3.178 \(\int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=156 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^8 d}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^3(c+d x)}{3 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}-\frac{2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7} \]

[Out]

ArcTanh[Sin[c + d*x]]/(a^8*d) + (((2*I)/7)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^7) - (((2*I)/5)*Sec[c +
d*x]^5)/(a^3*d*(a + I*a*Tan[c + d*x])^5) + (((2*I)/3)*Sec[c + d*x]^3)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3) -
((2*I)*Sec[c + d*x])/(d*(a^8 + I*a^8*Tan[c + d*x]))

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Rubi [A]  time = 0.216498, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3500, 3770} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^8 d}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^3(c+d x)}{3 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}-\frac{2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^8,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^8*d) + (((2*I)/7)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^7) - (((2*I)/5)*Sec[c +
d*x]^5)/(a^3*d*(a + I*a*Tan[c + d*x])^5) + (((2*I)/3)*Sec[c + d*x]^3)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3) -
((2*I)*Sec[c + d*x])/(d*(a^8 + I*a^8*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 ^9(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac{\int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{a^2}\\ &=\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{\int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx}{a^4}\\ &=\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^3(c+d x)}{3 a^5 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^6}\\ &=\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^3(c+d x)}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac{\int \sec (c+d x) \, dx}{a^8}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{a^8 d}+\frac{2 i \sec ^7(c+d x)}{7 a d (a+i a \tan (c+d x))^7}-\frac{2 i \sec ^5(c+d x)}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac{2 i \sec ^3(c+d x)}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac{2 i \sec (c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.03443, size = 304, normalized size = 1.95 \[ \frac{\sec ^8(c+d x) \left (\cos \left (\frac{9}{2} (c+d x)\right )+i \sin \left (\frac{9}{2} (c+d x)\right )\right ) \left (-70 \sin \left (\frac{1}{2} (c+d x)\right )-42 \sin \left (\frac{3}{2} (c+d x)\right )+210 \sin \left (\frac{5}{2} (c+d x)\right )+30 \sin \left (\frac{7}{2} (c+d x)\right )+70 i \cos \left (\frac{1}{2} (c+d x)\right )-42 i \cos \left (\frac{3}{2} (c+d x)\right )-210 i \cos \left (\frac{5}{2} (c+d x)\right )+30 i \cos \left (\frac{7}{2} (c+d x)\right )-105 \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+105 \cos \left (\frac{7}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-105 i \sin \left (\frac{7}{2} (c+d x)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+105 i \sin \left (\frac{7}{2} (c+d x)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{105 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c + d*x]^8*((70*I)*Cos[(c + d*x)/2] - (42*I)*Cos[(3*(c + d*x))/2] - (210*I)*Cos[(5*(c + d*x))/2] + (30*I)
*Cos[(7*(c + d*x))/2] - 105*Cos[(7*(c + d*x))/2]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 105*Cos[(7*(c + d*
x))/2]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 70*Sin[(c + d*x)/2] - 42*Sin[(3*(c + d*x))/2] + 210*Sin[(5*(
c + d*x))/2] + 30*Sin[(7*(c + d*x))/2] - (105*I)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[(7*(c + d*x))/2]
+ (105*I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[(7*(c + d*x))/2])*(Cos[(9*(c + d*x))/2] + I*Sin[(9*(c
+ d*x))/2]))/(105*a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]  time = 0.11, size = 176, normalized size = 1.1 \begin{align*}{\frac{1}{d{a}^{8}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{128\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-6}}+{\frac{16\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-2}}-{\frac{128\,i}{d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-4}}-{\frac{256}{7\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-7}}+{\frac{896}{5\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-5}}-{\frac{160}{3\,d{a}^{8}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -i \right ) ^{-3}}-{\frac{1}{d{a}^{8}}\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)^9/(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d/a^8*ln(tan(1/2*d*x+1/2*c)+1)+128*I/d/a^8/(tan(1/2*d*x+1/2*c)-I)^6+16*I/d/a^8/(tan(1/2*d*x+1/2*c)-I)^2-128*
I/d/a^8/(tan(1/2*d*x+1/2*c)-I)^4-256/7/d/a^8/(tan(1/2*d*x+1/2*c)-I)^7+896/5/d/a^8/(tan(1/2*d*x+1/2*c)-I)^5-160
/3/d/a^8/(tan(1/2*d*x+1/2*c)-I)^3-1/d/a^8*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.85278, size = 250, normalized size = 1.6 \begin{align*} \frac{-210 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 210 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 60 i \, \cos \left (7 \, d x + 7 \, c\right ) - 84 i \, \cos \left (5 \, d x + 5 \, c\right ) + 140 i \, \cos \left (3 \, d x + 3 \, c\right ) - 420 i \, \cos \left (d x + c\right ) + 105 \, \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 ) - 105 \, \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 ) + 60 \, \sin \left (7 \, d x + 7 \, c\right ) - 84 \, \sin \left (5 \, d x + 5 \, c\right ) + 140 \, \sin \left (3 \, d x + 3 \, c\right ) - 420 \, \sin \left (d x + c\right )}{210 \, a^{8} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/210*(-210*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 210*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 60*I*
cos(7*d*x + 7*c) - 84*I*cos(5*d*x + 5*c) + 140*I*cos(3*d*x + 3*c) - 420*I*cos(d*x + c) + 105*log(cos(d*x + c)^
2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - 105*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 60*
sin(7*d*x + 7*c) - 84*sin(5*d*x + 5*c) + 140*sin(3*d*x + 3*c) - 420*sin(d*x + c))/(a^8*d)

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Fricas [A]  time = 2.65172, size = 306, normalized size = 1.96 \begin{align*} \frac{{\left (105 \, e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 42 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 30 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{105 \, a^{8} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/105*(105*e^(7*I*d*x + 7*I*c)*log(e^(I*d*x + I*c) + I) - 105*e^(7*I*d*x + 7*I*c)*log(e^(I*d*x + I*c) - I) - 2
10*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) - 42*I*e^(2*I*d*x + 2*I*c) + 30*I)*e^(-7*I*d*x - 7*I*c)/(a
^8*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)**9/(a+I*a*tan(d*x+c))**8,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.23051, size = 169, normalized size = 1.08 \begin{align*} \frac{\frac{105 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{8}} - \frac{105 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{8}} - \frac{2 \,{\left (-840 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1400 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 3920 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2352 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1064 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 152\right )}}{a^{8}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{7}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/105*(105*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^8 - 105*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^8 - 2*(-840*I*tan
(1/2*d*x + 1/2*c)^5 - 1400*tan(1/2*d*x + 1/2*c)^4 + 3920*I*tan(1/2*d*x + 1/2*c)^3 + 2352*tan(1/2*d*x + 1/2*c)^
2 - 1064*I*tan(1/2*d*x + 1/2*c) - 152)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^7))/d