3.505 \(\int \sec ^8(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=439 \[ \frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (4 a^2-b^2\right ) \sin (c+d x)+3 a b\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (-39 a^2 b^2+32 a^4+7 b^4\right ) \sin (c+d x)\right )}{140 d \left (a^2-b^2\right )}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (-59 a^2 b^2+32 a^4+27 b^4\right )-\left (-272 a^4 b^2+165 a^2 b^4+128 a^6-21 b^6\right ) \sin (c+d x)\right )}{280 d \left (a^2-b^2\right )^2}+\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (-144 a^2 b^2+128 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{280 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{7 d} \]

[Out]

(Sec[c + d*x]^7*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(7*d) - ((128*a^4 - 144*a^2*b^2 + 21*b^4)*Ell
ipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(280*(a^2 - b^2)*d*Sqrt[(a + b*Sin[c + d*x
])/(a + b)]) + (2*a*(8*a^2 - 3*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a
+ b)])/(35*d*Sqrt[a + b*Sin[c + d*x]]) + (3*Sec[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]]*(3*a*b + (4*a^2 - b^2)*Sin
[c + d*x]))/(70*d) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(4*a*b*(a^2 - b^2) - (32*a^4 - 39*a^2*b^2 + 7*b^
4)*Sin[c + d*x]))/(140*(a^2 - b^2)*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b*(32*a^4 - 59*a^2*b^2 + 27*
b^4) - (128*a^6 - 272*a^4*b^2 + 165*a^2*b^4 - 21*b^6)*Sin[c + d*x]))/(280*(a^2 - b^2)^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.943791, antiderivative size = 439, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2691, 2861, 2866, 2752, 2663, 2661, 2655, 2653} \[ \frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (4 a^2-b^2\right ) \sin (c+d x)+3 a b\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (-39 a^2 b^2+32 a^4+7 b^4\right ) \sin (c+d x)\right )}{140 d \left (a^2-b^2\right )}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (-59 a^2 b^2+32 a^4+27 b^4\right )-\left (-272 a^4 b^2+165 a^2 b^4+128 a^6-21 b^6\right ) \sin (c+d x)\right )}{280 d \left (a^2-b^2\right )^2}+\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (-144 a^2 b^2+128 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{280 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^7*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(7*d) - ((128*a^4 - 144*a^2*b^2 + 21*b^4)*Ell
ipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(280*(a^2 - b^2)*d*Sqrt[(a + b*Sin[c + d*x
])/(a + b)]) + (2*a*(8*a^2 - 3*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a
+ b)])/(35*d*Sqrt[a + b*Sin[c + d*x]]) + (3*Sec[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]]*(3*a*b + (4*a^2 - b^2)*Sin
[c + d*x]))/(70*d) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(4*a*b*(a^2 - b^2) - (32*a^4 - 39*a^2*b^2 + 7*b^
4)*Sin[c + d*x]))/(140*(a^2 - b^2)*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b*(32*a^4 - 59*a^2*b^2 + 27*
b^4) - (128*a^6 - 272*a^4*b^2 + 165*a^2*b^4 - 21*b^6)*Sin[c + d*x]))/(280*(a^2 - b^2)^2*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 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
])

Rule 2866

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 + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 2752

