3.347 \(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d} \]

[Out]

-16/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^4/d+24/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^5/d-4/3*I*(a+I*a*tan(d*x+c))^(9/2)/a^
6/d+2/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^7/d

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Rubi [A]  time = 0.09, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3487, 43} \[ \frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-16*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^4*d) + (((24*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^5*d) - (((4*
I)/3)*(a + I*a*Tan[c + d*x])^(9/2))/(a^6*d) + (((2*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^7*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x)^3 (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (8 a^3 (a+x)^{3/2}-12 a^2 (a+x)^{5/2}+6 a (a+x)^{7/2}-(a+x)^{9/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d}\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 110, normalized size = 0.94 \[ \frac {2 i \sec ^6(c+d x) (\cos (4 (c+d x))+i \sin (4 (c+d x))) (494 i \cos (2 (c+d x))+110 \tan (c+d x)+215 \sin (3 (c+d x)) \sec (c+d x)+39 i)}{1155 a d (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((2*I)/1155)*Sec[c + d*x]^6*(Cos[4*(c + d*x)] + I*Sin[4*(c + d*x)])*(39*I + (494*I)*Cos[2*(c + d*x)] + 215*Se
c[c + d*x]*Sin[3*(c + d*x)] + 110*Tan[c + d*x]))/(a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.80, size = 149, normalized size = 1.27 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1024 i \, e^{\left (11 i \, d x + 11 i \, c\right )} - 5632 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 12672 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 14784 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{1155 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/1155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1024*I*e^(11*I*d*x + 11*I*c) - 5632*I*e^(9*I*d*x + 9*I*c) -
12672*I*e^(7*I*d*x + 7*I*c) - 14784*I*e^(5*I*d*x + 5*I*c))/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a^2*d*e^(8*I*d*x
+ 8*I*c) + 10*a^2*d*e^(6*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{8}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/(I*a*tan(d*x + c) + a)^(3/2), x)

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maple [A]  time = 1.37, size = 117, normalized size = 1.00 \[ -\frac {2 \left (256 i \left (\cos ^{5}\left (d x +c \right )\right )-256 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+32 i \left (\cos ^{3}\left (d x +c \right )\right )-160 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+245 i \cos \left (d x +c \right )+105 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{1155 d \cos \left (d x +c \right )^{5} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/1155/d*(256*I*cos(d*x+c)^5-256*sin(d*x+c)*cos(d*x+c)^4+32*I*cos(d*x+c)^3-160*cos(d*x+c)^2*sin(d*x+c)+245*I*
cos(d*x+c)+105*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^5/a^2

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maxima [A]  time = 0.37, size = 76, normalized size = 0.65 \[ \frac {2 i \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 770 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 1980 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 1848 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3}\right )}}{1155 \, a^{7} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/1155*I*(105*(I*a*tan(d*x + c) + a)^(11/2) - 770*(I*a*tan(d*x + c) + a)^(9/2)*a + 1980*(I*a*tan(d*x + c) + a)
^(7/2)*a^2 - 1848*(I*a*tan(d*x + c) + a)^(5/2)*a^3)/(a^7*d)

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mupad [B]  time = 7.64, size = 370, normalized size = 3.16 \[ -\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{1155\,a^2\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{1155\,a^2\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{385\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{231\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{33\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256i)/(33*a^2*d*(exp(c*2i + d*x*2i)
+ 1)^4) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(1155*a^2*d*(exp(c*2
i + d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(385*a^2*d
*(exp(c*2i + d*x*2i) + 1)^2) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/
(231*a^2*d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(
1/2)*1024i)/(1155*a^2*d) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(11*
a^2*d*(exp(c*2i + d*x*2i) + 1)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**8/(I*a*(tan(c + d*x) - I))**(3/2), x)

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