3.1096 \(\int \frac{(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 i a^2}{f (c-i d)^2 (c+d \tan (e+f x))}+\frac{a^2 (-d+i c)}{2 d f (d+i c) (c+d \tan (e+f x))^2}-\frac{2 a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f (d+i c)^3}+\frac{2 a^2 x}{(c-i d)^3} \]

[Out]

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

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Rubi [A]  time = 0.273193, antiderivative size = 125, 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 = {3542, 3529, 3531, 3530} \[ \frac{2 i a^2}{f (c-i d)^2 (c+d \tan (e+f x))}+\frac{a^2 (-d+i c)}{2 d f (d+i c) (c+d \tan (e+f x))^2}-\frac{2 a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{f (d+i c)^3}+\frac{2 a^2 x}{(c-i d)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2/(c + d*Tan[e + f*x])^3,x]

[Out]

(2*a^2*x)/(c - I*d)^3 - (2*a^2*Log[c*Cos[e + f*x] + d*Sin[e + f*x]])/((I*c + d)^3*f) + (a^2*(I*c - d))/(2*d*(I
*c + d)*f*(c + d*Tan[e + f*x])^2) + ((2*I)*a^2)/((c - I*d)^2*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 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{(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^3} \, dx &=\frac{a^2 (i c-d)}{2 d (i c+d) f (c+d \tan (e+f x))^2}+\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))^2} \, dx}{c^2+d^2}\\ &=\frac{a^2 (i c-d)}{2 d (i c+d) f (c+d \tan (e+f x))^2}+\frac{2 i a^2}{(c-i d)^2 f (c+d \tan (e+f x))}+\frac{\int \frac{2 a^2 (c+i d)^2+2 i a^2 (c+i d)^2 \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{\left (c^2+d^2\right )^2}\\ &=\frac{2 a^2 x}{(c-i d)^3}+\frac{a^2 (i c-d)}{2 d (i c+d) f (c+d \tan (e+f x))^2}+\frac{2 i a^2}{(c-i d)^2 f (c+d \tan (e+f x))}-\frac{\left (2 a^2\right ) \int \frac{d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{(i c+d)^3}\\ &=\frac{2 a^2 x}{(c-i d)^3}-\frac{2 a^2 \log (c \cos (e+f x)+d \sin (e+f x))}{(i c+d)^3 f}+\frac{a^2 (i c-d)}{2 d (i c+d) f (c+d \tan (e+f x))^2}+\frac{2 i a^2}{(c-i d)^2 f (c+d \tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 5.56139, size = 317, normalized size = 2.54 \[ \frac{a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\frac{2 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{c \left (c^2-3 d^2\right ) \sin (3 e+f x)+\left (d^3-3 c^2 d\right ) \cos (3 e+f x)}{c \left (c^2-3 d^2\right ) \cos (3 e+f x)-d \left (d^2-3 c^2\right ) \sin (3 e+f x)}\right )}{f}-\frac{(c-i d) (c+2 i d) (\cos (2 e)-i \sin (2 e)) \sin (f x)}{f (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{d (c-i d) (\cos (2 e)-i \sin (2 e))}{2 f (c \cos (e+f x)+d \sin (e+f x))^2}+\frac{(-\sin (2 e)-i \cos (2 e)) \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{f}+4 x (\cos (2 e)-i \sin (2 e))\right )}{(c-i d)^3 (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2/(c + d*Tan[e + f*x])^3,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[((-3*c^2*d + d^3)*Cos[3*e + f*x] + c*(c^2 - 3*d^2)*Sin[3*e + f*x])
/(c*(c^2 - 3*d^2)*Cos[3*e + f*x] - d*(-3*c^2 + d^2)*Sin[3*e + f*x])]*(Cos[2*e] - I*Sin[2*e]))/f + ((c - I*d)*d
*(Cos[2*e] - I*Sin[2*e]))/(2*f*(c*Cos[e + f*x] + d*Sin[e + f*x])^2) - ((c - I*d)*(c + (2*I)*d)*(Cos[2*e] - I*S
in[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)^3*(Cos[f*x] + I*Si
n[f*x])^2)