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \sec ^8(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}-\frac{1}{7} \int \sec ^6(c+d x) \sqrt{a+b \sin (c+d x)} \left (-6 a^2+\frac{3 b^2}{2}-\frac{9}{2} a b \sin (c+d x)\right ) \, dx\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}+\frac{1}{35} \int \frac{\sec ^4(c+d x) \left (\frac{3}{4} a \left (32 a^2-11 b^2\right )+\frac{21}{4} b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{140 \left (a^2-b^2\right ) d}-\frac{\int \frac{\sec ^2(c+d x) \left (-6 a \left (8 a^4-11 a^2 b^2+3 b^4\right )-\frac{9}{8} b \left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{105 \left (a^2-b^2\right )}\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{140 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (32 a^4-59 a^2 b^2+27 b^4\right )-\left (128 a^6-272 a^4 b^2+165 a^2 b^4-21 b^6\right ) \sin (c+d x)\right )}{280 \left (a^2-b^2\right )^2 d}+\frac{\int \frac{-\frac{3}{16} a b^2 \left (32 a^4-59 a^2 b^2+27 b^4\right )-\frac{3}{16} b \left (128 a^6-272 a^4 b^2+165 a^2 b^4-21 b^6\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{105 \left (a^2-b^2\right )^2}\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{140 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (32 a^4-59 a^2 b^2+27 b^4\right )-\left (128 a^6-272 a^4 b^2+165 a^2 b^4-21 b^6\right ) \sin (c+d x)\right )}{280 \left (a^2-b^2\right )^2 d}+\frac{1}{35} \left (a \left (8 a^2-3 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{\left (128 a^4-144 a^2 b^2+21 b^4\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{560 \left (a^2-b^2\right )}\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{140 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (32 a^4-59 a^2 b^2+27 b^4\right )-\left (128 a^6-272 a^4 b^2+165 a^2 b^4-21 b^6\right ) \sin (c+d x)\right )}{280 \left (a^2-b^2\right )^2 d}-\frac{\left (\left (128 a^4-144 a^2 b^2+21 b^4\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{560 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a \left (8 a^2-3 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{35 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{7 d}-\frac{\left (128 a^4-144 a^2 b^2+21 b^4\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{280 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 a \left (8 a^2-3 b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{35 d \sqrt{a+b \sin (c+d x)}}+\frac{3 \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{70 d}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b \left (a^2-b^2\right )-\left (32 a^4-39 a^2 b^2+7 b^4\right ) \sin (c+d x)\right )}{140 \left (a^2-b^2\right ) d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (32 a^4-59 a^2 b^2+27 b^4\right )-\left (128 a^6-272 a^4 b^2+165 a^2 b^4-21 b^6\right ) \sin (c+d x)\right )}{280 \left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 4.45949, size = 338, normalized size = 0.77 \[ \frac{\frac{\sec (c+d x) (a+b \sin (c+d x)) \left (-144 a^2 b^2 \sin (c+d x)+40 \left (a^2-b^2\right ) \sec ^6(c+d x) \left (\left (a^2+b^2\right ) \sin (c+d x)+2 a b\right )-4 \left (a^2-b^2\right ) \sec ^4(c+d x) \left (3 \left (b^2-4 a^2\right ) \sin (c+d x)+a b\right )+2 \left (a^2-b^2\right ) \sec ^2(c+d x) \left (\left (32 a^2-7 b^2\right ) \sin (c+d x)-4 a b\right )-32 a^3 b+128 a^4 \sin (c+d x)+27 a b^3+21 b^4 \sin (c+d x)\right )}{a^2-b^2}+\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \left (\left (-144 a^2 b^2+128 a^4+21 b^4\right ) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-16 a \left (-8 a^2 b+8 a^3-3 a b^2+3 b^3\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )}{a-b}}{280 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((((128*a^4 - 144*a^2*b^2 + 21*b^4)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] - 16*a*(8*a^3 - 8*a^2*b -
3*a*b^2 + 3*b^3)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)])*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(a - b)
+ (Sec[c + d*x]*(a + b*Sin[c + d*x])*(-32*a^3*b + 27*a*b^3 + 128*a^4*Sin[c + d*x] - 144*a^2*b^2*Sin[c + d*x] +
21*b^4*Sin[c + d*x] + 2*(a^2 - b^2)*Sec[c + d*x]^2*(-4*a*b + (32*a^2 - 7*b^2)*Sin[c + d*x]) - 4*(a^2 - b^2)*S
ec[c + d*x]^4*(a*b + 3*(-4*a^2 + b^2)*Sin[c + d*x]) + 40*(a^2 - b^2)*Sec[c + d*x]^6*(2*a*b + (a^2 + b^2)*Sin[c
+ d*x])))/(a^2 - b^2))/(280*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 6.417, size = 1888, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/cos(d*x+c)^9/(a+b*sin(d*x+c))^(3/2)/b/(a^2-b^2)*(2*cos(d*x+c)^4*
(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2*(4*a^4-5*a^2*b^2+b^4)+40*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos
(d*x+c)^2)^(1/2)*b^2*(3*a^4-2*a^2*b^2-b^4)+4*cos(d*x+c)^2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2
*(a^4-14*a^2*b^2+13*b^4)-cos(d*x+c)^8*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2*(128*a^4-144*a^2*b^
2+21*b^4)+16*cos(d*x+c)^2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(3*a^4-4*a^2*b^2+b^4)*sin(d*x+c
)+40*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(a^4+2*a^2*b^2-3*b^4)*sin(d*x+c)+16*cos(d*x+c)^6*(co
s(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(8*a^4-11*a^2*b^2+3*b^4)*sin(d*x+c)+2*cos(d*x+c)^4*(cos(d*x+
c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(32*a^4-43*a^2*b^2+11*b^4)*sin(d*x+c)-cos(d*x+c)^6*(cos(d*x+c)^2*s
in(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*(128*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2
)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^5*
b-96*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/
(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^4*b^2-176*(b/(a-b)*sin(d*x+c)+1/(a-b
)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1
/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^3*b^3+117*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)
+b/(a+b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-
b))^(1/2)*a^2*b^4+48*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*EllipticF((b/(a-
b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a*b^5-21*(b/(a-b)*sin(
d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a
-b)/(a+b))^(1/2))*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*b^6-128*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*s
in(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((
a-b)/(a+b))^(1/2))*a^6+272*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*
sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2-165*(b/(
a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*Ellip
ticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4+21*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)
*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)
*a)^(1/2),((a-b)/(a+b))^(1/2))*b^6-32*a^4*b^2+39*a^2*b^4-7*b^6))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (2 \, a b \sec \left (d x + c\right )^{8} \sin \left (d x + c\right ) -{\left (b^{2} \cos \left (d x + c\right )^{2} - a^{2} - b^{2}\right )} \sec \left (d x + c\right )^{8}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*a*b*sec(d*x + c)^8*sin(d*x + c) - (b^2*cos(d*x + c)^2 - a^2 - b^2)*sec(d*x + c)^8)*sqrt(b*sin(d*x
+ c) + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError