3.533 \(\int \frac{\cos ^8(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=384 \[ \frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-28 a b \sin (c+d x)-3 b^2\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)-144 a^2 b^2+128 a^4+15 b^4\right )}{99 b^7 d}+\frac{32 \left (-272 a^4 b^2+159 a^2 b^4+128 a^6-15 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{99 b^8 d \sqrt{a+b \sin (c+d x)}}-\frac{128 a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{99 b^8 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \]
[Out]
(-2*Cos[c + d*x]^7)/(3*b*d*(a + b*Sin[c + d*x])^(3/2)) - (128*a*(8*a^2 - 9*b^2)*(4*a^2 - 3*b^2)*EllipticE[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(99*b^8*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (32*(
128*a^6 - 272*a^4*b^2 + 159*a^2*b^4 - 15*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)])/(99*b^8*d*Sqrt[a + b*Sin[c + d*x]]) - (28*Cos[c + d*x]^5*(12*a + b*Sin[c + d*x]))/(33*b^3*d*S
qrt[a + b*Sin[c + d*x]]) + (40*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(32*a^2 - 3*b^2 - 28*a*b*Sin[c + d*x]))
/(99*b^5*d) - (16*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(128*a^4 - 144*a^2*b^2 + 15*b^4 - 3*a*b*(32*a^2 - 31*b
^2)*Sin[c + d*x]))/(99*b^7*d)
________________________________________________________________________________________
Rubi [A] time = 0.761083, antiderivative size = 384, 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 =
{2693, 2863, 2865, 2752, 2663, 2661, 2655, 2653} \[ \frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-28 a b \sin (c+d x)-3 b^2\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)-144 a^2 b^2+128 a^4+15 b^4\right )}{99 b^7 d}+\frac{32 \left (-272 a^4 b^2+159 a^2 b^4+128 a^6-15 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{99 b^8 d \sqrt{a+b \sin (c+d x)}}-\frac{128 a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{99 b^8 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^8/(a + b*Sin[c + d*x])^(5/2),x]
[Out]
(-2*Cos[c + d*x]^7)/(3*b*d*(a + b*Sin[c + d*x])^(3/2)) - (128*a*(8*a^2 - 9*b^2)*(4*a^2 - 3*b^2)*EllipticE[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(99*b^8*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (32*(
128*a^6 - 272*a^4*b^2 + 159*a^2*b^4 - 15*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)])/(99*b^8*d*Sqrt[a + b*Sin[c + d*x]]) - (28*Cos[c + d*x]^5*(12*a + b*Sin[c + d*x]))/(33*b^3*d*S
qrt[a + b*Sin[c + d*x]]) + (40*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(32*a^2 - 3*b^2 - 28*a*b*Sin[c + d*x]))
/(99*b^5*d) - (16*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(128*a^4 - 144*a^2*b^2 + 15*b^4 - 3*a*b*(32*a^2 - 31*b
^2)*Sin[c + d*x]))/(99*b^7*d)
Rule 2693
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]
Rule 2863
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]
Rule 2865
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && 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 \frac{\cos ^8(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{14 \int \frac{\cos ^6(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 b}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{280 \int \frac{\cos ^4(c+d x) \left (-\frac{b}{2}-6 a \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{33 b^3}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-3 b^2-28 a b \sin (c+d x)\right )}{99 b^5 d}+\frac{160 \int \frac{\cos ^2(c+d x) \left (\frac{3}{4} b \left (4 a^2-3 b^2\right )+\frac{3}{4} a \left (32 a^2-31 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{99 b^5}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-3 b^2-28 a b \sin (c+d x)\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (128 a^4-144 a^2 b^2+15 b^4-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)\right )}{99 b^7 d}+\frac{128 \int \frac{-\frac{3}{8} b \left (32 a^4-51 a^2 b^2+15 b^4\right )-\frac{3}{2} a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{297 b^7}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-3 b^2-28 a b \sin (c+d x)\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (128 a^4-144 a^2 b^2+15 b^4-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)\right )}{99 b^7 d}-\frac{\left (64 a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{99 b^8}+\frac{\left (16 \left (128 a^6-272 a^4 b^2+159 a^2 b^4-15 b^6\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{99 b^8}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-3 b^2-28 a b \sin (c+d x)\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (128 a^4-144 a^2 b^2+15 b^4-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)\right )}{99 b^7 d}-\frac{\left (64 a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{99 b^8 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (16 \left (128 a^6-272 a^4 b^2+159 a^2 b^4-15 b^6\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}{99 b^8 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 \cos ^7(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac{128 a \left (8 a^2-9 b^2\right ) \left (4 a^2-3 b^2\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)}}{99 b^8 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{32 \left (128 a^6-272 a^4 b^2+159 a^2 b^4-15 b^6\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}}}{99 b^8 d \sqrt{a+b \sin (c+d x)}}-\frac{28 \cos ^5(c+d x) (12 a+b \sin (c+d x))}{33 b^3 d \sqrt{a+b \sin (c+d x)}}+\frac{40 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (32 a^2-3 b^2-28 a b \sin (c+d x)\right )}{99 b^5 d}-\frac{16 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (128 a^4-144 a^2 b^2+15 b^4-3 a b \left (32 a^2-31 b^2\right ) \sin (c+d x)\right )}{99 b^7 d}\\ \end{align*}
