Integrand size = 25, antiderivative size = 135 \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {a \sqrt {a \sin (e+f x)}}{6 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}+\frac {a^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{12 b^2 f \sqrt {a \sin (e+f x)}} \]
1/3*(a*sin(f*x+e))^(5/2)/a/b/f/(b*sec(f*x+e))^(1/2)-1/6*a*(a*sin(f*x+e))^( 1/2)/b/f/(b*sec(f*x+e))^(1/2)-1/12*a^2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1 /4*Pi+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2 *f*x+2*e)^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\frac {a \sqrt {a \sin (e+f x)} \left (-2 \cos (2 (e+f x))+\csc ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{12 b f \sqrt {b \sec (e+f x)}} \]
(a*Sqrt[a*Sin[e + f*x]]*(-2*Cos[2*(e + f*x)] + Csc[e + f*x]^2*Hypergeometr ic2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(12*b*f*Sqr t[b*Sec[e + f*x]])
Time = 0.66 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3062, 3042, 3063, 3042, 3065, 3042, 3053, 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.
\(\displaystyle \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3062 |
\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3063 |
\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3065 |
\(\displaystyle \frac {\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\) |
(a*Sin[e + f*x])^(5/2)/(3*a*b*f*Sqrt[b*Sec[e + f*x]]) + (-((a*b*Sqrt[a*Sin [e + f*x]])/(f*Sqrt[b*Sec[e + f*x]])) + (a^2*EllipticF[e - Pi/4 + f*x, 2]* Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(2*f*Sqrt[a*Sin[e + f*x]]))/( 6*b^2)
3.5.78.3.1 Defintions of rubi rules used
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n + 1)/(a *b*f*(m - n))), x] - Simp[(n + 1)/(b^2*(m - n)) Int[(a*Sin[e + f*x])^m*(b *Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] & & NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*b*(a*Sin[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - n))), x] + Simp[a^2*((m - 1)/(m - n)) Int[(a*Sin[e + f*x])^( m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int egerQ[m - 1/2] && IntegerQ[n - 1/2]
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]
Result contains complex when optimal does not.
Time = 1.97 (sec) , antiderivative size = 1746, normalized size of antiderivative = 12.93
-1/48/f*2^(1/2)*(-6*I*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x +e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x +e)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(f*x+e)-6*I*(-cot(f*x+e)+csc(f*x+e) +1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*El lipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))+6*I*(-cot (f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-cs c(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2* 2^(1/2))-6*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2 )*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2 ),1/2+1/2*I,1/2*2^(1/2))*cos(f*x+e)+8*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(co t(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticF((-cot (f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*cos(f*x+e)+6*I*(-cot(f*x+e)+csc(f *x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/ 2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos( f*x+e)-6*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)* (cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2), 1/2-1/2*I,1/2*2^(1/2))*cos(f*x+e)+8*2^(1/2)*cos(f*x+e)^3*sin(f*x+e)-6*(-co t(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-c sc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2 *2^(1/2))+8*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^...
\[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]