Integrand size = 33, antiderivative size = 621 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^3 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}+\frac {a^3 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} f}-\frac {2 a^3 g \sqrt {g \cos (e+f x)}}{b^4 f}+\frac {2 a (g \cos (e+f x))^{5/2}}{5 b^2 f g}-\frac {2 a^4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^5 f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^3 f \sqrt {g \cos (e+f x)}}+\frac {4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{21 b f \sqrt {g \cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^3 f}+\frac {4 g \sqrt {g \cos (e+f x)} \sin (e+f x)}{21 b f}-\frac {2 (g \cos (e+f x))^{5/2} \sin (e+f x)}{7 b f g} \]
a^3*(-a^2+b^2)^(1/4)*g^(3/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2 )^(1/4)/g^(1/2))/b^(9/2)/f+a^3*(-a^2+b^2)^(1/4)*g^(3/2)*arctanh(b^(1/2)*(g *cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(9/2)/f+2/5*a*(g*cos(f*x+e) )^(5/2)/b^2/f/g-2/7*(g*cos(f*x+e))^(5/2)*sin(f*x+e)/b/f/g-2*a^4*g^2*(cos(1 /2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^( 1/2))*cos(f*x+e)^(1/2)/b^5/f/(g*cos(f*x+e))^(1/2)+2/3*a^2*g^2*(cos(1/2*f*x +1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))* cos(f*x+e)^(1/2)/b^3/f/(g*cos(f*x+e))^(1/2)+4/21*g^2*(cos(1/2*f*x+1/2*e)^2 )^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e )^(1/2)/b/f/(g*cos(f*x+e))^(1/2)+a^4*(a^2-b^2)*g^2*(cos(1/2*f*x+1/2*e)^2)^ (1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^( 1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^5/f/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos( f*x+e))^(1/2)+a^4*(a^2-b^2)*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1 /2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos( f*x+e)^(1/2)/b^5/f/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-2*a^3 *g*(g*cos(f*x+e))^(1/2)/b^4/f+2/3*a^2*g*sin(f*x+e)*(g*cos(f*x+e))^(1/2)/b^ 3/f+4/21*g*sin(f*x+e)*(g*cos(f*x+e))^(1/2)/b/f
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 22.99 (sec) , antiderivative size = 1991, normalized size of antiderivative = 3.21 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \]
-1/420*((g*Cos[e + f*x])^(3/2)*((-2*(70*a^3 - 19*a*b^2)*(a + b*Sqrt[1 - Co s[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2 , (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Co s[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/ 2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x ]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[ Cos[e + f*x]] + I*b*Cos[e + f*x]]))/(-a^2 + b^2)^(3/4))*Sin[e + f*x])/(Sqr t[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((210*a^3 - 21*a*b^2)*(a + b *Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*A rcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/ 2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ((1 + I)*S qrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4 )) + (4*Sqrt[Cos[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f* x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(5/2))/(5*(a^2 - ...
Time = 1.62 (sec) , antiderivative size = 621, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 2009}
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.
| \(\displaystyle \int \frac {\sin ^3(e+f x) (g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3 (g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (-\frac {a^3 (g \cos (e+f x))^{3/2}}{b^3 (a+b \sin (e+f x))}+\frac {a^2 (g \cos (e+f x))^{3/2}}{b^3}-\frac {a \sin (e+f x) (g \cos (e+f x))^{3/2}}{b^2}+\frac {\sin ^2(e+f x) (g \cos (e+f x))^{3/2}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^5 f \sqrt {g \cos (e+f x)}}-\frac {2 a^3 g \sqrt {g \cos (e+f x)}}{b^4 f}+\frac {2 a^2 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^3 f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^3 f}+\frac {a^4 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^4 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 g^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac {a^3 g^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac {2 a (g \cos (e+f x))^{5/2}}{5 b^2 f g}+\frac {4 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{21 b f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) (g \cos (e+f x))^{5/2}}{7 b f g}+\frac {4 g \sin (e+f x) \sqrt {g \cos (e+f x)}}{21 b f}\) |
(a^3*(-a^2 + b^2)^(1/4)*g^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a ^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*f) + (a^3*(-a^2 + b^2)^(1/4)*g^(3/2)*A rcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9 /2)*f) - (2*a^3*g*Sqrt[g*Cos[e + f*x]])/(b^4*f) + (2*a*(g*Cos[e + f*x])^(5 /2))/(5*b^2*f*g) - (2*a^4*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] )/(b^5*f*Sqrt[g*Cos[e + f*x]]) + (2*a^2*g^2*Sqrt[Cos[e + f*x]]*EllipticF[( e + f*x)/2, 2])/(3*b^3*f*Sqrt[g*Cos[e + f*x]]) + (4*g^2*Sqrt[Cos[e + f*x]] *EllipticF[(e + f*x)/2, 2])/(21*b*f*Sqrt[g*Cos[e + f*x]]) + (a^4*(a^2 - b^ 2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f* x)/2, 2])/(b^5*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) + (a^4*(a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^5*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos [e + f*x]]) + (2*a^2*g*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^3*f) + (4*g *Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(21*b*f) - (2*(g*Cos[e + f*x])^(5/2)*S in[e + f*x])/(7*b*f*g)
3.14.77.3.1 Defintions of rubi rules used
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.35 (sec) , antiderivative size = 1219, normalized size of antiderivative = 1.96
(1/20*g*a/b^4*(32*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2*e)^4 *b^2+10*a^2*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan((2^(1/2)*(-2*g*sin(1/ 2*f*x+1/2*e)^2+g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/((a^2*g^2-b^2*g^2)/b^2) ^(1/4))+5*a^2*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln(((g^2*(a^2-b^2)/b^2)^(1 /4)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*2^(1/2)-2*g*sin(1/2*f*x+1/2*e)^2+( g^2*(a^2-b^2)/b^2)^(1/2)+g)/(-(g^2*(a^2-b^2)/b^2)^(1/4)*(-2*g*sin(1/2*f*x+ 1/2*e)^2+g)^(1/2)*2^(1/2)-2*g*sin(1/2*f*x+1/2*e)^2+(g^2*(a^2-b^2)/b^2)^(1/ 2)+g))-10*a^2*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(1/((a^2*g^2-b^2*g^2 )/b^2)^(1/4)*(-2^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+(g^2*(a^2-b^2)/ b^2)^(1/4)))-32*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2*e)^2*b ^2-40*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*a^2+8*b^2*(-2*g*sin(1/2*f*x+1/2* e)^2+g)^(1/2))+32*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2 )*g^2*b*(-1/120/b^4*(-1+sin(1/2*f*x+1/2*e)^2)/(-2*g*sin(1/2*f*x+1/2*e)^4+g *sin(1/2*f*x+1/2*e)^2)^(1/2)*(24*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^6*b ^2-44*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^4*b^2+16*cos(1/2*f*x+1/2*e)*si n(1/2*f*x+1/2*e)^2*b^2+15*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f *x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*a^2-5*EllipticF(cos(1/ 2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2 -1)^(1/2)*b^2-15*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e) ^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*a^2+9*EllipticE(cos(1/2*f*x+...
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]