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]

35/8*arctanh(sin(d*x+c))/a^4/d+35/8*sec(d*x+c)*tan(d*x+c)/a^4/d+35/12*sec(d*x+c)^3*tan(d*x+c)/a^4/d-2*I*sec(d*
x+c)^7/a/d/(a+I*a*tan(d*x+c))^3-14/3*I*sec(d*x+c)^5/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 133, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 1.20, 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]

-1/192*(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)])))/(a^4*d)

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fricas [A]  time = 0.51, size = 230, normalized size = 1.73 \[ \frac {105 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 770 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 1022 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 558 i \, e^{\left (i \, d x + i \, c\right )}}{24 \, {\left (a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

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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giac [A]  time = 2.11, size = 151, normalized size = 1.14 \[ \frac {\frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} - \frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} - \frac {2 \, {\left (81 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 544 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 81 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 160 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a^{4}}}{24 \, d} \]

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(tan(1/2*d*x + 1/2*c) + 1)/a^4 - 105*log(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

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maple [B]  time = 0.42, size = 342, normalized size = 2.57 \[ \frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {4 i}{3 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {25}{8 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {27}{8 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {6 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{4 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}+\frac {25}{8 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {4 i}{3 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {6 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {27}{8 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}} \]

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]

1/2/a^4/d/(tan(1/2*d*x+1/2*c)-1)^3+4/3*I/a^4/d/(tan(1/2*d*x+1/2*c)+1)^3-25/8/a^4/d/(tan(1/2*d*x+1/2*c)-1)^2-2*
I/a^4/d/(tan(1/2*d*x+1/2*c)+1)^2-27/8/a^4/d/(tan(1/2*d*x+1/2*c)-1)-6*I/a^4/d/(tan(1/2*d*x+1/2*c)+1)+1/4/a^4/d/
(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)+25/8/a^4/d/(tan(1/2*d*x+1/2*c)+1)^2-4/3*I/a^4/d/(
tan(1/2*d*x+1/2*c)-1)^3+1/2/a^4/d/(tan(1/2*d*x+1/2*c)+1)^3+6*I/a^4/d/(tan(1/2*d*x+1/2*c)-1)-27/8/a^4/d/(tan(1/
2*d*x+1/2*c)+1)-2*I/a^4/d/(tan(1/2*d*x+1/2*c)-1)^2-1/4/a^4/d/(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 = 0.55, size = 295, normalized size = 2.22 \[ -\frac {\frac {2 \, {\left (\frac {81 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {544 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {105 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {480 i \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {96 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {81 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 160 i\right )}}{a^{4} - \frac {4 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{4} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{24 \, d} \]

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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mupad [B]  time = 6.87, size = 197, normalized size = 1.48 \[ \frac {35\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^4\,d}+\frac {\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4\,a^4}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4\,a^4}-\frac {27\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4\,a^4}-\frac {27\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,136{}\mathrm {i}}{3\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,40{}\mathrm {i}}{a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,8{}\mathrm {i}}{a^4}-\frac {40{}\mathrm {i}}{3\,a^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(35*atanh(tan(c/2 + (d*x)/2)))/(4*a^4*d) + ((tan(c/2 + (d*x)/2)^2*136i)/(3*a^4) + (35*tan(c/2 + (d*x)/2)^3)/(4
*a^4) - (tan(c/2 + (d*x)/2)^4*40i)/a^4 + (35*tan(c/2 + (d*x)/2)^5)/(4*a^4) + (tan(c/2 + (d*x)/2)^6*8i)/a^4 - (
27*tan(c/2 + (d*x)/2)^7)/(4*a^4) - 40i/(3*a^4) - (27*tan(c/2 + (d*x)/2))/(4*a^4))/(d*(6*tan(c/2 + (d*x)/2)^4 -
4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{9}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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]

Integral(sec(c + d*x)**9/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x
)/a**4

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