Integrand size = 35, antiderivative size = 366 \[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\frac {5 a \sec (e+f x) (b+a \sin (e+f x)) \sqrt {a+b \sin (e+f x)}}{6 f \sqrt {d \sin (e+f x)}}+\frac {\sec ^3(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}-\frac {5 a (a+b)^{3/2} \sqrt {-\frac {a (-1+\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (e+f x)}{6 \sqrt {d} f}-\frac {5 a b (a+b) \sqrt {-\frac {a (-1+\csc (e+f x))}{a+b}} \sqrt {\frac {b+a \csc (e+f x)}{-a+b}} E\left (\arcsin \left (\sqrt {-\frac {b+a \csc (e+f x)}{a-b}}\right )|\frac {-a+b}{a+b}\right ) (1+\sin (e+f x)) \tan (e+f x)}{6 f \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}} \]
1/3*sec(f*x+e)^3*(a+b*sin(f*x+e))^(5/2)*(d*sin(f*x+e))^(1/2)/d/f+5/6*a*sec (f*x+e)*(b+a*sin(f*x+e))*(a+b*sin(f*x+e))^(1/2)/f/(d*sin(f*x+e))^(1/2)-5/6 *a*(a+b)^(3/2)*EllipticF(d^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(d*sin (f*x+e))^(1/2),((-a-b)/(a-b))^(1/2))*(-a*(-1+csc(f*x+e))/(a+b))^(1/2)*(a*( 1+csc(f*x+e))/(a-b))^(1/2)*tan(f*x+e)/f/d^(1/2)-5/6*a*b*(a+b)*EllipticE((( -b-a*csc(f*x+e))/(a-b))^(1/2),((-a+b)/(a+b))^(1/2))*(1+sin(f*x+e))*(-a*(-1 +csc(f*x+e))/(a+b))^(1/2)*((b+a*csc(f*x+e))/(-a+b))^(1/2)*tan(f*x+e)/f/(a* (1+csc(f*x+e))/(a-b))^(1/2)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(4665\) vs. \(2(366)=732\).
Time = 28.72 (sec) , antiderivative size = 4665, normalized size of antiderivative = 12.75 \[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\text {Result too large to show} \]
(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((Sec[e + f*x]^3*(a^2 + b^2 + 2*a*b *Sin[e + f*x]))/3 + (Sec[e + f*x]*(5*a^2 - 2*b^2 + 5*a*b*Sin[e + f*x]))/6) )/(f*Sqrt[d*Sin[e + f*x]]) + (5*a*Csc[(e + f*x)/2]^4*Sec[(e + f*x)/2]^2*Si n[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]]*((5*a^2*Sqrt[a + b*Sin[e + f*x]])/(1 2*Sqrt[Sin[e + f*x]]) - (5*a*b*Sqrt[Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]] )/6)*(-2*b*Tan[(e + f*x)/2]^2 + (2*Sqrt[-a^2 + b^2]*Sqrt[(a*Sec[(e + f*x)/ 2]^2*(a + b*Sin[e + f*x]))/(a^2 - b^2)]*(-(b*EllipticE[ArcSin[Sqrt[(-b + S qrt[-a^2 + b^2] - a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[ -a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])]*Tan[(e + f*x)/2]) + a*EllipticF[ArcS in[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/Sqrt [2]], (2*Sqrt[-a^2 + b^2])/(b + Sqrt[-a^2 + b^2])]*Sqrt[(a*Tan[(e + f*x)/2 ])/(-b + Sqrt[-a^2 + b^2])]*Sqrt[-((a*Tan[(e + f*x)/2])/(b + Sqrt[-a^2 + b ^2]))]))/((a + b*Sin[e + f*x])*Sqrt[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])])))/(96*f*Sqrt[d*Sin[e + f*x]]*((5*a*b*Cos[e + f*x]*Csc[(e + f*x)/2 ]^4*Sec[(e + f*x)/2]^2*Sin[e + f*x]^(7/2)*(-2*b*Tan[(e + f*x)/2]^2 + (2*Sq rt[-a^2 + b^2]*Sqrt[(a*Sec[(e + f*x)/2]^2*(a + b*Sin[e + f*x]))/(a^2 - b^2 )]*(-(b*EllipticE[ArcSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f*x)/2]) /Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])] *Tan[(e + f*x)/2]) + a*EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan [(e + f*x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + Sq...
