3.1091 \(\int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac {a}{f (d+i c) (c+d \tan (e+f x))}-\frac {i a \log (c \cos (e+f x)+d \sin (e+f x))}{f (c-i d)^2}+\frac {a x}{(c-i d)^2} \]

[Out]

a*x/(c-I*d)^2-I*a*ln(c*cos(f*x+e)+d*sin(f*x+e))/(c-I*d)^2/f-a/(I*c+d)/f/(c+d*tan(f*x+e))

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Rubi [A]  time = 0.15, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3529, 3531, 3530} \[ -\frac {a}{f (d+i c) (c+d \tan (e+f x))}-\frac {i a \log (c \cos (e+f x)+d \sin (e+f x))}{f (c-i d)^2}+\frac {a x}{(c-i d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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]

Rubi steps

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

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Mathematica [B]  time = 2.93, size = 302, normalized size = 4.03 \[ \frac {(\cos (e)-i \sin (e)) \cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (4 \tan ^{-1}\left (\frac {\left (d^2-c^2\right ) \sin (2 e+f x)+2 c d \cos (2 e+f x)}{\left (c^2-d^2\right ) \cos (2 e+f x)+2 c d \sin (2 e+f x)}\right )+\frac {\left (c^2+d^2\right ) \cos (f x) \left (4 f x-i \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )\right )+\left (c^2-d^2\right ) \cos (2 e+f x) \left (4 f x-i \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )\right )-2 d \left (c \sin (2 e+f x) \left (-4 f x+i \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )\right )+2 (d+i c) \sin (f x)\right )}{(c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}\right )}{4 f (c-i d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 125, normalized size = 1.67 \[ \frac {-2 i \, a d - {\left (a c + i \, a d + {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{3} - c^{2} d - i \, c d^{2} - d^{3}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*a*d - (a*c + I*a*d + (a*c - I*a*d)*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)))/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - c^2*d - I*c*d^2 - d^3)*f)

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giac [B]  time = 0.51, size = 186, normalized size = 2.48 \[ \frac {2 \, {\left (\frac {a \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{2 i \, c^{2} + 4 \, c d - 2 i \, d^{2}} + \frac {a \log \left (-i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{-i \, c^{2} - 2 \, c d + i \, d^{2}} - \frac {a c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 i \, a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a c^{2}}{{\left (2 i \, c^{3} + 4 \, c^{2} d - 2 i \, c d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.27, size = 309, normalized size = 4.12 \[ -\frac {i a \ln \left (c +d \tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {i a \ln \left (c +d \tan \left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {2 a \ln \left (c +d \tan \left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {i a c}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}-\frac {a d}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}-\frac {a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {i a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2}}{2 f \left (c^{2}+d^{2}\right )^{2}}-\frac {i a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d^{2}}{2 f \left (c^{2}+d^{2}\right )^{2}}+\frac {2 i a \arctan \left (\tan \left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {a \arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {a \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/f*a/(c^2+d^2)^2*ln(c+d*tan(f*x+e))*c^2+I/f*a/(c^2+d^2)^2*ln(c+d*tan(f*x+e))*d^2+2/f*a/(c^2+d^2)^2*ln(c+d*ta
n(f*x+e))*c*d+I/f*a/(c^2+d^2)/(c+d*tan(f*x+e))*c-1/f*a/(c^2+d^2)/(c+d*tan(f*x+e))*d-1/f*a/(c^2+d^2)^2*ln(1+tan
(f*x+e)^2)*c*d+1/2*I/f*a/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*c^2-1/2*I/f*a/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*d^2+2*I/f
*a/(c^2+d^2)^2*arctan(tan(f*x+e))*c*d+1/f*a/(c^2+d^2)^2*arctan(tan(f*x+e))*c^2-1/f*a/(c^2+d^2)^2*arctan(tan(f*
x+e))*d^2

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maxima [B]  time = 0.76, size = 180, normalized size = 2.40 \[ \frac {\frac {2 \, {\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} {\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {2 \, {\left (-i \, a c^{2} + 2 \, a c d + i \, a d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {{\left (i \, a c^{2} - 2 \, a c d - i \, a d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {2 \, {\left (i \, a c - a d\right )}}{c^{3} + c d^{2} + {\left (c^{2} d + d^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.13, size = 135, normalized size = 1.80 \[ -\frac {2\,a\,\mathrm {atan}\left (\frac {\left (c^2+d^2\right )\,1{}\mathrm {i}}{{\left (d+c\,1{}\mathrm {i}\right )}^2}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,c^4\,d^2+4\,c^2\,d^4+2\,d^6\right )}{{\left (d+c\,1{}\mathrm {i}\right )}^2\,\left (c^3\,d\,1{}\mathrm {i}-c^2\,d^2+c\,d^3\,1{}\mathrm {i}-d^4\right )}\right )}{f\,{\left (d+c\,1{}\mathrm {i}\right )}^2}+\frac {a\,1{}\mathrm {i}}{d\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+\frac {c}{d}\right )\,\left (c-d\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*1i)/(d*f*(tan(e + f*x) + c/d)*(c - d*1i)) - (2*a*atan(((c^2 + d^2)*1i)/(c*1i + d)^2 - (tan(e + f*x)*(2*d^6
+ 4*c^2*d^4 + 2*c^4*d^2))/((c*1i + d)^2*(c*d^3*1i + c^3*d*1i - d^4 - c^2*d^2))))/(f*(c*1i + d)^2)

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sympy [B]  time = 3.56, size = 144, normalized size = 1.92 \[ \frac {2 i a d}{i c^{3} f + c^{2} d f + i c d^{2} f + d^{3} f + \left (i c^{3} f e^{2 i e} + 3 c^{2} d f e^{2 i e} - 3 i c d^{2} f e^{2 i e} - d^{3} f e^{2 i e}\right ) e^{2 i f x}} - \frac {i a \log {\left (\frac {i c - d}{i c e^{2 i e} + d e^{2 i e}} + e^{2 i f x} \right )}}{f \left (c - i d\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*a*d/(I*c**3*f + c**2*d*f + I*c*d**2*f + d**3*f + (I*c**3*f*exp(2*I*e) + 3*c**2*d*f*exp(2*I*e) - 3*I*c*d**2
*f*exp(2*I*e) - d**3*f*exp(2*I*e))*exp(2*I*f*x)) - I*a*log((I*c - d)/(I*c*exp(2*I*e) + d*exp(2*I*e)) + exp(2*I
*f*x))/(f*(c - I*d)**2)

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