3.1177 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=401 \[ -\frac{2 \left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (80 a^2-63 b^2\right ) \sin ^2(c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{16 \left (60 a^2-49 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 a \left (-199 a^2 b^2+160 a^4+39 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^6 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (-318 a^2 b^2+320 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^6 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \sin ^3(c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d} \]

[Out]

(-2*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(a*b^2*d*Sqrt[a + b*Sin[c + d*x]]) + (8*a*(160*a^2 - 139*b^2)*Cos
[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(315*b^5*d) - (16*(60*a^2 - 49*b^2)*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*S
in[c + d*x]])/(315*b^4*d) + (2*(80*a^2 - 63*b^2)*Cos[c + d*x]*Sin[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]])/(63*a*b
^3*d) - (2*Cos[c + d*x]*Sin[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(9*b^2*d) + (8*(320*a^4 - 318*a^2*b^2 + 21*b^
4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(315*b^6*d*Sqrt[(a + b*Sin[c + d*x])
/(a + b)]) - (16*a*(160*a^4 - 199*a^2*b^2 + 39*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/(315*b^6*d*Sqrt[a + b*Sin[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.903799, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {2892, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (80 a^2-63 b^2\right ) \sin ^2(c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{16 \left (60 a^2-49 b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 a \left (-199 a^2 b^2+160 a^4+39 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^6 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (-318 a^2 b^2+320 a^4+21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^6 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \sin ^3(c+d x) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(a*b^2*d*Sqrt[a + b*Sin[c + d*x]]) + (8*a*(160*a^2 - 139*b^2)*Cos
[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(315*b^5*d) - (16*(60*a^2 - 49*b^2)*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*S
in[c + d*x]])/(315*b^4*d) + (2*(80*a^2 - 63*b^2)*Cos[c + d*x]*Sin[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]])/(63*a*b
^3*d) - (2*Cos[c + d*x]*Sin[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(9*b^2*d) + (8*(320*a^4 - 318*a^2*b^2 + 21*b^
4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(315*b^6*d*Sqrt[(a + b*Sin[c + d*x])
/(a + b)]) - (16*a*(160*a^4 - 199*a^2*b^2 + 39*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)])/(315*b^6*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2892

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[((a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*b^2*d*
f*(m + 1)), x] + (-Dist[1/(a*b^2*(m + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Sim
p[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m
+ n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e +
f*x])^(n + 1))/(b^2*d*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2
*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && NeQ[m + n + 4, 0]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{4 \int \frac{\sin ^2(c+d x) \left (\frac{15}{4} \left (4 a^2-3 b^2\right )-\frac{1}{2} a b \sin (c+d x)-\frac{1}{4} \left (80 a^2-63 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{9 a b^2}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{8 \int \frac{\sin (c+d x) \left (-\frac{1}{2} a \left (80 a^2-63 b^2\right )+\frac{5}{2} a^2 b \sin (c+d x)+a \left (60 a^2-49 b^2\right ) \sin ^2(c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{63 a b^3}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{16 \left (60 a^2-49 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{16 \int \frac{a^2 \left (60 a^2-49 b^2\right )-\frac{1}{4} a b \left (40 a^2-21 b^2\right ) \sin (c+d x)-\frac{3}{4} a^2 \left (160 a^2-139 b^2\right ) \sin ^2(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 a b^4}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 \left (60 a^2-49 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{32 \int \frac{\frac{3}{8} a^2 b \left (80 a^2-57 b^2\right )+\frac{3}{8} a \left (320 a^4-318 a^2 b^2+21 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{945 a b^5}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 \left (60 a^2-49 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{\left (4 \left (320 a^4-318 a^2 b^2+21 b^4\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{315 b^6}-\frac{\left (8 a \left (160 a^4-199 a^2 b^2+39 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 b^6}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 \left (60 a^2-49 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{\left (4 \left (320 a^4-318 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}{315 b^6 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (8 a \left (160 a^4-199 a^2 b^2+39 b^4\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}{315 b^6 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 \left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{8 a \left (160 a^2-139 b^2\right ) \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^5 d}-\frac{16 \left (60 a^2-49 b^2\right ) \cos (c+d x) \sin (c+d x) \sqrt{a+b \sin (c+d x)}}{315 b^4 d}+\frac{2 \left (80 a^2-63 b^2\right ) \cos (c+d x) \sin ^2(c+d x) \sqrt{a+b \sin (c+d x)}}{63 a b^3 d}-\frac{2 \cos (c+d x) \sin ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{9 b^2 d}+\frac{8 \left (320 a^4-318 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)}}{315 b^6 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{16 a \left (160 a^4-199 a^2 b^2+39 b^4\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}}}{315 b^6 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.06707, size = 275, normalized size = 0.69 \[ \frac{-b \cos (c+d x) \left (-8 \left (40 a^2 b^2-21 b^4\right ) \cos (2 (c+d x))+4768 a^2 b^2-1280 a^3 b \sin (c+d x)-5120 a^4+1012 a b^3 \sin (c+d x)+100 a b^3 \sin (3 (c+d x))+35 b^4 \cos (4 (c+d x))-203 b^4\right )+64 a \left (-199 a^2 b^2+160 a^4+39 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-32 \left (-318 a^3 b^2-318 a^2 b^3+320 a^4 b+320 a^5+21 a b^4+21 b^5\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{1260 b^6 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-32*(320*a^5 + 320*a^4*b - 318*a^3*b^2 - 318*a^2*b^3 + 21*a*b^4 + 21*b^5)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2
*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + 64*a*(160*a^4 - 199*a^2*b^2 + 39*b^4)*EllipticF[(-2*c + Pi -
2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - b*Cos[c + d*x]*(-5120*a^4 + 4768*a^2*b^2 - 203*
b^4 - 8*(40*a^2*b^2 - 21*b^4)*Cos[2*(c + d*x)] + 35*b^4*Cos[4*(c + d*x)] - 1280*a^3*b*Sin[c + d*x] + 1012*a*b^
3*Sin[c + d*x] + 100*a*b^3*Sin[3*(c + d*x)]))/(1260*b^6*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 1.694, size = 1190, normalized size = 3. \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)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/315*(35*b^6*sin(d*x+c)^6-50*a*b^5*sin(d*x+c)^5+1280*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))
^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b-960
*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+
b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2-1592*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b
/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^
3*b^3+1044*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Elli
pticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^4+312*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*
x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^
(1/2))*a*b^5-84*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)
*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6-1280*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(
d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b)
)^(1/2))*a^6+2552*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/
2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b^2-1356*((a+b*sin(d*x+c))/(a-b))^(1/2)*(
-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)
/(a+b))^(1/2))*a^2*b^4+84*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a
-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6+80*a^2*b^4*sin(d*x+c)^4-112*b^6*s
in(d*x+c)^4-160*a^3*b^3*sin(d*x+c)^3+214*a*b^5*sin(d*x+c)^3-640*a^4*b^2*sin(d*x+c)^2+476*a^2*b^4*sin(d*x+c)^2+
77*b^6*sin(d*x+c)^2+160*a^3*b^3*sin(d*x+c)-164*a*b^5*sin(d*x+c)+640*a^4*b^2-556*a^2*b^4)/b^7/cos(d*x+c)/(a+b*s
in(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (\cos \left (d x + c\right )^{6} - \cos \left (d x + c\right )^{4}\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^6 - cos(d*x + c)^4)*sqrt(b*sin(d*x + c) + a)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) -
a^2 - b^2), 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)**4*sin(d*x+c)**2/(a+b*sin(d*x+c))**(3/2),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 )^{4} \sin \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)