Integrand size = 33, antiderivative size = 514 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f} \]
-a^2*(-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^(7/2)/f-a^2*(-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^(7/2)/f-2/5*(g*cos(f*x+e)) ^(5/2)/b/f/g+2*a^3*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ell ipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/b^4/f/(g*cos(f*x+e))^( 1/2)-2/3*a*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(s in(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/b^2/f/(g*cos(f*x+e))^(1/2)-a^3 *(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^4/ f/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-a^3*(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^4/f/(a^2-b*(b+(-a^2+b ^2)^(1/2)))/(g*cos(f*x+e))^(1/2)+2*a^2*g*(g*cos(f*x+e))^(1/2)/b^3/f-2/3*a* g*sin(f*x+e)*(g*cos(f*x+e))^(1/2)/b^2/f
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 24.59 (sec) , antiderivative size = 1953, normalized size of antiderivative = 3.80 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \]
((g*Cos[e + f*x])^(3/2)*Sec[e + f*x]*(-1/5*Cos[2*(e + f*x)]/b - (2*a*Sin[e + f*x])/(3*b^2)))/f + ((g*Cos[e + f*x])^(3/2)*((-2*(10*a^2 + 3*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, C os[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqr t[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*Cos[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)*App ellF1[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)] + L og[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])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((30*a^2 - 3*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)*ArcTan[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)*Sqrt[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...
Time = 1.40 (sec) , antiderivative size = 514, 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.
| \(\displaystyle \int \frac {\sin ^2(e+f x) (g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^2 (g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {a^2 (g \cos (e+f x))^{3/2}}{b^2 (a+b \sin (e+f x))}-\frac {a (g \cos (e+f x))^{3/2}}{b^2}+\frac {\sin (e+f x) (g \cos (e+f x))^{3/2}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^2 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}\) |
-((a^2*(-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^(7/2)*f)) - (a^2*(-a^2 + b^2)^(1/4)*g^(3/2 )*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b ^(7/2)*f) + (2*a^2*g*Sqrt[g*Cos[e + f*x]])/(b^3*f) - (2*(g*Cos[e + f*x])^( 5/2))/(5*b*f*g) + (2*a^3*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]) /(b^4*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*b^2*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^2 - b^2)*g^2*Sqrt[Co s[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4 *(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(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^4*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^2*f)
3.14.78.3.1 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 = 5.20 (sec) , antiderivative size = 1054, normalized size of antiderivative = 2.05
(-1/20*g/b^3*(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))-8*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)* a*g^2*(-1/12/b^4/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)* (4*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^4*b^2-2*cos(1/2*f*x+1/2*e)*sin(1/ 2*f*x+1/2*e)^2*b^2-3*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+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)-1/64*a^2*(a^2-b^2)/b^6*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/(g*(2* _alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b ^2)*(b^2*_alpha^2+4*a^2*cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2...
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]