3.181 \(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=213 \[ \frac {20 i \sec ^3(c+d x)}{3003 a^3 d (a+i a \tan (c+d x))^5}+\frac {8 i \sec ^3(c+d x)}{9009 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac {8 i \sec ^3(c+d x)}{3003 d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8} \]

[Out]

1/13*I*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^8+5/143*I*sec(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^7+20/1287*I*sec(d*x+c)^
3/a^2/d/(a+I*a*tan(d*x+c))^6+20/3003*I*sec(d*x+c)^3/a^3/d/(a+I*a*tan(d*x+c))^5+8/3003*I*sec(d*x+c)^3/d/(a^2+I*
a^2*tan(d*x+c))^4+8/9009*I*sec(d*x+c)^3/a^2/d/(a^2+I*a^2*tan(d*x+c))^3

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Rubi [A]  time = 0.28, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3502, 3488} \[ \frac {8 i \sec ^3(c+d x)}{9009 a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac {8 i \sec ^3(c+d x)}{3003 d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac {20 i \sec ^3(c+d x)}{3003 a^3 d (a+i a \tan (c+d x))^5}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/13)*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^8) + (((5*I)/143)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x]
)^7) + (((20*I)/1287)*Sec[c + d*x]^3)/(a^2*d*(a + I*a*Tan[c + d*x])^6) + (((20*I)/3003)*Sec[c + d*x]^3)/(a^3*d
*(a + I*a*Tan[c + d*x])^5) + (((8*I)/3003)*Sec[c + d*x]^3)/(d*(a^2 + I*a^2*Tan[c + d*x])^4) + (((8*I)/9009)*Se
c[c + d*x]^3)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3)

Rule 3488

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3502

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^7} \, dx}{13 a}\\ &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {20 \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{143 a^2}\\ &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {20 \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^5} \, dx}{429 a^3}\\ &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {20 i \sec ^3(c+d x)}{3003 a^3 d (a+i a \tan (c+d x))^5}+\frac {40 \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx}{3003 a^4}\\ &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {20 i \sec ^3(c+d x)}{3003 a^3 d (a+i a \tan (c+d x))^5}+\frac {8 i \sec ^3(c+d x)}{3003 d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac {8 \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{3003 a^5}\\ &=\frac {i \sec ^3(c+d x)}{13 d (a+i a \tan (c+d x))^8}+\frac {5 i \sec ^3(c+d x)}{143 a d (a+i a \tan (c+d x))^7}+\frac {20 i \sec ^3(c+d x)}{1287 a^2 d (a+i a \tan (c+d x))^6}+\frac {20 i \sec ^3(c+d x)}{3003 a^3 d (a+i a \tan (c+d x))^5}+\frac {8 i \sec ^3(c+d x)}{9009 a^5 d (a+i a \tan (c+d x))^3}+\frac {8 i \sec ^3(c+d x)}{3003 d \left (a^2+i a^2 \tan (c+d x)\right )^4}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 95, normalized size = 0.45 \[ \frac {i \sec ^8(c+d x) (1430 i \sin (c+d x)+2457 i \sin (3 (c+d x))+1155 i \sin (5 (c+d x))+11440 \cos (c+d x)+6552 \cos (3 (c+d x))+1848 \cos (5 (c+d x)))}{144144 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/144144)*Sec[c + d*x]^8*(11440*Cos[c + d*x] + 6552*Cos[3*(c + d*x)] + 1848*Cos[5*(c + d*x)] + (1430*I)*Sin[
c + d*x] + (2457*I)*Sin[3*(c + d*x)] + (1155*I)*Sin[5*(c + d*x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

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fricas [A]  time = 0.58, size = 74, normalized size = 0.35 \[ \frac {{\left (3003 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 9009 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 12870 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10010 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4095 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 693 i\right )} e^{\left (-13 i \, d x - 13 i \, c\right )}}{288288 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/288288*(3003*I*e^(10*I*d*x + 10*I*c) + 9009*I*e^(8*I*d*x + 8*I*c) + 12870*I*e^(6*I*d*x + 6*I*c) + 10010*I*e^
(4*I*d*x + 4*I*c) + 4095*I*e^(2*I*d*x + 2*I*c) + 693*I)*e^(-13*I*d*x - 13*I*c)/(a^8*d)

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giac [A]  time = 3.95, size = 177, normalized size = 0.83 \[ \frac {2 \, {\left (9009 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 45045 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 183183 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 435435 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 810810 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 1051050 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1076790 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 785070 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 451165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 171457 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 51675 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7111 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1240\right )}}{9009 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{13}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

