3.422 \(\int \sec ^6(c+d x) (a+b \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=381 \[ \frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}-\frac {\sec ^3(c+d x) \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^6}{15 d}-\frac {4 \sec (c+d x) \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^5}{15 d}-\frac {7}{2} b^6 x \left (8 a^2+b^2\right )+\frac {b \left (8 a^4-16 a^2 b^2+35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{15 d}+\frac {a b \left (8 a^4-32 a^2 b^2+87 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{15 d}+\frac {2 a b \left (8 a^6-48 a^4 b^2+163 a^2 b^4+192 b^6\right ) \cos (c+d x)}{15 d}+\frac {b^2 \left (16 a^6-88 a^4 b^2+282 a^2 b^4+105 b^6\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^7}{5 d} \]

[Out]

-7/2*b^6*(8*a^2+b^2)*x+2/15*a*b*(8*a^6-48*a^4*b^2+163*a^2*b^4+192*b^6)*cos(d*x+c)/d+1/30*b^2*(16*a^6-88*a^4*b^
2+282*a^2*b^4+105*b^6)*cos(d*x+c)*sin(d*x+c)/d+1/15*a*b*(8*a^4-32*a^2*b^2+87*b^4)*cos(d*x+c)*(a+b*sin(d*x+c))^
2/d+1/15*b*(8*a^4-16*a^2*b^2+35*b^4)*cos(d*x+c)*(a+b*sin(d*x+c))^3/d+4/15*a*b*(2*a^2+b^2)*cos(d*x+c)*(a+b*sin(
d*x+c))^4/d+1/5*sec(d*x+c)^5*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^7/d-1/15*sec(d*x+c)^3*(a+b*sin(d*x+c))^6*(3*a*b
-(4*a^2-7*b^2)*sin(d*x+c))/d-4/15*sec(d*x+c)*(a+b*sin(d*x+c))^5*(b*(4*a^2-7*b^2)-a*(2*a^2+b^2)*sin(d*x+c))/d

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Rubi [A]  time = 0.72, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2691, 2861, 2753, 2734} \[ \frac {2 a b \left (-48 a^4 b^2+163 a^2 b^4+8 a^6+192 b^6\right ) \cos (c+d x)}{15 d}+\frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}+\frac {b \left (-16 a^2 b^2+8 a^4+35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{15 d}+\frac {a b \left (-32 a^2 b^2+8 a^4+87 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{15 d}+\frac {b^2 \left (-88 a^4 b^2+282 a^2 b^4+16 a^6+105 b^6\right ) \sin (c+d x) \cos (c+d x)}{30 d}-\frac {\sec ^3(c+d x) \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^6}{15 d}-\frac {4 \sec (c+d x) \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^5}{15 d}-\frac {7}{2} b^6 x \left (8 a^2+b^2\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^7}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^8,x]

[Out]

(-7*b^6*(8*a^2 + b^2)*x)/2 + (2*a*b*(8*a^6 - 48*a^4*b^2 + 163*a^2*b^4 + 192*b^6)*Cos[c + d*x])/(15*d) + (b^2*(
16*a^6 - 88*a^4*b^2 + 282*a^2*b^4 + 105*b^6)*Cos[c + d*x]*Sin[c + d*x])/(30*d) + (a*b*(8*a^4 - 32*a^2*b^2 + 87
*b^4)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(15*d) + (b*(8*a^4 - 16*a^2*b^2 + 35*b^4)*Cos[c + d*x]*(a + b*Sin[c
 + d*x])^3)/(15*d) + (4*a*b*(2*a^2 + b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^4)/(15*d) + (Sec[c + d*x]^5*(b + a
*Sin[c + d*x])*(a + b*Sin[c + d*x])^7)/(5*d) - (Sec[c + d*x]^3*(a + b*Sin[c + d*x])^6*(3*a*b - (4*a^2 - 7*b^2)
*Sin[c + d*x]))/(15*d) - (4*Sec[c + d*x]*(a + b*Sin[c + d*x])^5*(b*(4*a^2 - 7*b^2) - a*(2*a^2 + b^2)*Sin[c + d
*x]))/(15*d)

Rule 2691

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(b + a*Sin[e + f*x]))/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1
)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[
2*m, 2*p] || IntegerQ[m])

