\(\int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx\) [143]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}+\frac {1}{6 a^2 b f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{6 a^2 b^2 f \sqrt {a \sin (e+f x)}} \] Output:

-1/3/b/f/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2)+1/6/a^2/b/f/(a*sin(f*x+ 
e))^(1/2)/(b*tan(f*x+e))^(1/2)-1/6*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f* 
x+1/2*e,2^(1/2))*(b*tan(f*x+e))^(1/2)/a^2/b^2/f/(a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt [4]{\cos ^2(e+f x)} \left (1-2 \csc ^2(e+f x)\right )-\operatorname {EllipticF}\left (\frac {1}{2} \arcsin (\sin (e+f x)),2\right ) \sin (e+f x)}{6 a^2 b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \] Input:

Integrate[1/((a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)),x]
 

Output:

((Cos[e + f*x]^2)^(1/4)*(1 - 2*Csc[e + f*x]^2) - EllipticF[ArcSin[Sin[e + 
f*x]]/2, 2]*Sin[e + f*x])/(6*a^2*b*f*(Cos[e + f*x]^2)^(1/4)*Sqrt[a*Sin[e + 
 f*x]]*Sqrt[b*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3077, 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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3077

\(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3079

\(\displaystyle -\frac {\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 a^2}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 a^2}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3081

\(\displaystyle -\frac {\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{2 a^2 \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a^2 \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}}{6 b^2}-\frac {1}{3 b f (a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}\)

Input:

Int[1/((a*Sin[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2)),x]
 

Output:

-1/3*1/(b*f*(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]) - (-(b/(a^2*f*Sqr 
t[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*b^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3077

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] 
)
 

rule 3079

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]
 

rule 3081

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])
 

rule 3120

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.47 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {i \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (\sin \left (f x +e \right )+\tan \left (f x +e \right )\right )-\cot \left (f x +e \right )^{2}-\csc \left (f x +e \right )^{2}}{6 f \sqrt {a \sin \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, a^{2} b}\)	\(121\)

Input:

int(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6/f/(a*sin(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)/a^2/b*(I*EllipticF(I*(csc( 
f*x+e)-cot(f*x+e)),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e))) 
^(1/2)*(sin(f*x+e)+tan(f*x+e))-cot(f*x+e)^2-csc(f*x+e)^2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {\frac {1}{2}} {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {\frac {1}{2}} {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{6 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} b^{2} f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \] Input:

integrate(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas 
")
 

Output:

-1/6*(sqrt(1/2)*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-a*b)*weierst 
rassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(1/2)*(cos(f*x + 
e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos(f*x 
 + e) - I*sin(f*x + e)) + (cos(f*x + e)^3 + cos(f*x + e))*sqrt(a*sin(f*x + 
 e))*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(a^3*b^2*f*cos(f*x + e)^4 - 2*a^3* 
b^2*f*cos(f*x + e)^2 + a^3*b^2*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a*sin(f*x+e))**(5/2)/(b*tan(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (a \sin \left (f x + e\right )\right )^{\frac {5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima 
")
 

Output:

integrate(1/((a*sin(f*x + e))^(5/2)*(b*tan(f*x + e))^(3/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int(1/((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2)),x)
 

Output:

int(1/((a*sin(e + f*x))^(5/2)*(b*tan(e + f*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(a \sin (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}}{\sin \left (f x +e \right )^{3} \tan \left (f x +e \right )^{2}}d x \right )}{a^{3} b^{2}} \] Input:

int(1/(a*sin(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x)
 

Output:

(sqrt(b)*sqrt(a)*int((sqrt(tan(e + f*x))*sqrt(sin(e + f*x)))/(sin(e + f*x) 
**3*tan(e + f*x)**2),x))/(a**3*b**2)