3.158 \(\int \frac{\sec ^9(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=133 \[ \frac{35 \tanh ^{-1}(\sin (c+d x))}{8 a^4 d}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{35 \tan (c+d x) \sec ^3(c+d x)}{12 a^4 d}+\frac{35 \tan (c+d x) \sec (c+d x)}{8 a^4 d}-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3} \]

[Out]

(35*ArcTanh[Sin[c + d*x]])/(8*a^4*d) + (35*Sec[c + d*x]*Tan[c + d*x])/(8*a^4*d) + (35*Sec[c + d*x]^3*Tan[c + d
*x])/(12*a^4*d) - ((2*I)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^3) - (((14*I)/3)*Sec[c + d*x]^5)/(d*(a^4
+ I*a^4*Tan[c + d*x]))

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Rubi [A]  time = 0.114347, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3500, 3768, 3770} \[ \frac{35 \tanh ^{-1}(\sin (c+d x))}{8 a^4 d}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{35 \tan (c+d x) \sec ^3(c+d x)}{12 a^4 d}+\frac{35 \tan (c+d x) \sec (c+d x)}{8 a^4 d}-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(35*ArcTanh[Sin[c + d*x]])/(8*a^4*d) + (35*Sec[c + d*x]*Tan[c + d*x])/(8*a^4*d) + (35*Sec[c + d*x]^3*Tan[c + d
*x])/(12*a^4*d) - ((2*I)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^3) - (((14*I)/3)*Sec[c + d*x]^5)/(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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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))^4} \, dx &=-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac{7 \int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{35 \int \sec ^5(c+d x) \, dx}{3 a^4}\\ &=\frac{35 \sec ^3(c+d x) \tan (c+d x)}{12 a^4 d}-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{35 \int \sec ^3(c+d x) \, dx}{4 a^4}\\ &=\frac{35 \sec (c+d x) \tan (c+d x)}{8 a^4 d}+\frac{35 \sec ^3(c+d x) \tan (c+d x)}{12 a^4 d}-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{35 \int \sec (c+d x) \, dx}{8 a^4}\\ &=\frac{35 \tanh ^{-1}(\sin (c+d x))}{8 a^4 d}+\frac{35 \sec (c+d x) \tan (c+d x)}{8 a^4 d}+\frac{35 \sec ^3(c+d x) \tan (c+d x)}{12 a^4 d}-\frac{2 i \sec ^7(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac{14 i \sec ^5(c+d x)}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.961373, size = 237, normalized size = 1.78 \[ -\frac{\sec ^4(c+d x) \left (896 i \cos (c+d x)+3 \left (42 \sin (c+d x)+58 \sin (3 (c+d x))+128 i \cos (3 (c+d x))+35 \cos (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+105 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+140 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-35 \cos (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-105 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{192 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-(Sec[c + d*x]^4*((896*I)*Cos[c + d*x] + 3*((128*I)*Cos[3*(c + d*x)] + 105*Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]] + 35*Cos[4*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 140*Cos[2*(c + d*x)]*(Log[Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 105*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]] - 35*Cos[4*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 42*Sin[c + d*x] + 58*Sin[3*(c + d*x)])))
/(192*a^4*d)

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Maple [B]  time = 0.099, size = 342, normalized size = 2.6 \begin{align*}{\frac{25}{8\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{2\,i}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{{\frac{4\,i}{3}}}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{27}{8\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{6\,i}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{4\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{35}{8\,{a}^{4}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{{\frac{4\,i}{3}}}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{25}{8\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{2\,i}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{27}{8\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{6\,i}{{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{4\,{a}^{4}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}-{\frac{35}{8\,{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)^9/(a+I*a*tan(d*x+c))^4,x)

[Out]

25/8/d/a^4/(tan(1/2*d*x+1/2*c)+1)^2-2*I/d/a^4/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/a^4/(tan(1/2*d*x+1/2*c)+1)^3+4/3*
I/d/a^4/(tan(1/2*d*x+1/2*c)+1)^3-27/8/d/a^4/(tan(1/2*d*x+1/2*c)+1)-6*I/d/a^4/(tan(1/2*d*x+1/2*c)+1)-1/4/d/a^4/
(tan(1/2*d*x+1/2*c)+1)^4+35/8/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)+1/2/d/a^4/(tan(1/2*d*x+1/2*c)-1)^3-4/3*I/d/a^4/(t
an(1/2*d*x+1/2*c)-1)^3-25/8/d/a^4/(tan(1/2*d*x+1/2*c)-1)^2-2*I/d/a^4/(tan(1/2*d*x+1/2*c)-1)^2-27/8/d/a^4/(tan(
1/2*d*x+1/2*c)-1)+6*I/d/a^4/(tan(1/2*d*x+1/2*c)-1)+1/4/d/a^4/(tan(1/2*d*x+1/2*c)-1)^4-35/8/d/a^4*ln(tan(1/2*d*
x+1/2*c)-1)

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Maxima [B]  time = 1.0538, size = 398, normalized size = 2.99 \begin{align*} -\frac{\frac{2 \,{\left (\frac{81 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{544 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{105 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{480 i \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{105 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{96 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{81 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 160 i\right )}}{a^{4} - \frac{4 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{4} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{105 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{105 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{24 \, 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))^4,x, algorithm="maxima")

[Out]

-1/24*(2*(81*sin(d*x + c)/(cos(d*x + c) + 1) - 544*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 105*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 + 480*I*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 105*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9
6*I*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 81*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 160*I)/(a^4 - 4*a^4*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + 6*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^4*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 + a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 105*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 105*log
(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d

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Fricas [A]  time = 2.51169, size = 674, normalized size = 5.07 \begin{align*} \frac{105 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 770 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 1022 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 558 i \, e^{\left (i \, d x + i \, c\right )}}{24 \,{\left (a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \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))^4,x, algorithm="fricas")

[Out]

1/24*(105*(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x + 2*I*c) + 1)*lo
g(e^(I*d*x + I*c) + I) - 105*(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d
*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 210*I*e^(7*I*d*x + 7*I*c) - 770*I*e^(5*I*d*x + 5*I*c) - 1022*I*e^(
3*I*d*x + 3*I*c) - 558*I*e^(I*d*x + I*c))/(a^4*d*e^(8*I*d*x + 8*I*c) + 4*a^4*d*e^(6*I*d*x + 6*I*c) + 6*a^4*d*e
^(4*I*d*x + 4*I*c) + 4*a^4*d*e^(2*I*d*x + 2*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)**9/(a+I*a*tan(d*x+c))**4,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [A]  time = 1.23231, size = 207, normalized size = 1.56 \begin{align*} \frac{\frac{105 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{105 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{2 \,{\left (81 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 96 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 480 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 544 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 81 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 160 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4} a^{4}}}{24 \, 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))^4,x, algorithm="giac")

[Out]

1/24*(105*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 105*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 2*(81*tan(1/2*
d*x + 1/2*c)^7 - 96*I*tan(1/2*d*x + 1/2*c)^6 - 105*tan(1/2*d*x + 1/2*c)^5 + 480*I*tan(1/2*d*x + 1/2*c)^4 - 105
*tan(1/2*d*x + 1/2*c)^3 - 544*I*tan(1/2*d*x + 1/2*c)^2 + 81*tan(1/2*d*x + 1/2*c) + 160*I)/((tan(1/2*d*x + 1/2*
c)^2 - 1)^4*a^4))/d