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
 FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2861

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
 + (f_.)*(x_)]), x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x]))/(f*
g*(p + 1)), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(p +
 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 && GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

Rubi steps

\begin {align*} \int \sec ^6(c+d x) (a+b \sin (c+d x))^8 \, dx &=\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {1}{5} \int \sec ^4(c+d x) (a+b \sin (c+d x))^6 \left (-4 a^2+7 b^2+3 a b \sin (c+d x)\right ) \, dx\\ &=\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}+\frac {1}{15} \int \sec ^2(c+d x) (a+b \sin (c+d x))^5 \left (4 a \left (2 a^2+b^2\right )-4 b \left (4 a^2-7 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {4 \sec (c+d x) (a+b \sin (c+d x))^5 \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {1}{15} \int (a+b \sin (c+d x))^4 \left (-20 b^2 \left (4 a^2-7 b^2\right )+20 a b \left (2 a^2+b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}+\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {4 \sec (c+d x) (a+b \sin (c+d x))^5 \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {1}{75} \int (a+b \sin (c+d x))^3 \left (-60 a b^2 \left (4 a^2-13 b^2\right )+20 b \left (8 a^4-16 a^2 b^2+35 b^4\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {b \left (8 a^4-16 a^2 b^2+35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{15 d}+\frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}+\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {4 \sec (c+d x) (a+b \sin (c+d x))^5 \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {1}{300} \int (a+b \sin (c+d x))^2 \left (-60 b^2 \left (8 a^4-36 a^2 b^2-35 b^4\right )+60 a b \left (8 a^4-32 a^2 b^2+87 b^4\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {a b \left (8 a^4-32 a^2 b^2+87 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{15 d}+\frac {b \left (8 a^4-16 a^2 b^2+35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{15 d}+\frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}+\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {4 \sec (c+d x) (a+b \sin (c+d x))^5 \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {1}{900} \int (a+b \sin (c+d x)) \left (-60 a b^2 \left (8 a^4-44 a^2 b^2-279 b^4\right )+60 b \left (16 a^6-88 a^4 b^2+282 a^2 b^4+105 b^6\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {7}{2} b^6 \left (8 a^2+b^2\right ) x+\frac {2 a b \left (8 a^6-48 a^4 b^2+163 a^2 b^4+192 b^6\right ) \cos (c+d x)}{15 d}+\frac {b^2 \left (16 a^6-88 a^4 b^2+282 a^2 b^4+105 b^6\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac {a b \left (8 a^4-32 a^2 b^2+87 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{15 d}+\frac {b \left (8 a^4-16 a^2 b^2+35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{15 d}+\frac {4 a b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{15 d}+\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{5 d}-\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^6 \left (3 a b-\left (4 a^2-7 b^2\right ) \sin (c+d x)\right )}{15 d}-\frac {4 \sec (c+d x) (a+b \sin (c+d x))^5 \left (b \left (4 a^2-7 b^2\right )-a \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 472, normalized size = 1.24 \[ \frac {\sec ^5(c+d x) \left (640 a^8 \sin (c+d x)+320 a^8 \sin (3 (c+d x))+64 a^8 \sin (5 (c+d x))+3072 a^7 b+8960 a^6 b^2 \sin (c+d x)-2240 a^6 b^2 \sin (3 (c+d x))-448 a^6 b^2 \sin (5 (c+d x))-17920 a^5 b^3 \cos (2 (c+d x))+3584 a^5 b^3+16800 a^4 b^4 \sin (c+d x)-8400 a^4 b^4 \sin (3 (c+d x))+1680 a^4 b^4 \sin (5 (c+d x))+17920 a^3 b^5 \cos (2 (c+d x))+13440 a^3 b^5 \cos (4 (c+d x))+25984 a^3 b^5+11200 a^2 b^6 \sin (c+d x)+5600 a^2 b^6 \sin (3 (c+d x))+5152 a^2 b^6 \sin (5 (c+d x))-33600 a^2 b^6 (c+d x) \cos (c+d x)-16800 a^2 b^6 (c+d x) \cos (3 (c+d x))-3360 a^2 b^6 (c+d x) \cos (5 (c+d x))+22560 a b^7 \cos (2 (c+d x))+8640 a b^7 \cos (4 (c+d x))+480 a b^7 \cos (6 (c+d x))+17472 a b^7+875 b^8 \sin (c+d x)+1015 b^8 \sin (3 (c+d x))+539 b^8 \sin (5 (c+d x))+15 b^8 \sin (7 (c+d x))-4200 b^8 (c+d x) \cos (c+d x)-2100 b^8 (c+d x) \cos (3 (c+d x))-420 b^8 (c+d x) \cos (5 (c+d x))\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^8,x]

