (f x + e\right ) - 15 i \, c d^{2} \tan \left (f x + e\right ) - 13 \, d^{3} \tan \left (f x + e\right ) + 5 \, c^{3} - 6 i \, c^{2} d - 13 \, c d^{2} + 8 i \, d^{3}}{{\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )}{\left (d \tan \left (f x + e\right )^{2} + c \tan \left (f x + e\right ) - i \, d \tan \left (f x + e\right ) - i \, c\right )}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/8*(16*(3*c*d^3 - I*d^4)*log(I*d*tan(f*x + e) + I*c)/(-2*I*a*c^5*d + 2*a*c^4*d^2 - 4*I*a*c^3*d^3 + 4*a*c^2*d^
4 - 2*I*a*c*d^5 + 2*a*d^6) - 16*(I*c - 5*d)*log(tan(f*x + e) - I)/(8*a*c^3 + 24*I*a*c^2*d - 24*a*c*d^2 - 8*I*a
*d^3) - 16*log(-I*tan(f*x + e) + 1)/(8*I*a*c^2 + 16*a*c*d - 8*I*a*d^2) + (I*c^2*d*tan(f*x + e)^2 - 2*c*d^2*tan
(f*x + e)^2 - I*d^3*tan(f*x + e)^2 + I*c^3*tan(f*x + e) + 3*c^2*d*tan(f*x + e) - 15*I*c*d^2*tan(f*x + e) - 13*
d^3*tan(f*x + e) + 5*c^3 - 6*I*c^2*d - 13*c*d^2 + 8*I*d^3)/((a*c^4 + 2*a*c^2*d^2 + a*d^4)*(d*tan(f*x + e)^2 +
c*tan(f*x + e) - I*d*tan(f*x + e) - I*c)))/f