Integrand size = 25, antiderivative size = 87 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=-\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \] Output:
-b*(a*sin(f*x+e))^(1/2)/a^2/f/(b*tan(f*x+e))^(3/2)-EllipticE(sin(1/2*f*x+1 /2*e),2^(1/2))*(a*sin(f*x+e))^(1/2)/a^2/f/cos(f*x+e)^(1/2)/(b*tan(f*x+e))^ (1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.74 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=-\frac {b \sqrt {a \sin (e+f x)} \left (2 \cos ^2(e+f x)^{3/4}+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\sin ^2(e+f x)\right ) \sin ^2(e+f x)\right )}{2 a^2 f \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2}} \] Input:
Integrate[1/((a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]),x]
Output:
-1/2*(b*Sqrt[a*Sin[e + f*x]]*(2*(Cos[e + f*x]^2)^(3/4) + Hypergeometric2F1 [1/4, 1/2, 3/2, Sin[e + f*x]^2]*Sin[e + f*x]^2))/(a^2*f*(Cos[e + f*x]^2)^( 3/4)*(b*Tan[e + f*x])^(3/2))
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3079, 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.
\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 3079 |
\(\displaystyle -\frac {\int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{2 a^2}-\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{2 a^2}-\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle -\frac {\sqrt {a \sin (e+f x)} \int \sqrt {\cos (e+f x)}dx}{2 a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {a \sin (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{2 a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {b \sqrt {a \sin (e+f x)}}{a^2 f (b \tan (e+f x))^{3/2}}-\frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
Input:
Int[1/((a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]),x]
Output:
-((b*Sqrt[a*Sin[e + f*x]])/(a^2*f*(b*Tan[e + f*x])^(3/2))) - (EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(a^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])
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[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 = 1.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.97
method | result | size |
default | \(-\frac {i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (\sin \left (f x +e \right )+\tan \left (f x +e \right )\right )+i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (-\sin \left (f x +e \right )-\tan \left (f x +e \right )\right )+1}{f \sqrt {b \tan \left (f x +e \right )}\, \sqrt {a \sin \left (f x +e \right )}\, a}\) | \(171\) |
Input:
int(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/f/(b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2)/a*(I*(cos(f*x+e)/(1+cos(f*x +e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I )*(sin(f*x+e)+tan(f*x+e))+I*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f* x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(-sin(f*x+e)-tan(f*x+e ))+1)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {\sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + \sqrt {\frac {1}{2}} \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {\frac {1}{2}} \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{a^{2} b f \cos \left (f x + e\right )^{2} - a^{2} b f} \] Input:
integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas ")
Output:
(sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)^2 + s qrt(1/2)*sqrt(-a*b)*(cos(f*x + e)^2 - 1)*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + sqrt(1/2)*sqrt(-a*b)*(c os(f*x + e)^2 - 1)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f *x + e) - I*sin(f*x + e))))/(a^2*b*f*cos(f*x + e)^2 - a^2*b*f)
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\text {Timed out} \] Input:
integrate(1/(a*sin(f*x+e))**(3/2)/(b*tan(f*x+e))**(1/2),x)
Output:
Timed out
\[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima ")
Output:
integrate(1/((a*sin(f*x + e))^(3/2)*sqrt(b*tan(f*x + e))), x)
\[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate(1/((a*sin(f*x + e))^(3/2)*sqrt(b*tan(f*x + e))), x)
Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \] Input:
int(1/((a*sin(e + f*x))^(3/2)*(b*tan(e + f*x))^(1/2)),x)
Output:
int(1/((a*sin(e + f*x))^(3/2)*(b*tan(e + f*x))^(1/2)), x)
\[ \int \frac {1}{(a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}}{\sin \left (f x +e \right )^{2} \tan \left (f x +e \right )}d x \right )}{a^{2} b} \] Input:
int(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x)
Output:
(sqrt(b)*sqrt(a)*int((sqrt(tan(e + f*x))*sqrt(sin(e + f*x)))/(sin(e + f*x) **2*tan(e + f*x)),x))/(a**2*b)