Integrand size = 25, antiderivative size = 167 \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}+\frac {1}{30 a^2 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{12 a^4 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{12 a^4 b^2 f \sqrt {a \sin (e+f x)}} \]
-1/5/b/f/(a*sin(f*x+e))^(9/2)/(b*tan(f*x+e))^(1/2)+1/30/a^2/b/f/(a*sin(f*x +e))^(5/2)/(b*tan(f*x+e))^(1/2)+1/12/a^4/b/f/(a*sin(f*x+e))^(1/2)/(b*tan(f *x+e))^(1/2)-1/12*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Elliptic F(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*tan(f*x+e))^(1/2)/a^4/b^ 2/f/(a*sin(f*x+e))^(1/2)
Time = 0.95 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt [4]{\cos ^2(e+f x)} \left (5+2 \csc ^2(e+f x)-12 \csc ^4(e+f x)\right )-5 \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (\sin (e+f x)),2\right ) \sin (e+f x)}{60 a^4 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \]
((Cos[e + f*x]^2)^(1/4)*(5 + 2*Csc[e + f*x]^2 - 12*Csc[e + f*x]^4) - 5*Ell ipticF[ArcSin[Sin[e + f*x]]/2, 2]*Sin[e + f*x])/(60*a^4*b*f*(Cos[e + f*x]^ 2)^(1/4)*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])
Time = 0.74 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3077, 3042, 3079, 3042, 3079, 3042, 3081, 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.
\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3077 |
\(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{9/2}}dx}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{9/2}}dx}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle -\frac {\frac {5 \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5 \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle -\frac {\frac {5 \left (\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 a^2}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}\right )}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5 \left (\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 a^2}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}\right )}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle -\frac {\frac {5 \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{2 a^2 \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}\right )}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {5 \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a^2 \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}\right )}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {5 \left (\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}\right )}{6 a^2}-\frac {b}{3 a^2 f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}}{10 b^2}-\frac {1}{5 b f (a \sin (e+f x))^{9/2} \sqrt {b \tan (e+f x)}}\) |
-1/5*1/(b*f*(a*Sin[e + f*x])^(9/2)*Sqrt[b*Tan[e + f*x]]) - (-1/3*b/(a^2*f* (a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]) + (5*(-(b/(a^2*f*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/ 2, 2]*Sqrt[b*Tan[e + f*x]])/(a^2*f*Sqrt[a*Sin[e + f*x]])))/(6*a^2))/(10*b^ 2)
3.2.44.3.1 Defintions of rubi rules used
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n + 1))), x] - Simp[(n + 1)/(b^2*(m + n + 1)) Int[(a*Sin[e + f*x])^m*( b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n] && !(EqQ[n, -3/2] && EqQ[m, 1] )
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1) /(a^2*f*(m + n + 1))), x] + Simp[(m + 2)/(a^2*(m + n + 1)) Int[(a*Sin[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && L tQ[m, -1] && NeQ[m + n + 1, 0] && 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[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 = 2.54 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \sin \left (f x +e \right )+5 i \tan \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )-5 \left (\cot ^{4}\left (f x +e \right )\right )+12 \left (\cot ^{2}\left (f x +e \right )\right ) \left (\csc ^{2}\left (f x +e \right )\right )+5 \left (\csc ^{4}\left (f x +e \right )\right )}{60 f \sqrt {\sin \left (f x +e \right ) a}\, \sqrt {b \tan \left (f x +e \right )}\, a^{4} b}\) | \(193\) |
-1/60/f/(sin(f*x+e)*a)^(1/2)/(b*tan(f*x+e))^(1/2)/a^4/b*(5*I*EllipticF(I*( cot(f*x+e)-csc(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e) +1))^(1/2)*sin(f*x+e)+5*I*tan(f*x+e)*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I )*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)-5*cot(f*x+e)^ 4+12*cot(f*x+e)^2*csc(f*x+e)^2+5*csc(f*x+e)^4)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {5 \, {\left (\sqrt {2} \cos \left (f x + e\right )^{6} - 3 \, \sqrt {2} \cos \left (f x + e\right )^{4} + 3 \, \sqrt {2} \cos \left (f x + e\right )^{2} - \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, {\left (\sqrt {2} \cos \left (f x + e\right )^{6} - 3 \, \sqrt {2} \cos \left (f x + e\right )^{4} + 3 \, \sqrt {2} \cos \left (f x + e\right )^{2} - \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, \cos \left (f x + e\right )^{5} - 12 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{120 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{6} - 3 \, a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 3 \, a^{5} b^{2} f \cos \left (f x + e\right )^{2} - a^{5} b^{2} f\right )}} \]
-1/120*(5*(sqrt(2)*cos(f*x + e)^6 - 3*sqrt(2)*cos(f*x + e)^4 + 3*sqrt(2)*c os(f*x + e)^2 - sqrt(2))*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos(f*x + e ) + I*sin(f*x + e)) + 5*(sqrt(2)*cos(f*x + e)^6 - 3*sqrt(2)*cos(f*x + e)^4 + 3*sqrt(2)*cos(f*x + e)^2 - sqrt(2))*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(5*cos(f*x + e)^5 - 12*cos(f*x + e)^ 3 - 5*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e)) )/(a^5*b^2*f*cos(f*x + e)^6 - 3*a^5*b^2*f*cos(f*x + e)^4 + 3*a^5*b^2*f*cos (f*x + e)^2 - a^5*b^2*f)
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {9}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{9/2} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{9/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]