3.1084 \(\int \frac{(a+i a \tan (e+f x))^2}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{a^2 c x (c+i d)}{d^2 (c-i d)}+\frac{a^2 x (c+2 i d)}{d^2}-\frac{a^2 (-d+i c) \log (c \cos (e+f x)+d \sin (e+f x))}{d f (d+i c)}+\frac{a^2 \log (\cos (e+f x))}{d f} \]

[Out]

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

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Rubi [A]  time = 0.123487, antiderivative size = 106, 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 = {3541, 3475, 3484, 3530} \[ -\frac{a^2 c x (c+i d)}{d^2 (c-i d)}+\frac{a^2 x (c+2 i d)}{d^2}-\frac{a^2 (-d+i c) \log (c \cos (e+f x)+d \sin (e+f x))}{d f (d+i c)}+\frac{a^2 \log (\cos (e+f x))}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3541

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*(2
*b*c - a*d)*x)/b^2, x] + (Dist[d^2/b, Int[Tan[e + f*x], x], x] + Dist[(b*c - a*d)^2/b^2, Int[1/(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]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3484

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2),
Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 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)} \, dx &=\frac{a^2 (c+2 i d) x}{d^2}-\frac{a^2 \int \tan (e+f x) \, dx}{d}+\frac{(-i a c+a d)^2 \int \frac{1}{c+d \tan (e+f x)} \, dx}{d^2}\\ &=-\frac{a^2 c (c+i d) x}{(c-i d) d^2}+\frac{a^2 (c+2 i d) x}{d^2}+\frac{a^2 \log (\cos (e+f x))}{d f}+\frac{(-i a c+a d)^2 \int \frac{d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{d \left (c^2+d^2\right )}\\ &=-\frac{a^2 c (c+i d) x}{(c-i d) d^2}+\frac{a^2 (c+2 i d) x}{d^2}+\frac{a^2 \log (\cos (e+f x))}{d f}-\frac{a^2 (i c-d) \log (c \cos (e+f x)+d \sin (e+f x))}{d (i c+d) f}\\ \end{align*}

Mathematica [A]  time = 2.39091, size = 176, normalized size = 1.66 \[ \frac{a^2 \left ((-2 d-2 i c) \tan ^{-1}(\tan (3 e+f x))-c \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )-i d \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )+2 (d-i c) \tan ^{-1}\left (\frac{d \cos (3 e+f x)-c \sin (3 e+f x)}{c \cos (3 e+f x)+d \sin (3 e+f x)}\right )+c \log \left (\cos ^2(e+f x)\right )-i d \log \left (\cos ^2(e+f x)\right )+8 d f x\right )}{2 d f (c-i d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*d*f*x + 2*((-I)*c + d)*ArcTan[(d*Cos[3*e + f*x] - c*Sin[3*e + f*x])/(c*Cos[3*e + f*x] + d*Sin[3*e + f*
x])] + ((-2*I)*c - 2*d)*ArcTan[Tan[3*e + f*x]] + c*Log[Cos[e + f*x]^2] - I*d*Log[Cos[e + f*x]^2] - c*Log[(c*Co
s[e + f*x] + d*Sin[e + f*x])^2] - I*d*Log[(c*Cos[e + f*x] + d*Sin[e + f*x])^2]))/(2*(c - I*d)*d*f)

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Maple [B]  time = 0.023, size = 204, normalized size = 1.9 \begin{align*}{\frac{i{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{2\,i{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ) d}{f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{f \left ({c}^{2}+{d}^{2} \right ) }}+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{2\,i{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{{a}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) d}}+{\frac{{a}^{2}d\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{f \left ({c}^{2}+{d}^{2} \right ) }} \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)),x)

[Out]

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

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Maxima [A]  time = 1.7395, size = 155, normalized size = 1.46 \begin{align*} \frac{\frac{4 \,{\left (a^{2} c + i \, a^{2} d\right )}{\left (f x + e\right )}}{c^{2} + d^{2}} - \frac{2 \,{\left (a^{2} c^{2} + 2 i \, a^{2} c d - a^{2} d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{2} d + d^{3}} + \frac{{\left (2 i \, a^{2} c - 2 \, a^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75817, size = 197, normalized size = 1.86 \begin{align*} \frac{{\left (-i \, a^{2} c + a^{2} d\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 \, a^{2} c + a^{2} d\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (i \, c d + d^{2}\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)),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 4.32593, size = 181, normalized size = 1.71 \begin{align*} -\frac{-\frac{4 i \, a^{2} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c - i \, d} - \frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d} - \frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d} + \frac{2 \,{\left (a^{2} c + i \, a^{2} d\right )} \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 )}{2 \, c d - 2 i \, d^{2}}}{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)),x, algorithm="giac")

[Out]

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