[Out]

(Sec[c + d*x]^5*(3072*a^7*b + 3584*a^5*b^3 + 25984*a^3*b^5 + 17472*a*b^7 - 33600*a^2*b^6*(c + d*x)*Cos[c + d*x
] - 4200*b^8*(c + d*x)*Cos[c + d*x] - 17920*a^5*b^3*Cos[2*(c + d*x)] + 17920*a^3*b^5*Cos[2*(c + d*x)] + 22560*
a*b^7*Cos[2*(c + d*x)] - 16800*a^2*b^6*(c + d*x)*Cos[3*(c + d*x)] - 2100*b^8*(c + d*x)*Cos[3*(c + d*x)] + 1344
0*a^3*b^5*Cos[4*(c + d*x)] + 8640*a*b^7*Cos[4*(c + d*x)] - 3360*a^2*b^6*(c + d*x)*Cos[5*(c + d*x)] - 420*b^8*(
c + d*x)*Cos[5*(c + d*x)] + 480*a*b^7*Cos[6*(c + d*x)] + 640*a^8*Sin[c + d*x] + 8960*a^6*b^2*Sin[c + d*x] + 16
800*a^4*b^4*Sin[c + d*x] + 11200*a^2*b^6*Sin[c + d*x] + 875*b^8*Sin[c + d*x] + 320*a^8*Sin[3*(c + d*x)] - 2240
*a^6*b^2*Sin[3*(c + d*x)] - 8400*a^4*b^4*Sin[3*(c + d*x)] + 5600*a^2*b^6*Sin[3*(c + d*x)] + 1015*b^8*Sin[3*(c
+ d*x)] + 64*a^8*Sin[5*(c + d*x)] - 448*a^6*b^2*Sin[5*(c + d*x)] + 1680*a^4*b^4*Sin[5*(c + d*x)] + 5152*a^2*b^
6*Sin[5*(c + d*x)] + 539*b^8*Sin[5*(c + d*x)] + 15*b^8*Sin[7*(c + d*x)]))/(1920*d)

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fricas [A]  time = 0.50, size = 281, normalized size = 0.74 \[ \frac {240 \, a b^{7} \cos \left (d x + c\right )^{6} + 48 \, a^{7} b + 336 \, a^{5} b^{3} + 336 \, a^{3} b^{5} + 48 \, a b^{7} - 105 \, {\left (8 \, a^{2} b^{6} + b^{8}\right )} d x \cos \left (d x + c\right )^{5} + 240 \, {\left (7 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} - 80 \, {\left (7 \, a^{5} b^{3} + 14 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (15 \, b^{8} \cos \left (d x + c\right )^{6} + 6 \, a^{8} + 168 \, a^{6} b^{2} + 420 \, a^{4} b^{4} + 168 \, a^{2} b^{6} + 6 \, b^{8} + 4 \, {\left (4 \, a^{8} - 28 \, a^{6} b^{2} + 105 \, a^{4} b^{4} + 322 \, a^{2} b^{6} + 29 \, b^{8}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (a^{8} - 7 \, a^{6} b^{2} - 105 \, a^{4} b^{4} - 77 \, a^{2} b^{6} - 4 \, b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/30*(240*a*b^7*cos(d*x + c)^6 + 48*a^7*b + 336*a^5*b^3 + 336*a^3*b^5 + 48*a*b^7 - 105*(8*a^2*b^6 + b^8)*d*x*c
os(d*x + c)^5 + 240*(7*a^3*b^5 + 3*a*b^7)*cos(d*x + c)^4 - 80*(7*a^5*b^3 + 14*a^3*b^5 + 3*a*b^7)*cos(d*x + c)^
2 + (15*b^8*cos(d*x + c)^6 + 6*a^8 + 168*a^6*b^2 + 420*a^4*b^4 + 168*a^2*b^6 + 6*b^8 + 4*(4*a^8 - 28*a^6*b^2 +
 105*a^4*b^4 + 322*a^2*b^6 + 29*b^8)*cos(d*x + c)^4 + 8*(a^8 - 7*a^6*b^2 - 105*a^4*b^4 - 77*a^2*b^6 - 4*b^8)*c
os(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 1.19, size = 663, normalized size = 1.74 \[ -\frac {105 \, {\left (8 \, a^{2} b^{6} + b^{8}\right )} {\left (d x + c\right )} + \frac {30 \, {\left (b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 16 \, a b^{7}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {4 \, {\left (15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 20 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 560 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2240 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 220 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1680 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 720 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 58 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 224 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3360 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4984 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 398 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 240 \, a^{7} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 560 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4480 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1920 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 20 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, a^{6} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 220 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, a^{5} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2240 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1200 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 420 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{7} b - 112 \, a^{5} b^{3} + 448 \, a^{3} b^{5} + 264 \, a b^{7}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c))^8,x, algorithm="giac")

