Integrand size = 25, antiderivative size = 123 \[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=-\frac {4 a^2 b (a \sin (e+f x))^{5/2}}{15 f (b \tan (e+f x))^{3/2}}-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}+\frac {8 a^4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{15 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \]
8/15*a^4*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2 *f*x+1/2*e),2^(1/2))*(a*sin(f*x+e))^(1/2)/f/cos(f*x+e)^(1/2)/(b*tan(f*x+e) )^(1/2)-4/15*a^2*b*(a*sin(f*x+e))^(5/2)/f/(b*tan(f*x+e))^(3/2)-2/9*b*(a*si n(f*x+e))^(9/2)/f/(b*tan(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=\frac {a^4 \left (\cos ^2(e+f x)^{3/4} (-17+5 \cos (2 (e+f x)))+12 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\sin ^2(e+f x)\right )\right ) \sqrt {a \sin (e+f x)} \sin (2 (e+f x))}{90 f \cos ^2(e+f x)^{3/4} \sqrt {b \tan (e+f x)}} \]
(a^4*((Cos[e + f*x]^2)^(3/4)*(-17 + 5*Cos[2*(e + f*x)]) + 12*Hypergeometri c2F1[1/4, 1/2, 3/2, Sin[e + f*x]^2])*Sqrt[a*Sin[e + f*x]]*Sin[2*(e + f*x)] )/(90*f*(Cos[e + f*x]^2)^(3/4)*Sqrt[b*Tan[e + f*x]])
Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3078, 3042, 3078, 3042, 3081, 3042, 3119}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 3078 |
| \(\displaystyle \frac {2}{3} a^2 \int \frac {(a \sin (e+f x))^{5/2}}{\sqrt {b \tan (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} a^2 \int \frac {(a \sin (e+f x))^{5/2}}{\sqrt {b \tan (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3078 |
| \(\displaystyle \frac {2}{3} a^2 \left (\frac {2}{5} a^2 \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{3/2}}\right )-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} a^2 \left (\frac {2}{5} a^2 \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{3/2}}\right )-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3081 |
| \(\displaystyle \frac {2}{3} a^2 \left (\frac {2 a^2 \sqrt {a \sin (e+f x)} \int \sqrt {\cos (e+f x)}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{3/2}}\right )-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} a^2 \left (\frac {2 a^2 \sqrt {a \sin (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{3/2}}\right )-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2}{3} a^2 \left (\frac {4 a^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{5 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 b (a \sin (e+f x))^{5/2}}{5 f (b \tan (e+f x))^{3/2}}\right )-\frac {2 b (a \sin (e+f x))^{9/2}}{9 f (b \tan (e+f x))^{3/2}}\) |
(-2*b*(a*Sin[e + f*x])^(9/2))/(9*f*(b*Tan[e + f*x])^(3/2)) + (2*a^2*((-2*b *(a*Sin[e + f*x])^(5/2))/(5*f*(b*Tan[e + f*x])^(3/2)) + (4*a^2*EllipticE[( e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])))/3
3.2.26.3.1 Defintions of rubi rules used
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( f*m)), x] + Simp[a^2*((m + n - 1)/m) Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 ] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Result contains complex when optimal does not.
Time = 5.08 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.77
| method | result | size |
| default | \(-\frac {2 \sqrt {\sin \left (f x +e \right ) a}\, a^{4} \left (12 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-12 i \cos \left (f x +e \right ) E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-5 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+24 i F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-24 i E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-5 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+12 i \sec \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-12 i \sec \left (f x +e \right ) E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+11 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+11 \sin \left (f x +e \right ) \cos \left (f x +e \right )-12 \sin \left (f x +e \right )\right )}{45 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {b \tan \left (f x +e \right )}}\) | \(464\) |
-2/45/f*(sin(f*x+e)*a)^(1/2)*a^4/(cos(f*x+e)+1)/(b*tan(f*x+e))^(1/2)*(12*I *cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e) +1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)-12*I*cos(f*x+e)*EllipticE(I*(csc(f*x+e )-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2 )-5*cos(f*x+e)^4*sin(f*x+e)+24*I*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(c os(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)-24*I*EllipticE(I* (csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e )+1))^(1/2)-5*cos(f*x+e)^3*sin(f*x+e)+12*I*sec(f*x+e)*EllipticF(I*(csc(f*x +e)-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1 /2)-12*I*sec(f*x+e)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(co s(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)+11*sin(f*x+e)*cos(f*x+e)^2+11* sin(f*x+e)*cos(f*x+e)-12*sin(f*x+e))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=-\frac {2 \, {\left (6 \, \sqrt {2} \sqrt {-a b} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 6 \, \sqrt {2} \sqrt {-a b} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (5 \, a^{4} \cos \left (f x + e\right )^{4} - 11 \, a^{4} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}\right )}}{45 \, b f} \]
-2/45*(6*sqrt(2)*sqrt(-a*b)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse (-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 6*sqrt(2)*sqrt(-a*b)*a^4*weierst rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - (5*a^4*cos(f*x + e)^4 - 11*a^4*cos(f*x + e)^2)*sqrt(a*sin(f*x + e))*sqr t(b*sin(f*x + e)/cos(f*x + e)))/(b*f)
Timed out. \[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {9}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]
\[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {9}{2}}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {(a \sin (e+f x))^{9/2}}{\sqrt {b \tan (e+f x)}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{9/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]