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Maple [B]  time = 0.038, size = 562, normalized size = 4.5 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

6*I/f*a^2/(c^2+d^2)^3*ln(c+d*tan(f*x+e))*c*d^2+I/f*a^2/(c^2+d^2)/(c+d*tan(f*x+e))^2*c-2*I/f*a^2/(c^2+d^2)^3*ln
(c+d*tan(f*x+e))*c^3-3*I/f*a^2/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*c*d^2-3/f*a^2/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*c^2
*d+1/f*a^2/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*d^3+2/f*a^2/(c^2+d^2)^3*arctan(tan(f*x+e))*c^3-6/f*a^2/(c^2+d^2)^3*a
rctan(tan(f*x+e))*c*d^2-2*I/f*a^2/(c^2+d^2)^3*arctan(tan(f*x+e))*d^3+1/2/f*a^2/(c^2+d^2)/d/(c+d*tan(f*x+e))^2*
c^2-1/2/f*a^2/(c^2+d^2)*d/(c+d*tan(f*x+e))^2-2*I/f*a^2/(c^2+d^2)^2/(c+d*tan(f*x+e))*d^2+2*I/f*a^2/(c^2+d^2)^2/
(c+d*tan(f*x+e))*c^2-4/f*a^2/(c^2+d^2)^2/(c+d*tan(f*x+e))*c*d+I/f*a^2/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*c^3+6*I/f
*a^2/(c^2+d^2)^3*arctan(tan(f*x+e))*c^2*d+6/f*a^2/(c^2+d^2)^3*ln(c+d*tan(f*x+e))*c^2*d-2/f*a^2/(c^2+d^2)^3*ln(
c+d*tan(f*x+e))*d^3

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Maxima [B]  time = 1.77287, size = 514, normalized size = 4.11 \begin{align*} \frac{\frac{2 \,{\left (2 \, a^{2} c^{3} + 6 i \, a^{2} c^{2} d - 6 \, a^{2} c d^{2} - 2 i \, a^{2} d^{3}\right )}{\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{2 \,{\left (-2 i \, a^{2} c^{3} + 6 \, a^{2} c^{2} d + 6 i \, a^{2} c d^{2} - 2 \, a^{2} d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{{\left (2 i \, a^{2} c^{3} - 6 \, a^{2} c^{2} d - 6 i \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{a^{2} c^{4} + 6 i \, a^{2} c^{3} d - 8 \, a^{2} c^{2} d^{2} - 2 i \, a^{2} c d^{3} - a^{2} d^{4} -{\left (-4 i \, a^{2} c^{2} d^{2} + 8 \, a^{2} c d^{3} + 4 i \, a^{2} d^{4}\right )} \tan \left (f x + e\right )}{c^{6} d + 2 \, c^{4} d^{3} + c^{2} d^{5} +{\left (c^{4} d^{3} + 2 \, c^{2} d^{5} + d^{7}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (c^{5} d^{2} + 2 \, c^{3} d^{4} + c d^{6}\right )} \tan \left (f x + e\right )}}{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))^3,x, algorithm="maxima")

[Out]

1/2*(2*(2*a^2*c^3 + 6*I*a^2*c^2*d - 6*a^2*c*d^2 - 2*I*a^2*d^3)*(f*x + e)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6) +
2*(-2*I*a^2*c^3 + 6*a^2*c^2*d + 6*I*a^2*c*d^2 - 2*a^2*d^3)*log(d*tan(f*x + e) + c)/(c^6 + 3*c^4*d^2 + 3*c^2*d
^4 + d^6) + (2*I*a^2*c^3 - 6*a^2*c^2*d - 6*I*a^2*c*d^2 + 2*a^2*d^3)*log(tan(f*x + e)^2 + 1)/(c^6 + 3*c^4*d^2 +
3*c^2*d^4 + d^6) + (a^2*c^4 + 6*I*a^2*c^3*d - 8*a^2*c^2*d^2 - 2*I*a^2*c*d^3 - a^2*d^4 - (-4*I*a^2*c^2*d^2 + 8
*a^2*c*d^3 + 4*I*a^2*d^4)*tan(f*x + e))/(c^6*d + 2*c^4*d^3 + c^2*d^5 + (c^4*d^3 + 2*c^2*d^5 + d^7)*tan(f*x + e
)^2 + 2*(c^5*d^2 + 2*c^3*d^4 + c*d^6)*tan(f*x + e)))/f