[Out]

-1/30*(105*(8*a^2*b^6 + b^8)*(d*x + c) + 30*(b^8*tan(1/2*d*x + 1/2*c)^3 - 16*a*b^7*tan(1/2*d*x + 1/2*c)^2 - b^
8*tan(1/2*d*x + 1/2*c) - 16*a*b^7)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + 4*(15*a^8*tan(1/2*d*x + 1/2*c)^9 + 420*a^2
*b^6*tan(1/2*d*x + 1/2*c)^9 + 45*b^8*tan(1/2*d*x + 1/2*c)^9 + 120*a^7*b*tan(1/2*d*x + 1/2*c)^8 + 120*a*b^7*tan
(1/2*d*x + 1/2*c)^8 - 20*a^8*tan(1/2*d*x + 1/2*c)^7 + 560*a^6*b^2*tan(1/2*d*x + 1/2*c)^7 - 2240*a^2*b^6*tan(1/
2*d*x + 1/2*c)^7 - 220*b^8*tan(1/2*d*x + 1/2*c)^7 + 1680*a^5*b^3*tan(1/2*d*x + 1/2*c)^6 - 720*a*b^7*tan(1/2*d*
x + 1/2*c)^6 + 58*a^8*tan(1/2*d*x + 1/2*c)^5 + 224*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 + 3360*a^4*b^4*tan(1/2*d*x +
 1/2*c)^5 + 4984*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 398*b^8*tan(1/2*d*x + 1/2*c)^5 + 240*a^7*b*tan(1/2*d*x + 1/2
*c)^4 + 560*a^5*b^3*tan(1/2*d*x + 1/2*c)^4 + 4480*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 + 1920*a*b^7*tan(1/2*d*x + 1/
2*c)^4 - 20*a^8*tan(1/2*d*x + 1/2*c)^3 + 560*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 - 2240*a^2*b^6*tan(1/2*d*x + 1/2*c
)^3 - 220*b^8*tan(1/2*d*x + 1/2*c)^3 + 560*a^5*b^3*tan(1/2*d*x + 1/2*c)^2 - 2240*a^3*b^5*tan(1/2*d*x + 1/2*c)^
2 - 1200*a*b^7*tan(1/2*d*x + 1/2*c)^2 + 15*a^8*tan(1/2*d*x + 1/2*c) + 420*a^2*b^6*tan(1/2*d*x + 1/2*c) + 45*b^
8*tan(1/2*d*x + 1/2*c) + 24*a^7*b - 112*a^5*b^3 + 448*a^3*b^5 + 264*a*b^7)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.52, size = 544, normalized size = 1.43 \[ \frac {-a^{8} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {8 a^{7} b}{5 \cos \left (d x +c \right )^{5}}+28 a^{6} b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+56 a^{5} b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {14 a^{4} b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{5}}+56 a^{3} b^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{6}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}\right )+28 a^{2} b^{6} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )+8 a \,b^{7} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+b^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+b*sin(d*x+c))^8,x)

