3.1152 \(\int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=394 \[ -\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (-113 a^2 b^2+32 a^4+177 b^4\right )-3 b \left (-27 a^2 b^2+8 a^4-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}-\frac{8 a \left (-145 a^4 b^2+290 a^2 b^4+32 a^6-177 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15015 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (-137 a^4 b^2+258 a^2 b^4+32 a^6+231 b^6\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15015 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d} \]

[Out]

(-6*a*Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^(3/2))/(13*d)
+ (8*(32*a^6 - 137*a^4*b^2 + 258*a^2*b^4 + 231*b^6)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Si
n[c + d*x]])/(15015*b^5*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (8*a*(32*a^6 - 145*a^4*b^2 + 290*a^2*b^4 - 177
*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(15015*b^5*d*Sqrt[a + b
*Sin[c + d*x]]) - (2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(4*a*(2*a^2 - 5*b^2) - 7*b*(a^2 + 11*b^2)*Sin[c +
d*x]))/(3003*b^2*d) + (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*(32*a^4 - 113*a^2*b^2 + 177*b^4) - 3*b*(8*a
^4 - 27*a^2*b^2 - 77*b^4)*Sin[c + d*x]))/(15015*b^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.858637, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2862, 2865, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (-113 a^2 b^2+32 a^4+177 b^4\right )-3 b \left (-27 a^2 b^2+8 a^4-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}-\frac{8 a \left (-145 a^4 b^2+290 a^2 b^4+32 a^6-177 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15015 b^5 d \sqrt{a+b \sin (c+d x)}}+\frac{8 \left (-137 a^4 b^2+258 a^2 b^4+32 a^6+231 b^6\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15015 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^(3/2))/(13*d)
+ (8*(32*a^6 - 137*a^4*b^2 + 258*a^2*b^4 + 231*b^6)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Si
n[c + d*x]])/(15015*b^5*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (8*a*(32*a^6 - 145*a^4*b^2 + 290*a^2*b^4 - 177
*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(15015*b^5*d*Sqrt[a + b
*Sin[c + d*x]]) - (2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(4*a*(2*a^2 - 5*b^2) - 7*b*(a^2 + 11*b^2)*Sin[c +
d*x]))/(3003*b^2*d) + (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*(32*a^4 - 113*a^2*b^2 + 177*b^4) - 3*b*(8*a
^4 - 27*a^2*b^2 - 77*b^4)*Sin[c + d*x]))/(15015*b^4*d)

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[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 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 \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac{2}{13} \int \cos ^4(c+d x) \left (\frac{3 b}{2}+\frac{3}{2} a \sin (c+d x)\right ) \sqrt{a+b \sin (c+d x)} \, dx\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac{4}{143} \int \frac{\cos ^4(c+d x) \left (9 a b+\frac{3}{4} \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{16 \int \frac{\cos ^2(c+d x) \left (-\frac{3}{8} a b \left (a^2-97 b^2\right )-\frac{3}{8} \left (8 a^4-27 a^2 b^2-77 b^4\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{3003 b^2}\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (32 a^4-113 a^2 b^2+177 b^4\right )-3 b \left (8 a^4-27 a^2 b^2-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}+\frac{64 \int \frac{\frac{3}{2} a b \left (a^4-4 a^2 b^2+51 b^4\right )+\frac{3}{16} \left (32 a^6-137 a^4 b^2+258 a^2 b^4+231 b^6\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{45045 b^4}\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (32 a^4-113 a^2 b^2+177 b^4\right )-3 b \left (8 a^4-27 a^2 b^2-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}-\frac{\left (4 a \left (32 a^6-145 a^4 b^2+290 a^2 b^4-177 b^6\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{15015 b^5}+\frac{\left (4 \left (32 a^6-137 a^4 b^2+258 a^2 b^4+231 b^6\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{15015 b^5}\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}-\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (32 a^4-113 a^2 b^2+177 b^4\right )-3 b \left (8 a^4-27 a^2 b^2-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}+\frac{\left (4 \left (32 a^6-137 a^4 b^2+258 a^2 b^4+231 b^6\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{15015 b^5 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (4 a \left (32 a^6-145 a^4 b^2+290 a^2 b^4-177 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}{15015 b^5 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{6 a \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2}}{13 d}+\frac{8 \left (32 a^6-137 a^4 b^2+258 a^2 b^4+231 b^6\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)}}{15015 b^5 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{8 a \left (32 a^6-145 a^4 b^2+290 a^2 b^4-177 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}}}{15015 b^5 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (2 a^2-5 b^2\right )-7 b \left (a^2+11 b^2\right ) \sin (c+d x)\right )}{3003 b^2 d}+\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (a \left (32 a^4-113 a^2 b^2+177 b^4\right )-3 b \left (8 a^4-27 a^2 b^2-77 b^4\right ) \sin (c+d x)\right )}{15015 b^4 d}\\ \end{align*}