2/9009*(9009*tan(1/2*d*x + 1/2*c)^12 - 45045*I*tan(1/2*d*x + 1/2*c)^11 - 183183*tan(1/2*d*x + 1/2*c)^10 + 4354
35*I*tan(1/2*d*x + 1/2*c)^9 + 810810*tan(1/2*d*x + 1/2*c)^8 - 1051050*I*tan(1/2*d*x + 1/2*c)^7 - 1076790*tan(1
/2*d*x + 1/2*c)^6 + 785070*I*tan(1/2*d*x + 1/2*c)^5 + 451165*tan(1/2*d*x + 1/2*c)^4 - 171457*I*tan(1/2*d*x + 1
/2*c)^3 - 51675*tan(1/2*d*x + 1/2*c)^2 + 7111*I*tan(1/2*d*x + 1/2*c) + 1240)/(a^8*d*(tan(1/2*d*x + 1/2*c) - I)
^13)

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maple [A]  time = 0.54, size = 222, normalized size = 1.04 \[ \frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}+\frac {864 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{10}}+\frac {14 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {480}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {4544}{11 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{11}}-\frac {200 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {2672 i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}+\frac {256}{13 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{13}}-\frac {128 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{12}}-\frac {188}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}-\frac {1472 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{8}}-\frac {9056}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {11680}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{9}}}{a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^8,x)

[Out]

2/d/a^8*(1/(tan(1/2*d*x+1/2*c)-I)+432*I/(tan(1/2*d*x+1/2*c)-I)^10+7*I/(tan(1/2*d*x+1/2*c)-I)^2+240/(tan(1/2*d*
x+1/2*c)-I)^5-2272/11/(tan(1/2*d*x+1/2*c)-I)^11-100*I/(tan(1/2*d*x+1/2*c)-I)^4+1336/3*I/(tan(1/2*d*x+1/2*c)-I)
^6+128/13/(tan(1/2*d*x+1/2*c)-I)^13-64*I/(tan(1/2*d*x+1/2*c)-I)^12-94/3/(tan(1/2*d*x+1/2*c)-I)^3-736*I/(tan(1/
2*d*x+1/2*c)-I)^8-4528/7/(tan(1/2*d*x+1/2*c)-I)^7+5840/9/(tan(1/2*d*x+1/2*c)-I)^9)

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maxima [A]  time = 0.42, size = 141, normalized size = 0.66 \[ \frac {693 i \, \cos \left (13 \, d x + 13 \, c\right ) + 4095 i \, \cos \left (11 \, d x + 11 \, c\right ) + 10010 i \, \cos \left (9 \, d x + 9 \, c\right ) + 12870 i \, \cos \left (7 \, d x + 7 \, c\right ) + 9009 i \, \cos \left (5 \, d x + 5 \, c\right ) + 3003 i \, \cos \left (3 \, d x + 3 \, c\right ) + 693 \, \sin \left (13 \, d x + 13 \, c\right ) + 4095 \, \sin \left (11 \, d x + 11 \, c\right ) + 10010 \, \sin \left (9 \, d x + 9 \, c\right ) + 12870 \, \sin \left (7 \, d x + 7 \, c\right ) + 9009 \, \sin \left (5 \, d x + 5 \, c\right ) + 3003 \, \sin \left (3 \, d x + 3 \, c\right )}{288288 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/288288*(693*I*cos(13*d*x + 13*c) + 4095*I*cos(11*d*x + 11*c) + 10010*I*cos(9*d*x + 9*c) + 12870*I*cos(7*d*x
+ 7*c) + 9009*I*cos(5*d*x + 5*c) + 3003*I*cos(3*d*x + 3*c) + 693*sin(13*d*x + 13*c) + 4095*sin(11*d*x + 11*c)
+ 10010*sin(9*d*x + 9*c) + 12870*sin(7*d*x + 7*c) + 9009*sin(5*d*x + 5*c) + 3003*sin(3*d*x + 3*c))/(a^8*d)

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mupad [B]  time = 4.22, size = 159, normalized size = 0.75 \[ \frac {\frac {{\cos \left (3\,c+3\,d\,x\right )}^3\,5{}\mathrm {i}}{36}+\frac {5\,\sin \left (3\,c+3\,d\,x\right )\,{\cos \left (3\,c+3\,d\,x\right )}^2}{36}-\frac {\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{32}+\frac {\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}}{32}+\frac {\cos \left (7\,c+7\,d\,x\right )\,5{}\mathrm {i}}{112}+\frac {\cos \left (11\,c+11\,d\,x\right )\,5{}\mathrm {i}}{352}+\frac {\cos \left (13\,c+13\,d\,x\right )\,1{}\mathrm {i}}{416}-\frac {7\,\sin \left (3\,c+3\,d\,x\right )}{288}+\frac {\sin \left (5\,c+5\,d\,x\right )}{32}+\frac {5\,\sin \left (7\,c+7\,d\,x\right )}{112}+\frac {5\,\sin \left (11\,c+11\,d\,x\right )}{352}+\frac {\sin \left (13\,c+13\,d\,x\right )}{416}}{a^8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