[Out]

1/d*(-a^8*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+8/5*a^7*b/cos(d*x+c)^5+28*a^6*b^2*(1/5*sin(d*x
+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+56*a^5*b^3*(1/5*sin(d*x+c)^4/cos(d*x+c)^5+1/15*sin(d*x+c)^4
/cos(d*x+c)^3-1/15*sin(d*x+c)^4/cos(d*x+c)-1/15*(2+sin(d*x+c)^2)*cos(d*x+c))+14*a^4*b^4*sin(d*x+c)^5/cos(d*x+c
)^5+56*a^3*b^5*(1/5*sin(d*x+c)^6/cos(d*x+c)^5-1/15*sin(d*x+c)^6/cos(d*x+c)^3+1/5*sin(d*x+c)^6/cos(d*x+c)+1/5*(
8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+28*a^2*b^6*(1/5*tan(d*x+c)^5-1/3*tan(d*x+c)^3+tan(d*x+c)-d*x-c)
+8*a*b^7*(1/5*sin(d*x+c)^8/cos(d*x+c)^5-1/5*sin(d*x+c)^8/cos(d*x+c)^3+sin(d*x+c)^8/cos(d*x+c)+(16/5+sin(d*x+c)
^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+b^8*(1/5*sin(d*x+c)^9/cos(d*x+c)^5-4/15*sin(d*x+c)^9/cos(d*x
+c)^3+8/5*sin(d*x+c)^9/cos(d*x+c)+8/5*(sin(d*x+c)^7+7/6*sin(d*x+c)^5+35/24*sin(d*x+c)^3+35/16*sin(d*x+c))*cos(
d*x+c)-7/2*d*x-7/2*c))

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maxima [A]  time = 0.44, size = 315, normalized size = 0.83 \[ \frac {420 \, a^{4} b^{4} \tan \left (d x + c\right )^{5} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{8} + 56 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{6} b^{2} + 56 \, {\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{2} b^{6} + {\left (6 \, \tan \left (d x + c\right )^{5} - 20 \, \tan \left (d x + c\right )^{3} - 105 \, d x - 105 \, c + \frac {15 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} + 90 \, \tan \left (d x + c\right )\right )} b^{8} + 48 \, a b^{7} {\left (\frac {15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )} - \frac {112 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{5} b^{3}}{\cos \left (d x + c\right )^{5}} + \frac {112 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} a^{3} b^{5}}{\cos \left (d x + c\right )^{5}} + \frac {48 \, a^{7} b}{\cos \left (d x + c\right )^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/30*(420*a^4*b^4*tan(d*x + c)^5 + 2*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^8 + 56*(3*tan(
d*x + c)^5 + 5*tan(d*x + c)^3)*a^6*b^2 + 56*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x
+ c))*a^2*b^6 + (6*tan(d*x + c)^5 - 20*tan(d*x + c)^3 - 105*d*x - 105*c + 15*tan(d*x + c)/(tan(d*x + c)^2 + 1)
 + 90*tan(d*x + c))*b^8 + 48*a*b^7*((15*cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 1)/cos(d*x + c)^5 + 5*cos(d*x + c)
) - 112*(5*cos(d*x + c)^2 - 3)*a^5*b^3/cos(d*x + c)^5 + 112*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*a^3*b^
5/cos(d*x + c)^5 + 48*a^7*b/cos(d*x + c)^5)/d

