3.125 \(\int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{2 i \sec (c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

-(ArcTanh[Sin[c + d*x]]/(a^2*d)) + ((2*I)*Sec[c + d*x])/(d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.0471635, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3500, 3770} \[ -\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{2 i \sec (c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sin[c + d*x]]/(a^2*d)) + ((2*I)*Sec[c + d*x])/(d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{2 i \sec (c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\int \sec (c+d x) \, dx}{a^2}\\ &=-\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{2 i \sec (c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.177561, size = 184, normalized size = 3.83 \[ -\frac{\sec ^2(c+d x) \left (\cos \left (\frac{3}{2} (c+d x)\right )+i \sin \left (\frac{3}{2} (c+d x)\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 i\right )+\sin \left (\frac{1}{2} (c+d x)\right ) \left (i \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-i \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2\right )\right )}{a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.073, size = 63, normalized size = 1.3 \begin{align*} -{\frac{1}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+4\,{\frac{1}{{a}^{2}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.49095, size = 158, normalized size = 3.29 \begin{align*} -\frac{-2 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 2 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - 4 i \, \cos \left (d x + c\right ) + \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right )}{2 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(-2*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 2*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - 4*I*cos(d*
x + c) + log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*s
in(d*x + c) + 1) - 4*sin(d*x + c))/(a^2*d)

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Fricas [A]  time = 2.30481, size = 161, normalized size = 3.35 \begin{align*} -\frac{{\left (e^{\left (i \, d x + i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - e^{\left (i \, d x + i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 2 i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.26314, size = 80, normalized size = 1.67 \begin{align*} -\frac{\frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{4}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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