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Fricas [B]  time = 2.10253, size = 729, normalized size = 5.83 \begin{align*} \frac{2 \, a^{2} c^{2} + 6 i \, a^{2} c d - 4 \, a^{2} d^{2} +{\left (2 \, a^{2} c^{2} + 4 i \, a^{2} c d + 6 \, a^{2} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 \, a^{2} c^{2} + 4 i \, a^{2} c d - 2 \, a^{2} d^{2} +{\left (2 \, a^{2} c^{2} - 4 i \, a^{2} c d - 2 \, a^{2} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \,{\left (a^{2} c^{2} + a^{2} d^{2}\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^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (2 i \, c^{5} + 6 \, c^{4} d - 4 i \, c^{3} d^{2} + 4 \, c^{2} d^{3} - 6 i \, c d^{4} - 2 \, d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, c^{5} + c^{4} d + 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} + i \, c d^{4} + d^{5}\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))^3,x, algorithm="fricas")

[Out]

(2*a^2*c^2 + 6*I*a^2*c*d - 4*a^2*d^2 + (2*a^2*c^2 + 4*I*a^2*c*d + 6*a^2*d^2)*e^(2*I*f*x + 2*I*e) + (2*a^2*c^2
+ 4*I*a^2*c*d - 2*a^2*d^2 + (2*a^2*c^2 - 4*I*a^2*c*d - 2*a^2*d^2)*e^(4*I*f*x + 4*I*e) + 4*(a^2*c^2 + a^2*d^2)*
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^5 + 5*c^4*d - 10*I*c^3*d^
2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f*e^(4*I*f*x + 4*I*e) + (2*I*c^5 + 6*c^4*d - 4*I*c^3*d^2 + 4*c^2*d^3 - 6*I*c
*d^4 - 2*d^5)*f*e^(2*I*f*x + 2*I*e) + (I*c^5 + c^4*d + 2*I*c^3*d^2 + 2*c^2*d^3 + I*c*d^4 + d^5)*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.52827, size = 641, normalized size = 5.13 \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^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}} - \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^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}} - \frac{3 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2 i \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 10 \, a^{2} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 i \, a^{2} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 i \, a^{2} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 6 \, a^{2} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 10 i \, a^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, a^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 i \, a^{2} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 10 \, a^{2} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 i \, a^{2} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, a^{2} c^{4}}{{\left (-2 i \, c^{5} - 6 \, c^{4} d + 6 i \, c^{3} d^{2} + 2 \, c^{2} d^{3}\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 )}^{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))^3,x, algorithm="giac")

[Out]

-2*(2*a^2*log(-I*tan(1/2*f*x + 1/2*e) + 1)/(I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3) - 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^3 + 3*c^2*d - 3*I*c*d^2 - d^3) - (3*a^2*c^4*tan(1/2*f*x + 1/2*e)
^4 - 2*I*a^2*c^4*tan(1/2*f*x + 1/2*e)^3 - 10*a^2*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 6*I*a^2*c^2*d^2*tan(1/2*f*x +
1/2*e)^3 - 2*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*a^2*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*I*a^2*c^3*d*tan(1/2*f*x +
1/2*e)^2 + 6*a^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 + 10*I*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*a^2*d^4*tan(1/2*f
*x + 1/2*e)^2 + 2*I*a^2*c^4*tan(1/2*f*x + 1/2*e) + 10*a^2*c^3*d*tan(1/2*f*x + 1/2*e) + 6*I*a^2*c^2*d^2*tan(1/2
*f*x + 1/2*e) + 2*a^2*c*d^3*tan(1/2*f*x + 1/2*e) + 3*a^2*c^4)/((-2*I*c^5 - 6*c^4*d + 6*I*c^3*d^2 + 2*c^2*d^3)*
(c*tan(1/2*f*x + 1/2*e)^2 - 2*d*tan(1/2*f*x + 1/2*e) - c)^2))/f