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mupad [B]  time = 7.60, size = 665, normalized size = 1.75 \[ -\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (2\,a^8+56\,a^2\,b^6+7\,b^8\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (48\,a^7\,b+\frac {1568\,a^5\,b^3}{3}+\frac {1792\,a^3\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (32\,a^7\,b+224\,a^5\,b^3\right )+\frac {256\,a\,b^7}{5}+\frac {16\,a^7\,b}{5}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^8+56\,a^2\,b^6+7\,b^8\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (64\,a^7\,b+448\,a^5\,b^3+896\,a^3\,b^5+256\,a\,b^7\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-\frac {32\,a^7\,b}{5}-\frac {224\,a^5\,b^3}{5}+\frac {896\,a^3\,b^5}{5}+\frac {768\,a\,b^7}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {176\,a^7\,b}{5}+\frac {3136\,a^5\,b^3}{15}+\frac {896\,a^3\,b^5}{15}+\frac {256\,a\,b^7}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {22\,a^8}{5}+\frac {896\,a^6\,b^2}{5}+448\,a^4\,b^4+\frac {616\,a^2\,b^6}{5}+\frac {77\,b^8}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {22\,a^8}{5}+\frac {896\,a^6\,b^2}{5}+448\,a^4\,b^4+\frac {616\,a^2\,b^6}{5}+\frac {77\,b^8}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {152\,a^8}{15}+\frac {3136\,a^6\,b^2}{15}+896\,a^4\,b^4+\frac {10976\,a^2\,b^6}{15}+\frac {412\,b^8}{15}\right )+\frac {896\,a^3\,b^5}{15}-\frac {224\,a^5\,b^3}{15}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {4\,a^8}{3}+\frac {224\,a^6\,b^2}{3}-\frac {560\,a^2\,b^6}{3}-\frac {70\,b^8}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {4\,a^8}{3}+\frac {224\,a^6\,b^2}{3}-\frac {560\,a^2\,b^6}{3}-\frac {70\,b^8}{3}\right )+16\,a^7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {7\,b^6\,\mathrm {atan}\left (\frac {7\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^2+b^2\right )}{56\,a^2\,b^6+7\,b^8}\right )\,\left (8\,a^2+b^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^8/cos(c + d*x)^6,x)

[Out]

- (tan(c/2 + (d*x)/2)^13*(2*a^8 + 7*b^8 + 56*a^2*b^6) + tan(c/2 + (d*x)/2)^8*(48*a^7*b + (1792*a^3*b^5)/3 + (1
568*a^5*b^3)/3) + tan(c/2 + (d*x)/2)^10*(32*a^7*b + 224*a^5*b^3) + (256*a*b^7)/5 + (16*a^7*b)/5 + tan(c/2 + (d
*x)/2)*(2*a^8 + 7*b^8 + 56*a^2*b^6) + tan(c/2 + (d*x)/2)^6*(256*a*b^7 + 64*a^7*b + 896*a^3*b^5 + 448*a^5*b^3)
- tan(c/2 + (d*x)/2)^2*((768*a*b^7)/5 - (32*a^7*b)/5 + (896*a^3*b^5)/5 - (224*a^5*b^3)/5) + tan(c/2 + (d*x)/2)
^4*((256*a*b^7)/5 + (176*a^7*b)/5 + (896*a^3*b^5)/15 + (3136*a^5*b^3)/15) + tan(c/2 + (d*x)/2)^5*((22*a^8)/5 +
 (77*b^8)/5 + (616*a^2*b^6)/5 + 448*a^4*b^4 + (896*a^6*b^2)/5) + tan(c/2 + (d*x)/2)^9*((22*a^8)/5 + (77*b^8)/5
 + (616*a^2*b^6)/5 + 448*a^4*b^4 + (896*a^6*b^2)/5) + tan(c/2 + (d*x)/2)^7*((152*a^8)/15 + (412*b^8)/15 + (109
76*a^2*b^6)/15 + 896*a^4*b^4 + (3136*a^6*b^2)/15) + (896*a^3*b^5)/15 - (224*a^5*b^3)/15 + tan(c/2 + (d*x)/2)^3
*((4*a^8)/3 - (70*b^8)/3 - (560*a^2*b^6)/3 + (224*a^6*b^2)/3) + tan(c/2 + (d*x)/2)^11*((4*a^8)/3 - (70*b^8)/3
- (560*a^2*b^6)/3 + (224*a^6*b^2)/3) + 16*a^7*b*tan(c/2 + (d*x)/2)^12)/(d*(3*tan(c/2 + (d*x)/2)^2 - tan(c/2 +
(d*x)/2)^4 - 5*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 3*tan(c/2 + (d*x)/2)^12
 + tan(c/2 + (d*x)/2)^14 - 1)) - (7*b^6*atan((7*b^6*tan(c/2 + (d*x)/2)*(8*a^2 + b^2))/(7*b^8 + 56*a^2*b^6))*(8
*a^2 + b^2))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+b*sin(d*x+c))**8,x)

[Out]

Timed out

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