Integrand size = 25, antiderivative size = 157 \[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 a \left (a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {6 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{5 d e}+\frac {2 b (a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}}{5 d e} \]
-2*a*(a^2+2*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2* d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/(e*si n(d*x+c))^(1/2)+2/5*b*(11*a^2+4*b^2)*(e*sin(d*x+c))^(1/2)/d/e+6/5*a*b*(a+b *cos(d*x+c))*(e*sin(d*x+c))^(1/2)/d/e+2/5*b*(a+b*cos(d*x+c))^2*(e*sin(d*x+ c))^(1/2)/d/e
Time = 1.51 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\frac {-10 a \left (a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sqrt {\sin (c+d x)}+b \left (30 a^2+9 b^2+10 a b \cos (c+d x)+b^2 \cos (2 (c+d x))\right ) \sin (c+d x)}{5 d \sqrt {e \sin (c+d x)}} \]
(-10*a*(a^2 + 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sqrt[Sin[c + d*x] ] + b*(30*a^2 + 9*b^2 + 10*a*b*Cos[c + d*x] + b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(5*d*Sqrt[e*Sin[c + d*x]])
Time = 0.79 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3171, 27, 3042, 3341, 27, 3042, 3148, 3042, 3121, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3171 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (5 a^2+9 b \cos (c+d x) a+4 b^2\right )}{2 \sqrt {e \sin (c+d x)}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \cos (c+d x)) \left (5 a^2+9 b \cos (c+d x) a+4 b^2\right )}{\sqrt {e \sin (c+d x)}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \left (5 a^2-9 b \sin \left (c+d x-\frac {\pi }{2}\right ) a+4 b^2\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 a \left (a^2+2 b^2\right )+b \left (11 a^2+4 b^2\right ) \cos (c+d x)\right )}{2 \sqrt {e \sin (c+d x)}}dx+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2+2 b^2\right )+b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{\sqrt {e \sin (c+d x)}}dx+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2+2 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{5} \left (5 a \left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 a \left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {1}{5} \left (\frac {5 a \left (a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5 a \left (a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}+\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}+\frac {10 a \left (a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}+\frac {6 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))^2}{5 d e}\) |
(2*b*(a + b*Cos[c + d*x])^2*Sqrt[e*Sin[c + d*x]])/(5*d*e) + ((10*a*(a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[ c + d*x]]) + (2*b*(11*a^2 + 4*b^2)*Sqrt[e*Sin[c + d*x]])/(d*e) + (6*a*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(d*e))/5
3.1.53.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[1/(m + p) Int[(g*Cos[e + f*x])^p* (a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1) *Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m])
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] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[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] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Time = 3.48 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.34
| method | result | size |
| default | \(-\frac {5 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{3}+10 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}-2 b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-30 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(210\) |
| parts | \(-\frac {a^{3} \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{3} \left (\frac {\left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\sqrt {e \sin \left (d x +c \right )}\, e^{2}\right )}{d \,e^{3}}+\frac {3 a \,b^{2} \left (-\frac {2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {6 a^{2} b \sqrt {e \sin \left (d x +c \right )}}{e d}\) | \(235\) |
-1/5/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(5*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c) +2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^3 +10*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF ((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b^2-2*b^3*cos(d*x+c)^3*sin(d*x+c)-10* a*b^2*cos(d*x+c)^2*sin(d*x+c)-30*a^2*b*cos(d*x+c)*sin(d*x+c)-8*b^3*cos(d*x +c)*sin(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\frac {5 \, \sqrt {2} {\left (a^{3} + 2 \, a b^{2}\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (a^{3} + 2 \, a b^{2}\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} + 5 \, a b^{2} \cos \left (d x + c\right ) + 15 \, a^{2} b + 4 \, b^{3}\right )} \sqrt {e \sin \left (d x + c\right )}}{5 \, d e} \]
1/5*(5*sqrt(2)*(a^3 + 2*a*b^2)*sqrt(-I*e)*weierstrassPInverse(4, 0, cos(d* x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(a^3 + 2*a*b^2)*sqrt(I*e)*weierstrass PInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(b^3*cos(d*x + c)^2 + 5* a*b^2*cos(d*x + c) + 15*a^2*b + 4*b^3)*sqrt(e*sin(d*x + c)))/(d*e)
\[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{\sqrt {e \sin (c+d x)}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{\sqrt {e\,\sin \left (c+d\,x\right )}} \,d x \]