3.160 \(\int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=82 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3} \]

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) + (((2*I)/3)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3) - ((2*I)*Sec[c + d*x
])/(d*(a^4 + I*a^4*Tan[c + d*x]))

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Rubi [A]  time = 0.0866693, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3500, 3770} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) + (((2*I)/3)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3) - ((2*I)*Sec[c + d*x
])/(d*(a^4 + I*a^4*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 ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac{2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \sec (c+d x) \, dx}{a^4}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac{2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.241428, size = 247, normalized size = 3.01 \[ \frac{\sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4 \left (-6 i \sin (3 c) \sin (d x)+2 i \sin (c) \sin (3 d x)-2 \sin (c) \cos (3 d x)+6 \sin (3 c) \cos (d x)+\cos (3 c) (-6 \sin (d x)-6 i \cos (d x))+2 \cos (c) (\sin (3 d x)+i \cos (3 d x))-3 \cos (4 c) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 i \sin (4 c) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \cos (4 c) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 i \sin (4 c) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{3 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])^4*(-3*Cos[4*c]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[4*c]*L
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*Cos[3*d*x]*Sin[c] + 6*Cos[d*x]*Sin[3*c] - (3*I)*Log[Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]]*Sin[4*c] + (3*I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*c] + Cos[3*c]*((-6*I)*C
os[d*x] - 6*Sin[d*x]) - (6*I)*Sin[3*c]*Sin[d*x] + (2*I)*Sin[c]*Sin[3*d*x] + 2*Cos[c]*(I*Cos[3*d*x] + Sin[3*d*x
])))/(3*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.09, size = 86, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{4}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{8\,i}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-2}}-{\frac{16}{3\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-3}}-{\frac{1}{{a}^{4}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x)

[Out]

1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)+8*I/d/a^4/(tan(1/2*d*x+1/2*c)-I)^2-16/3/d/a^4/(tan(1/2*d*x+1/2*c)-I)^3-1/d/a^
4*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.51308, size = 190, normalized size = 2.32 \begin{align*} \frac{-6 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 6 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 4 i \, \cos \left (3 \, d x + 3 \, c\right ) - 12 i \, \cos \left (d x + c\right ) + 3 \, \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 ) - 3 \, \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 (3 \, d x + 3 \, c\right ) - 12 \, \sin \left (d x + c\right )}{6 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/6*(-6*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 6*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 4*I*cos(3*d
*x + 3*c) - 12*I*cos(d*x + c) + 3*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - 3*log(cos(d*x +
c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 4*sin(3*d*x + 3*c) - 12*sin(d*x + c))/(a^4*d)

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Fricas [A]  time = 2.51307, size = 221, normalized size = 2.7 \begin{align*} \frac{{\left (3 \, e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.20956, size = 99, normalized size = 1.21 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{8 \,{\left (3 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{a^{4}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

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