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*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]))

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Rubi [A]  time = 0.192524, antiderivative size = 93, normalized size of antiderivative = 1., 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 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]

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{(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*}

Mathematica [B]  time = 2.537, 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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Maple [B]  time = 0.032, size = 366, normalized size = 3.9 \begin{align*}{\frac{i{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{i{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{4\,i{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-2\,{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{2\,i{a}^{2}c}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{{a}^{2}{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) d \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{{a}^{2}d}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{2\,i{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{2\,i{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+4\,{\frac{{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}} \end{align*}

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]

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

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Maxima [B]  time = 1.5247, size = 288, normalized size = 3.1 \begin{align*} \frac{\frac{2 \,{\left (a^{2} c^{2} + 2 i \, a^{2} c d - 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} \end{align*}

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 + 2*I*a^2*c*d - 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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Fricas [A]  time = 1.53818, size = 333, normalized size = 3.58 \begin{align*} -\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} \end{align*}

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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Sympy [B]  time = 19.625, size = 530, normalized size = 5.7 \begin{align*} \frac{2 a^{2} \left (c^{18} - 18 i c^{17} d - 153 c^{16} d^{2} + 816 i c^{15} d^{3} + 3060 c^{14} d^{4} - 8568 i c^{13} d^{5} - 18564 c^{12} d^{6} + 31824 i c^{11} d^{7} + 43758 c^{10} d^{8} - 48620 i c^{9} d^{9} - 43758 c^{8} d^{10} + 31824 i c^{7} d^{11} + 18564 c^{6} d^{12} - 8568 i c^{5} d^{13} - 3060 c^{4} d^{14} + 816 i c^{3} d^{15} + 153 c^{2} d^{16} - 18 i c d^{17} - d^{18}\right ) \log{\left (\frac{c^{2} + d^{2}}{c^{2} e^{2 i e} - 2 i c d e^{2 i e} - d^{2} e^{2 i e}} + e^{2 i f x} \right )}}{f \left (i c^{20} + 20 c^{19} d - 190 i c^{18} d^{2} - 1140 c^{17} d^{3} + 4845 i c^{16} d^{4} + 15504 c^{15} d^{5} - 38760 i c^{14} d^{6} - 77520 c^{13} d^{7} + 125970 i c^{12} d^{8} + 167960 c^{11} d^{9} - 184756 i c^{10} d^{10} - 167960 c^{9} d^{11} + 125970 i c^{8} d^{12} + 77520 c^{7} d^{13} - 38760 i c^{6} d^{14} - 15504 c^{5} d^{15} + 4845 i c^{4} d^{16} + 1140 c^{3} d^{17} - 190 i c^{2} d^{18} - 20 c d^{19} + i d^{20}\right )} + \frac{2 a^{2} c^{2} + 4 i a^{2} c d - 2 a^{2} d^{2}}{\left (e^{2 i f x} + \frac{c^{2} + 2 i c d - d^{2}}{c^{2} e^{2 i e} + d^{2} e^{2 i e}}\right ) \left (i c^{4} f e^{2 i e} + 2 c^{3} d f e^{2 i e} + 2 c d^{3} f e^{2 i e} - i d^{4} f e^{2 i e}\right )} \end{align*}

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*a**2*(c**18 - 18*I*c**17*d - 153*c**16*d**2 + 816*I*c**15*d**3 + 3060*c**14*d**4 - 8568*I*c**13*d**5 - 18564
*c**12*d**6 + 31824*I*c**11*d**7 + 43758*c**10*d**8 - 48620*I*c**9*d**9 - 43758*c**8*d**10 + 31824*I*c**7*d**1
1 + 18564*c**6*d**12 - 8568*I*c**5*d**13 - 3060*c**4*d**14 + 816*I*c**3*d**15 + 153*c**2*d**16 - 18*I*c*d**17
- d**18)*log((c**2 + d**2)/(c**2*exp(2*I*e) - 2*I*c*d*exp(2*I*e) - d**2*exp(2*I*e)) + exp(2*I*f*x))/(f*(I*c**2
0 + 20*c**19*d - 190*I*c**18*d**2 - 1140*c**17*d**3 + 4845*I*c**16*d**4 + 15504*c**15*d**5 - 38760*I*c**14*d**
6 - 77520*c**13*d**7 + 125970*I*c**12*d**8 + 167960*c**11*d**9 - 184756*I*c**10*d**10 - 167960*c**9*d**11 + 12
5970*I*c**8*d**12 + 77520*c**7*d**13 - 38760*I*c**6*d**14 - 15504*c**5*d**15 + 4845*I*c**4*d**16 + 1140*c**3*d
**17 - 190*I*c**2*d**18 - 20*c*d**19 + I*d**20)) + (2*a**2*c**2 + 4*I*a**2*c*d - 2*a**2*d**2)/((exp(2*I*f*x) +
(c**2 + 2*I*c*d - d**2)/(c**2*exp(2*I*e) + d**2*exp(2*I*e)))*(I*c**4*f*exp(2*I*e) + 2*c**3*d*f*exp(2*I*e) + 2
*c*d**3*f*exp(2*I*e) - I*d**4*f*exp(2*I*e)))

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Giac [B]  time = 1.46086, size = 315, normalized size = 3.39 \begin{align*} \frac{2 \,{\left (\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} \log \left ({\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 )}{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} \end{align*}

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*(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*log(abs(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) - (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