3.1083 \(\int \frac{(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=115 \[ -\frac{a^3 (-3 d+i c) \log (\cos (e+f x))}{d^2 f}-\frac{a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 f (d+i c)}+\frac{4 a^3 x}{c-i d}-\frac{a^3+i a^3 \tan (e+f x)}{d f} \]

[Out]

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

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Rubi [A]  time = 0.360146, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3556, 3589, 3475, 3531, 3530} \[ -\frac{a^3 (-3 d+i c) \log (\cos (e+f x))}{d^2 f}-\frac{a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 f (d+i c)}+\frac{4 a^3 x}{c-i d}-\frac{a^3+i a^3 \tan (e+f x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 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 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))^3}{c+d \tan (e+f x)} \, dx &=-\frac{a^3+i a^3 \tan (e+f x)}{d f}+\frac{a \int \frac{(a+i a \tan (e+f x)) (a (i c+d)+a (c+3 i d) \tan (e+f x))}{c+d \tan (e+f x)} \, dx}{d}\\ &=-\frac{a^3+i a^3 \tan (e+f x)}{d f}+\frac{a \int \frac{a^2 d (i c+d)-a^2 \left (i c^2-3 c d-4 i d^2\right ) \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{d^2}+\frac{\left (a^3 (i c-3 d)\right ) \int \tan (e+f x) \, dx}{d^2}\\ &=\frac{4 a^3 x}{c-i d}-\frac{a^3 (i c-3 d) \log (\cos (e+f x))}{d^2 f}-\frac{a^3+i a^3 \tan (e+f x)}{d f}-\frac{\left (a^3 (c+i d)^2\right ) \int \frac{d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{d^2 (i c+d)}\\ &=\frac{4 a^3 x}{c-i d}-\frac{a^3 (i c-3 d) \log (\cos (e+f x))}{d^2 f}-\frac{a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 (i c+d) f}-\frac{a^3+i a^3 \tan (e+f x)}{d f}\\ \end{align*}

Mathematica [A]  time = 5.41203, size = 229, normalized size = 1.99 \[ \frac{a^3 \sec (e+f x) \left (\cos (f x) \left (-i \left (c^2+2 i c d+3 d^2\right ) \log \left (\cos ^2(e+f x)\right )+i (c+i d)^2 \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )+8 d^2 f x\right )+\cos (2 e+f x) \left (-i \left (c^2+2 i c d+3 d^2\right ) \log \left (\cos ^2(e+f x)\right )+i (c+i d)^2 \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )+8 d^2 f x\right )-4 i d (c-i d) \sin (f x)\right )}{4 d^2 f (c-i d) \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e + f*x]*(Cos[f*x]*(8*d^2*f*x - I*(c^2 + (2*I)*c*d + 3*d^2)*Log[Cos[e + f*x]^2] + I*(c + I*d)^2*Log[(
c*Cos[e + f*x] + d*Sin[e + f*x])^2]) + Cos[2*e + f*x]*(8*d^2*f*x - I*(c^2 + (2*I)*c*d + 3*d^2)*Log[Cos[e + f*x
]^2] + I*(c + I*d)^2*Log[(c*Cos[e + f*x] + d*Sin[e + f*x])^2]) - (4*I)*(c - I*d)*d*Sin[f*x]))/(4*(c - I*d)*d^2
*f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2]))

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Maple [B]  time = 0.027, size = 257, normalized size = 2.2 \begin{align*}{\frac{-i{a}^{3}\tan \left ( fx+e \right ) }{fd}}+{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{4\,i{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) d}{f \left ({c}^{2}+{d}^{2} \right ) }}-2\,{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{f \left ({c}^{2}+{d}^{2} \right ) }}+4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{i{a}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{3}}{f{d}^{2} \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{3\,i{a}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) }}-3\,{\frac{{a}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}}{fd \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{a}^{3}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))^3/(c+d*tan(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.81895, size = 193, normalized size = 1.68 \begin{align*} -\frac{\frac{2 i \, a^{3} \tan \left (f x + e\right )}{d} - \frac{8 \,{\left (a^{3} c + i \, a^{3} d\right )}{\left (f x + e\right )}}{c^{2} + d^{2}} - \frac{2 \,{\left (i \, a^{3} c^{3} - 3 \, a^{3} c^{2} d - 3 i \, a^{3} c d^{2} + a^{3} d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{2} d^{2} + d^{4}} - \frac{{\left (4 i \, a^{3} c - 4 \, a^{3} 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))^3/(c+d*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.93819, size = 487, normalized size = 4.23 \begin{align*} \frac{2 i \, a^{3} c d + 2 \, a^{3} d^{2} -{\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2} +{\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} 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 (a^{3} c^{2} + 2 i \, a^{3} c d + 3 \, a^{3} d^{2} +{\left (a^{3} c^{2} + 2 i \, a^{3} c d + 3 \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (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))^3/(c+d*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.80749, size = 340, normalized size = 2.96 \begin{align*} -\frac{-\frac{8 i \, a^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c - i \, d} + \frac{2 \,{\left (-i \, a^{3} c^{2} + 2 \, a^{3} c d + i \, a^{3} d^{2}\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} - 2 i \, d^{3}} - \frac{{\left (-i \, a^{3} c + 3 \, a^{3} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d^{2}} + \frac{{\left (i \, a^{3} c - 3 \, a^{3} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d^{2}} + \frac{-i \, a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 i \, a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i \, a^{3} c - 3 \, a^{3} d}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} d^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e)),x, algorithm="giac")

[Out]

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