((cos(5*c + 5*d*x)*1i)/32 - (cos(3*c + 3*d*x)*3i)/32 + (cos(7*c + 7*d*x)*5i)/112 + (cos(11*c + 11*d*x)*5i)/352
+ (cos(13*c + 13*d*x)*1i)/416 - (7*sin(3*c + 3*d*x))/288 + sin(5*c + 5*d*x)/32 + (5*sin(7*c + 7*d*x))/112 + (
5*sin(11*c + 11*d*x))/352 + sin(13*c + 13*d*x)/416 + (cos(3*c + 3*d*x)^3*5i)/36 + (5*cos(3*c + 3*d*x)^2*sin(3*
c + 3*d*x))/36)/(a^8*d)

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sympy [A]  time = 36.21, size = 928, normalized size = 4.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**8,x)

[Out]

Piecewise((-8*tan(c + d*x)**5*sec(c + d*x)**3/(9009*a**8*d*tan(c + d*x)**8 - 72072*I*a**8*d*tan(c + d*x)**7 -
252252*a**8*d*tan(c + d*x)**6 + 504504*I*a**8*d*tan(c + d*x)**5 + 630630*a**8*d*tan(c + d*x)**4 - 504504*I*a**
8*d*tan(c + d*x)**3 - 252252*a**8*d*tan(c + d*x)**2 + 72072*I*a**8*d*tan(c + d*x) + 9009*a**8*d) + 64*I*tan(c
+ d*x)**4*sec(c + d*x)**3/(9009*a**8*d*tan(c + d*x)**8 - 72072*I*a**8*d*tan(c + d*x)**7 - 252252*a**8*d*tan(c
+ d*x)**6 + 504504*I*a**8*d*tan(c + d*x)**5 + 630630*a**8*d*tan(c + d*x)**4 - 504504*I*a**8*d*tan(c + d*x)**3
- 252252*a**8*d*tan(c + d*x)**2 + 72072*I*a**8*d*tan(c + d*x) + 9009*a**8*d) + 236*tan(c + d*x)**3*sec(c + d*x
)**3/(9009*a**8*d*tan(c + d*x)**8 - 72072*I*a**8*d*tan(c + d*x)**7 - 252252*a**8*d*tan(c + d*x)**6 + 504504*I*
a**8*d*tan(c + d*x)**5 + 630630*a**8*d*tan(c + d*x)**4 - 504504*I*a**8*d*tan(c + d*x)**3 - 252252*a**8*d*tan(c
+ d*x)**2 + 72072*I*a**8*d*tan(c + d*x) + 9009*a**8*d) - 544*I*tan(c + d*x)**2*sec(c + d*x)**3/(9009*a**8*d*t
an(c + d*x)**8 - 72072*I*a**8*d*tan(c + d*x)**7 - 252252*a**8*d*tan(c + d*x)**6 + 504504*I*a**8*d*tan(c + d*x)
**5 + 630630*a**8*d*tan(c + d*x)**4 - 504504*I*a**8*d*tan(c + d*x)**3 - 252252*a**8*d*tan(c + d*x)**2 + 72072*
I*a**8*d*tan(c + d*x) + 9009*a**8*d) - 911*tan(c + d*x)*sec(c + d*x)**3/(9009*a**8*d*tan(c + d*x)**8 - 72072*I
*a**8*d*tan(c + d*x)**7 - 252252*a**8*d*tan(c + d*x)**6 + 504504*I*a**8*d*tan(c + d*x)**5 + 630630*a**8*d*tan(
c + d*x)**4 - 504504*I*a**8*d*tan(c + d*x)**3 - 252252*a**8*d*tan(c + d*x)**2 + 72072*I*a**8*d*tan(c + d*x) +
9009*a**8*d) + 1240*I*sec(c + d*x)**3/(9009*a**8*d*tan(c + d*x)**8 - 72072*I*a**8*d*tan(c + d*x)**7 - 252252*a
**8*d*tan(c + d*x)**6 + 504504*I*a**8*d*tan(c + d*x)**5 + 630630*a**8*d*tan(c + d*x)**4 - 504504*I*a**8*d*tan(
c + d*x)**3 - 252252*a**8*d*tan(c + d*x)**2 + 72072*I*a**8*d*tan(c + d*x) + 9009*a**8*d), Ne(d, 0)), (x*sec(c)
**3/(I*a*tan(c) + a)**8, True))

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