3.363 \(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d} \]

[Out]

-16/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^4/d+24/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^5/d-12/7*I*(a+I*a*tan(d*x+c))^(7/2)/a
^6/d+2/9*I*(a+I*a*tan(d*x+c))^(9/2)/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} \[ \frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.70, size = 108, normalized size = 0.92 \[ \frac {2 \sec ^6(c+d x) (\cos (4 (c+d x))+i \sin (4 (c+d x))) (242 i \cos (2 (c+d x))+54 \tan (c+d x)+89 \sin (3 (c+d x)) \sec (c+d x)+77 i)}{315 a^2 d (\tan (c+d x)-i)^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*Sec[c + d*x]^6*(Cos[4*(c + d*x)] + I*Sin[4*(c + d*x)])*(77*I + (242*I)*Cos[2*(c + d*x)] + 89*Sec[c + d*x]*S
in[3*(c + d*x)] + 54*Tan[c + d*x]))/(315*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.65, size = 134, normalized size = 1.15 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-512 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 2304 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 4032 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 3360 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{315 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-512*I*e^(9*I*d*x + 9*I*c) - 2304*I*e^(7*I*d*x + 7*I*c) - 403
2*I*e^(5*I*d*x + 5*I*c) - 3360*I*e^(3*I*d*x + 3*I*c))/(a^3*d*e^(8*I*d*x + 8*I*c) + 4*a^3*d*e^(6*I*d*x + 6*I*c)
+ 6*a^3*d*e^(4*I*d*x + 4*I*c) + 4*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{8}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/(I*a*tan(d*x + c) + a)^(5/2), x)

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maple [A]  time = 1.27, size = 100, normalized size = 0.85 \[ -\frac {2 \left (128 i \left (\cos ^{4}\left (d x +c \right )\right )-128 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+226 i \left (\cos ^{2}\left (d x +c \right )\right )+130 \cos \left (d x +c \right ) \sin \left (d x +c \right )-35 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{315 d \cos \left (d x +c \right )^{4} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/315/d*(128*I*cos(d*x+c)^4-128*cos(d*x+c)^3*sin(d*x+c)+226*I*cos(d*x+c)^2+130*cos(d*x+c)*sin(d*x+c)-35*I)*(a
*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4/a^3

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maxima [A]  time = 0.60, size = 76, normalized size = 0.65 \[ \frac {2 i \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 270 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 756 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 840 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}\right )}}{315 \, a^{7} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/315*I*(35*(I*a*tan(d*x + c) + a)^(9/2) - 270*(I*a*tan(d*x + c) + a)^(7/2)*a + 756*(I*a*tan(d*x + c) + a)^(5/
2)*a^2 - 840*(I*a*tan(d*x + c) + a)^(3/2)*a^3)/(a^7*d)

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mupad [B]  time = 6.68, size = 306, normalized size = 2.62 \[ -\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{315\,a^3\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{315\,a^3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{105\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{63\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

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)^(5/2)),x)

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(9*a^3*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)*256i)/(315*a^3*d*(exp(c*2i +
d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(105*a^3*d*(ex
p(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)*32i)/(63*
a^3*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)*
512i)/(315*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**8/(I*a*(tan(c + d*x) - I))**(5/2), x)

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