Integrand size = 25, antiderivative size = 109 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {10 a b \sqrt {e \cos (c+d x)}}{3 d e}+\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e} \]
2/3*(3*a^2+2*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(e*cos(d*x+c))^(1/2)-10/3 *a*b*(e*cos(d*x+c))^(1/2)/d/e-2/3*b*(a+b*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/ d/e
Time = 0.79 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-2 b \cos (c+d x) (6 a+b \sin (c+d x))}{3 d \sqrt {e \cos (c+d x)}} \]
(2*(3*a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - 2*b*Cos[ c + d*x]*(6*a + b*Sin[c + d*x]))/(3*d*Sqrt[e*Cos[c + d*x]])
Time = 0.52 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3171, 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.
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3171 |
| \(\displaystyle \frac {2}{3} \int \frac {3 a^2+5 b \sin (c+d x) a+2 b^2}{2 \sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {3 a^2+5 b \sin (c+d x) a+2 b^2}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {3 a^2+5 b \sin (c+d x) a+2 b^2}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}}dx-\frac {10 a b \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a b \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {e \cos (c+d x)}}-\frac {10 a b \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {e \cos (c+d x)}}-\frac {10 a b \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}-\frac {10 a b \sqrt {e \cos (c+d x)}}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e}\) |
((-10*a*b*Sqrt[e*Cos[c + d*x]])/(d*e) + (2*(3*a^2 + 2*b^2)*Sqrt[Cos[c + d* x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]))/3 - (2*b*Sqrt[e*C os[c + d*x]]*(a + b*Sin[c + d*x]))/(3*d*e)
3.6.51.3.1 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])
Time = 3.52 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.93
| method | result | size |
| default | \(\frac {\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{3}-\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{3}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}}{3}+8 a b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a b}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(210\) |
| parts | \(\frac {2 a^{2} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}}+\frac {4 b^{2} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}-\frac {4 a b \sqrt {e \cos \left (d x +c \right )}}{d e}\) | \(268\) |
2/3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(4*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2* b^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt icF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+12*a*b *sin(1/2*d*x+1/2*c)^3-6*sin(1/2*d*x+1/2*c)*a*b)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (b^{2} \sin \left (d x + c\right ) + 6 \, a b\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, d e} \]
1/3*(sqrt(2)*(-3*I*a^2 - 2*I*b^2)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*a^2 + 2*I*b^2)*sqrt(e)*weierstras sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(b^2*sin(d*x + c) + 6* a*b)*sqrt(e*cos(d*x + c)))/(d*e)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]