Integrand size = 25, antiderivative size = 152 \[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{5 d e}+\frac {2 a \left (a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{5 d e}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e} \]
2*a*(a^2+2*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF( 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)-2/5*b* (11*a^2+4*b^2)*(e*cos(d*x+c))^(1/2)/d/e-6/5*a*b*(a+b*sin(d*x+c))*(e*cos(d* x+c))^(1/2)/d/e-2/5*b*(a+b*sin(d*x+c))^2*(e*cos(d*x+c))^(1/2)/d/e
Time = 1.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\frac {10 a \left (a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b \cos (c+d x) \left (-30 a^2-9 b^2+b^2 \cos (2 (c+d x))-10 a b \sin (c+d x)\right )}{5 d \sqrt {e \cos (c+d x)}} \]
(10*a*(a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + b*Cos[c + d*x]*(-30*a^2 - 9*b^2 + b^2*Cos[2*(c + d*x)] - 10*a*b*Sin[c + d*x]))/(5 *d*Sqrt[e*Cos[c + d*x]])
Time = 0.75 (sec) , antiderivative size = 153, 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.
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3171 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \sin (c+d x)) \left (5 a^2+9 b \sin (c+d x) a+4 b^2\right )}{2 \sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \sin (c+d x)) \left (5 a^2+9 b \sin (c+d x) a+4 b^2\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \sin (c+d x)) \left (5 a^2+9 b \sin (c+d x) a+4 b^2\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 a \left (a^2+2 b^2\right )+b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \sqrt {e \cos (c+d x)}}dx-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2+2 b^2\right )+b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2+2 b^2\right )+b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{5} \left (5 a \left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 a \left (a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {1}{5} \left (\frac {5 a \left (a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {e \cos (c+d x)}}-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5 a \left (a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {e \cos (c+d x)}}-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (-\frac {2 b \left (11 a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}+\frac {10 a \left (a^2+2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}-\frac {6 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\) |
(-2*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(5*d*e) + ((-2*b*(11*a^ 2 + 4*b^2)*Sqrt[e*Cos[c + d*x]])/(d*e) + (10*a*(a^2 + 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]) - (6*a*b*Sqrt[e* Cos[c + d*x]]*(a + b*Sin[c + d*x]))/(d*e))/5
3.6.59.3.1 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 = 3.50 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.84
| method | result | size |
| default | \(\frac {-\frac {16 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{5}+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+\frac {24 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{5}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}+12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b +\frac {8 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{5}-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b -\frac {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{5}}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(279\) |
| parts | \(\frac {2 a^{3} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}}+\frac {2 b^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-e^{2} \sqrt {e \cos \left (d x +c \right )}\right )}{d \,e^{3}}+\frac {4 a \,b^{2} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}-\frac {6 a^{2} b \sqrt {e \cos \left (d x +c \right )}}{e d}\) | \(310\) |
2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-8*sin(1/2*d*x +1/2*c)^7*b^3+20*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+12*sin(1/2* d*x+1/2*c)^5*b^3-10*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2-5*(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))*a^3-10*(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))*a*b^2+30*sin(1/2*d*x+ 1/2*c)^3*a^2*b+4*sin(1/2*d*x+1/2*c)^3*b^3-15*sin(1/2*d*x+1/2*c)*a^2*b-4*si n(1/2*d*x+1/2*c)*b^3)/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 = 0.88 \[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 5 \, a b^{2} \sin \left (d x + c\right ) - 15 \, a^{2} b - 5 \, b^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, d e} \]
-1/5*(5*sqrt(2)*(I*a^3 + 2*I*a*b^2)*sqrt(e)*weierstrassPInverse(-4, 0, cos (d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*a^3 - 2*I*a*b^2)*sqrt(e)*weier strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(b^3*cos(d*x + c) ^2 - 5*a*b^2*sin(d*x + c) - 15*a^2*b - 5*b^3)*sqrt(e*cos(d*x + c)))/(d*e)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]