3.1090 \(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=93 \[ \frac {a^2 (-d+i c)}{d f (d+i c) (c+d \tan (e+f x))}-\frac {2 i a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f (c-i d)^2}+\frac {2 a^2 x}{(c-i d)^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3542, 3531, 3530} \[ \frac {a^2 (-d+i c)}{d f (d+i c) (c+d \tan (e+f x))}-\frac {2 i a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f (c-i d)^2}+\frac {2 a^2 x}{(c-i d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3542

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

Rubi steps

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

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Mathematica [B]  time = 2.95, size = 253, normalized size = 2.72 \[ \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (\frac {2 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {\left (d^2-c^2\right ) \sin (3 e+f x)+2 c d \cos (3 e+f x)}{\left (c^2-d^2\right ) \cos (3 e+f x)+2 c d \sin (3 e+f x)}\right )}{f}-\frac {(c-i d) (c+i d) (\cos (2 e)-i \sin (2 e)) \sin (f x)}{f (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac {(-\sin (2 e)-i \cos (2 e)) \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{f}+4 x (\cos (2 e)-i \sin (2 e))\right )}{(c-i d)^2 (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 143, normalized size = 1.54 \[ -\frac {2 \, a^{2} c + 2 i \, a^{2} d + {\left (2 \, a^{2} c + 2 i \, a^{2} d + {\left (2 \, a^{2} c - 2 i \, a^{2} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{3} - c^{2} d - i \, c d^{2} - d^{3}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*c + 2*I*a^2*d + (2*a^2*c + 2*I*a^2*d + (2*a^2*c - 2*I*a^2*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 = 0.97, size = 232, normalized size = 2.49 \[ \frac {2 \, {\left (\frac {a^{2} \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{i \, c^{2} + 2 \, c d - i \, d^{2}} + \frac {2 \, a^{2} \log \left (-i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{-i \, c^{2} - 2 \, c d + i \, d^{2}} - \frac {a^{2} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - i \, a^{2} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i \, a^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a^{2} c^{2}}{{\left (i \, c^{3} + 2 \, c^{2} d - i \, c d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^2*log(c*tan(1/2*f*x + 1/2*e)^2 - 2*d*tan(1/2*f*x + 1/2*e) - c)/(I*c^2 + 2*c*d - I*d^2) + 2*a^2*log(-I*tan
(1/2*f*x + 1/2*e) + 1)/(-I*c^2 - 2*c*d + I*d^2) - (a^2*c^2*tan(1/2*f*x + 1/2*e)^2 - I*a^2*c^2*tan(1/2*f*x + 1/
2*e) - 2*a^2*c*d*tan(1/2*f*x + 1/2*e) - I*a^2*d^2*tan(1/2*f*x + 1/2*e) - a^2*c^2)/((I*c^3 + 2*c^2*d - 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 = 0.24, size = 366, normalized size = 3.94 \[ \frac {2 i a^{2} c}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}+\frac {a^{2} c^{2}}{f \left (c^{2}+d^{2}\right ) d \left (c +d \tan \left (f x +e \right )\right )}-\frac {a^{2} d}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}-\frac {2 i a^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {2 i a^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {4 a^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {4 i a^{2} \arctan \left (\tan \left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {2 a^{2} \arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {2 a^{2} \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 213, normalized size = 2.29 \[ \frac {\frac {{\left (2 \, a^{2} c^{2} + 4 i \, a^{2} c d - 2 \, a^{2} d^{2}\right )} {\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {{\left (-2 i \, a^{2} c^{2} + 4 \, a^{2} c d + 2 i \, a^{2} d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {{\left (i \, a^{2} c^{2} - 2 \, a^{2} c d - i \, a^{2} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}}{c^{3} d + c d^{3} + {\left (c^{2} d^{2} + d^{4}\right )} \tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.18, size = 139, normalized size = 1.49 \[ -\frac {a^2\,\mathrm {atanh}\left (\frac {c^2+d^2}{{\left (d+c\,1{}\mathrm {i}\right )}^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,c^4\,d^2+4\,c^2\,d^4+2\,d^6\right )}{{\left (d+c\,1{}\mathrm {i}\right )}^2\,\left (c^3\,d+c^2\,d^2\,1{}\mathrm {i}+c\,d^3+d^4\,1{}\mathrm {i}\right )}\right )\,4{}\mathrm {i}}{f\,{\left (d+c\,1{}\mathrm {i}\right )}^2}+\frac {a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d^2\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+\frac {c}{d}\right )\,\left (c-d\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 3.53, size = 156, normalized size = 1.68 \[ - \frac {2 i a^{2} \log {\left (\frac {i c - d}{i c e^{2 i e} + d e^{2 i e}} + e^{2 i f x} \right )}}{f \left (c - i d\right )^{2}} + \frac {2 a^{2} c + 2 i a^{2} d}{i c^{3} f + c^{2} d f + i c d^{2} f + d^{3} f + \left (i c^{3} f e^{2 i e} + 3 c^{2} d f e^{2 i e} - 3 i c d^{2} f e^{2 i e} - d^{3} f e^{2 i e}\right ) e^{2 i f x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*a**2*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) + (2*a**2*c + 2*I*a**
2*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))

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