Mathematica [A]  time = 11.3914, size = 382, normalized size = 0.97 \[ \frac{-3 b \cos (c+d x) \left (2088 a^3 b^3 \sin (c+d x)+40 a^3 b^3 \sin (3 (c+d x))+\left (24512 a^2 b^4-128 a^4 b^2+8547 b^6\right ) \cos (2 (c+d x))+70 \left (86 a^2 b^4-11 b^6\right ) \cos (4 (c+d x))+8640 a^4 b^2+1980 a^2 b^4-512 a^5 b \sin (c+d x)-2048 a^6-19492 a b^5 \sin (c+d x)+11870 a b^5 \sin (3 (c+d x))+5250 a b^5 \sin (5 (c+d x))-1155 b^6 \cos (6 (c+d x))-6622 b^6\right )+384 a \left (-145 a^4 b^2+290 a^2 b^4+32 a^6-177 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-384 \left (-137 a^5 b^2-137 a^4 b^3+258 a^3 b^4+258 a^2 b^5+32 a^6 b+32 a^7+231 a b^6+231 b^7\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{720720 b^5 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-384*(32*a^7 + 32*a^6*b - 137*a^5*b^2 - 137*a^4*b^3 + 258*a^3*b^4 + 258*a^2*b^5 + 231*a*b^6 + 231*b^7)*Ellipt
icE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + 384*a*(32*a^6 - 145*a^4*b^2 + 2
90*a^2*b^4 - 177*b^6)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - 3*b
*Cos[c + d*x]*(-2048*a^6 + 8640*a^4*b^2 + 1980*a^2*b^4 - 6622*b^6 + (-128*a^4*b^2 + 24512*a^2*b^4 + 8547*b^6)*
Cos[2*(c + d*x)] + 70*(86*a^2*b^4 - 11*b^6)*Cos[4*(c + d*x)] - 1155*b^6*Cos[6*(c + d*x)] - 512*a^5*b*Sin[c + d
*x] + 2088*a^3*b^3*Sin[c + d*x] - 19492*a*b^5*Sin[c + d*x] + 40*a^3*b^3*Sin[3*(c + d*x)] + 11870*a*b^5*Sin[3*(
c + d*x)] + 5250*a*b^5*Sin[5*(c + d*x)]))/(720720*b^5*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 1.836, size = 1619, normalized size = 4.1 \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)*(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/15015*(-5*a^3*b^5*sin(d*x+c)^5-7390*a*b^7*sin(d*x+c)^5+8*a^4*b^4*sin(d*x+c)^4-4542*a^2*b^6*sin(d*x+c)^4-16*a
^5*b^3*sin(d*x+c)^3+74*a^3*b^5*sin(d*x+c)^3+6089*a*b^7*sin(d*x+c)^3-64*a^6*b^2*sin(d*x+c)^2+258*a^4*b^4*sin(d*
x+c)^2+4053*a^2*b^6*sin(d*x+c)^2+16*a^5*b^3*sin(d*x+c)-69*a^3*b^5*sin(d*x+c)-1324*a*b^7*sin(d*x+c)+2625*a*b^7*
sin(d*x+c)^7-266*a^4*b^4-1016*a^2*b^6+64*a^6*b^2+600*((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^2*b^6+15
05*a^2*b^6*sin(d*x+c)^6+1155*b^8*sin(d*x+c)^8-3080*b^8*sin(d*x+c)^6+2233*b^8*sin(d*x+c)^4-308*b^8*sin(d*x+c)^2
-708*((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^7-1580*((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^4-96*((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^6*b^2-580*((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^3+676*((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*b^2+128*((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^7*b+108*((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^6+420*((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^4+1160*((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^5-128*((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^8+924*((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^8-924*((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^8)/b^6/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/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{3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\,{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)*(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b \cos \left (d x + c\right )^{6} - a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - b \cos \left (d x + c\right )^{4}\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(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-(b*cos(d*x + c)^6 - a*cos(d*x + c)^4*sin(d*x + c) - b*cos(d*x + c)^4)*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(cos(d*x+c)**4*sin(d*x+c)*(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{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )\,{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)*(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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