Mathematica [A] time = 1.76154, size = 356, normalized size = 0.93 \[ \frac{\frac{1}{2} b \cos (c+d x) \left (74112 a^3 b^3 \sin (c+d x)-384 a^3 b^3 \sin (3 (c+d x))+\left (-3648 a^2 b^4+2048 a^4 b^2+1383 b^6\right ) \cos (2 (c+d x))+\left (126 b^6-96 a^2 b^4\right ) \cos (4 (c+d x))+55296 a^4 b^2-18144 a^2 b^4-40960 a^5 b \sin (c+d x)-32768 a^6-30920 a b^5 \sin (c+d x)+596 a b^5 \sin (3 (c+d x))+28 a b^5 \sin (5 (c+d x))+9 b^6 \cos (6 (c+d x))-2574 b^6\right )+256 (a+b) \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} \left (b \left (-51 a^2 b^3+32 a^4 b+15 b^5\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+4 \left (-60 a^3 b^2+32 a^5+27 a b^4\right ) \left ((a+b) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )\right )}{792 b^8 d (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^8/(a + b*Sin[c + d*x])^(5/2),x]
[Out]
(256*(a + b)*(b*(32*a^4*b - 51*a^2*b^3 + 15*b^5)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + 4*(32*a^5 -
60*a^3*b^2 + 27*a*b^4)*((a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] - a*EllipticF[(-2*c + Pi - 2*
d*x)/4, (2*b)/(a + b)]))*((a + b*Sin[c + d*x])/(a + b))^(3/2) + (b*Cos[c + d*x]*(-32768*a^6 + 55296*a^4*b^2 -
18144*a^2*b^4 - 2574*b^6 + (2048*a^4*b^2 - 3648*a^2*b^4 + 1383*b^6)*Cos[2*(c + d*x)] + (-96*a^2*b^4 + 126*b^6)
*Cos[4*(c + d*x)] + 9*b^6*Cos[6*(c + d*x)] - 40960*a^5*b*Sin[c + d*x] + 74112*a^3*b^3*Sin[c + d*x] - 30920*a*b
^5*Sin[c + d*x] - 384*a^3*b^3*Sin[3*(c + d*x)] + 596*a*b^5*Sin[3*(c + d*x)] + 28*a*b^5*Sin[5*(c + d*x)]))/2)/(
792*b^8*d*(a + b*Sin[c + d*x])^(3/2))
________________________________________________________________________________________
Maple [B] time = 0.596, size = 2253, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^8/(a+b*sin(d*x+c))^(5/2),x)
[Out]
-2/99*(-14*a*b^7*sin(d*x+c)*cos(d*x+c)^6+(48*a^3*b^5-64*a*b^7)*cos(d*x+c)^4*sin(d*x+c)+(1280*a^5*b^3-2328*a^3*
b^5+984*a*b^7)*cos(d*x+c)^2*sin(d*x+c)+16*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1
/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*b*(128*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(
1/2))*a^6*b-96*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^5*b^2-272*EllipticF((b/(a
-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^3+189*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2
),((a-b)/(a+b))^(1/2))*a^3*b^4+159*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^5
-93*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^6-15*EllipticF((b/(a-b)*sin(d*x+c)
+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b^7-128*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1
/2))*a^7+368*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^5*b^2-348*EllipticE((b/(a-b
)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^4+108*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),
((a-b)/(a+b))^(1/2))*a*b^6)*sin(d*x+c)-9*b^8*cos(d*x+c)^8+(24*a^2*b^6-18*b^8)*cos(d*x+c)^6+(-128*a^4*b^4+204*a
^2*b^6-60*b^8)*cos(d*x+c)^4+(1024*a^6*b^2-1664*a^4*b^4+456*a^2*b^6+120*b^8)*cos(d*x+c)^2-2048*(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^8+5888*(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^6*b^2-5568*(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^4+1728*
(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)*E
llipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^6+2048*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1
/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^7*b-1536*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1
/2))*a^6*b^2-4352*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^5*b^3+3024*(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)*EllipticF((b/
(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^4+2544*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1
/2),((a-b)/(a+b))^(1/2))*a^3*b^5-1488*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b
^6-240*(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^7)/(a+b*sin(d*x+c))^(3/2)/b^9/co
s(d*x+c)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{8}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^8/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate(cos(d*x + c)^8/(b*sin(d*x + c) + a)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{8}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^8/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral(-sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^8/(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^
2 - 3*a^2*b - b^3)*sin(d*x + c)), 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(cos(d*x+c)**8/(a+b*sin(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{8}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^8/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)^8/(b*sin(d*x + c) + a)^(5/2), x)