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.
| \(\displaystyle \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^{5/2}}{\cos (e+f x)^4 \sqrt {d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3367 |
| \(\displaystyle \frac {5}{6} a \int \frac {\sec ^2(e+f x) (a+b \sin (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)}}dx+\frac {\sec ^3(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{6} a \int \frac {(a+b \sin (e+f x))^{3/2}}{\cos (e+f x)^2 \sqrt {d \sin (e+f x)}}dx+\frac {\sec ^3(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}\) |
| \(\Big \downarrow \) 3404 |
| \(\displaystyle \frac {5}{6} a \int \frac {\sec ^2(e+f x) (a+b \sin (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)}}dx+\frac {\sec ^3(e+f x) \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}}{3 d f}\) |
3.15.80.3.1 Defintions of rubi rules used
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*(g*Cos [e + f*x])^(p + 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^m/(d*f*g*(2*m + 1))), x] + Simp[2*a*(m/(g^2*(2*m + 1))) Int[(g*Cos[e + f*x])^(p + 2)*( (a + b*Sin[e + f*x])^(m - 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && EqQ[m + p + 3/2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Unin tegrable[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1987\) vs. \(2(329)=658\).
Time = 3.80 (sec) , antiderivative size = 1988, normalized size of antiderivative = 5.43
-1/12/f/(a+b*sin(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)*(10*cos(f*x+e)*(-a^2+b ^2)^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*x+e)+(-a^2+b^2)^(1 /2)+b))^(1/2)*((a*cot(f*x+e)-a*csc(f*x+e)+(-a^2+b^2)^(1/2)-b)/(-a^2+b^2)^( 1/2))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot(f*x+e)))^(1/2)*Ellipt icE((1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*x+e)+(-a^2+b^2)^(1/2)+b) )^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*b^2-5*c os(f*x+e)*(-a^2+b^2)^(1/2)*(1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*x +e)+(-a^2+b^2)^(1/2)+b))^(1/2)*((a*cot(f*x+e)-a*csc(f*x+e)+(-a^2+b^2)^(1/2 )-b)/(-a^2+b^2)^(1/2))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot(f*x+ e)))^(1/2)*EllipticF((1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*x+e)+(- a^2+b^2)^(1/2)+b))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2 ))^(1/2))*a^2-10*cos(f*x+e)*(1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f* x+e)+(-a^2+b^2)^(1/2)+b))^(1/2)*((a*cot(f*x+e)-a*csc(f*x+e)+(-a^2+b^2)^(1/ 2)-b)/(-a^2+b^2)^(1/2))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot(f*x +e)))^(1/2)*EllipticE((1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*x+e)+( -a^2+b^2)^(1/2)+b))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/ 2))^(1/2))*a^2*b+10*cos(f*x+e)*(1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot (f*x+e)+(-a^2+b^2)^(1/2)+b))^(1/2)*((a*cot(f*x+e)-a*csc(f*x+e)+(-a^2+b^2)^ (1/2)-b)/(-a^2+b^2)^(1/2))^(1/2)*(-a/(b+(-a^2+b^2)^(1/2))*(csc(f*x+e)-cot( f*x+e)))^(1/2)*EllipticE((1/(b+(-a^2+b^2)^(1/2))*(a*csc(f*x+e)-a*cot(f*...
\[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt {d \sin \left (f x + e\right )}} \,d x } \]
integral((2*a*b*sec(f*x + e)^4*sin(f*x + e) - (b^2*cos(f*x + e)^2 - a^2 - b^2)*sec(f*x + e)^4)*sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e))/(d*sin( f*x + e)), x)
Timed out. \[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt {d \sin \left (f x + e\right )}} \,d x } \]
\[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int { \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )^{4}}{\sqrt {d \sin \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(e+f x) (a+b \sin (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\cos \left (e+f\,x\right )}^4\,\sqrt {d\,\sin \left (e+f\,x\right )}} \,d x \]