\(\int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [139]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {4 a^4 (a \sin (e+f x))^{3/2}}{77 b f \sqrt {b \tan (e+f x)}}-\frac {2 a^2 (a \sin (e+f x))^{7/2}}{77 b f \sqrt {b \tan (e+f x)}}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}+\frac {8 a^6 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{77 b^2 f \sqrt {a \sin (e+f x)}} \] Output:

-4/77*a^4*(a*sin(f*x+e))^(3/2)/b/f/(b*tan(f*x+e))^(1/2)-2/77*a^2*(a*sin(f* 
x+e))^(7/2)/b/f/(b*tan(f*x+e))^(1/2)+2/11*(a*sin(f*x+e))^(11/2)/b/f/(b*tan 
(f*x+e))^(1/2)+8/77*a^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)/b^2/f/(a*sin(f*x+e))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.71 \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {a^5 \left (\sqrt [4]{\cos ^2(e+f x)} (-22 \cos (e+f x)-17 \cos (3 (e+f x))+7 \cos (5 (e+f x)))+64 \cot (e+f x) \operatorname {EllipticF}\left (\frac {1}{2} \arcsin (\sin (e+f x)),2\right )\right ) \sqrt {a \sin (e+f x)} \tan ^2(e+f x)}{616 f \sqrt [4]{\cos ^2(e+f x)} (b \tan (e+f x))^{3/2}} \] Input:

Integrate[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(a^5*((Cos[e + f*x]^2)^(1/4)*(-22*Cos[e + f*x] - 17*Cos[3*(e + f*x)] + 7*C 
os[5*(e + f*x)]) + 64*Cot[e + f*x]*EllipticF[ArcSin[Sin[e + f*x]]/2, 2])*S 
qrt[a*Sin[e + f*x]]*Tan[e + f*x]^2)/(616*f*(Cos[e + f*x]^2)^(1/4)*(b*Tan[e 
 + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3076, 3042, 3078, 3042, 3078, 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.

\(\displaystyle \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3076

\(\displaystyle \frac {a^2 \int (a \sin (e+f x))^{7/2} \sqrt {b \tan (e+f x)}dx}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \int (a \sin (e+f x))^{7/2} \sqrt {b \tan (e+f x)}dx}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \int (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}dx-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \int (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}dx-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \left (\frac {2}{3} a^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{3/2}}{3 f \sqrt {b \tan (e+f x)}}\right )-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \left (\frac {2}{3} a^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {2 b (a \sin (e+f x))^{3/2}}{3 f \sqrt {b \tan (e+f x)}}\right )-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \left (\frac {2 a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 \sqrt {a \sin (e+f x)}}-\frac {2 b (a \sin (e+f x))^{3/2}}{3 f \sqrt {b \tan (e+f x)}}\right )-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \left (\frac {2 a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {a \sin (e+f x)}}-\frac {2 b (a \sin (e+f x))^{3/2}}{3 f \sqrt {b \tan (e+f x)}}\right )-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a^2 \left (\frac {6}{7} a^2 \left (\frac {4 a^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{3 f \sqrt {a \sin (e+f x)}}-\frac {2 b (a \sin (e+f x))^{3/2}}{3 f \sqrt {b \tan (e+f x)}}\right )-\frac {2 b (a \sin (e+f x))^{7/2}}{7 f \sqrt {b \tan (e+f x)}}\right )}{11 b^2}+\frac {2 (a \sin (e+f x))^{11/2}}{11 b f \sqrt {b \tan (e+f x)}}\)

Input:

Int[(a*Sin[e + f*x])^(11/2)/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(2*(a*Sin[e + f*x])^(11/2))/(11*b*f*Sqrt[b*Tan[e + f*x]]) + (a^2*((-2*b*(a 
*Sin[e + f*x])^(7/2))/(7*f*Sqrt[b*Tan[e + f*x]]) + (6*a^2*((-2*b*(a*Sin[e 
+ f*x])^(3/2))/(3*f*Sqrt[b*Tan[e + f*x]]) + (4*a^2*Sqrt[Cos[e + f*x]]*Elli 
pticF[(e + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(3*f*Sqrt[a*Sin[e + f*x]])))/7 
))/(11*b^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3076

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)) 
, x] - Simp[a^2*((n + 1)/(b^2*m))   Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[e + 
 f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && GtQ[m, 1] 
 && IntegersQ[2*m, 2*n]
 

rule 3078

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]
 

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 = 9.55 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {2 \sqrt {a \sin \left (f x +e \right )}\, a^{5} \left (\sin \left (f x +e \right ) \left (7 \cos \left (f x +e \right )^{4}-13 \cos \left (f x +e \right )^{2}+4\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (-4-4 \sec \left (f x +e \right )\right )\right )}{77 f \sqrt {b \tan \left (f x +e \right )}\, b}\)	\(127\)

Input:

int((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/77/f*(a*sin(f*x+e))^(1/2)*a^5/(b*tan(f*x+e))^(1/2)/b*(sin(f*x+e)*(7*cos( 
f*x+e)^4-13*cos(f*x+e)^2+4)+I*(1/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f* 
x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-4-4*sec(f*x+e)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {2 \, {\left (4 \, \sqrt {\frac {1}{2}} \sqrt {-a b} a^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 4 \, \sqrt {\frac {1}{2}} \sqrt {-a b} a^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (7 \, a^{5} \cos \left (f x + e\right )^{5} - 13 \, a^{5} \cos \left (f x + e\right )^{3} + 4 \, a^{5} \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}\right )}}{77 \, b^{2} f} \] Input:

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas" 
)
 

Output:

2/77*(4*sqrt(1/2)*sqrt(-a*b)*a^5*weierstrassPInverse(-4, 0, cos(f*x + e) + 
 I*sin(f*x + e)) + 4*sqrt(1/2)*sqrt(-a*b)*a^5*weierstrassPInverse(-4, 0, c 
os(f*x + e) - I*sin(f*x + e)) + (7*a^5*cos(f*x + e)^5 - 13*a^5*cos(f*x + e 
)^3 + 4*a^5*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x 
 + e)))/(b^2*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a*sin(f*x+e))**(11/2)/(b*tan(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {11}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima" 
)
 

Output:

integrate((a*sin(f*x + e))^(11/2)/(b*tan(f*x + e))^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a*sin(f*x+e))^(11/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{11/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a*sin(e + f*x))^(11/2)/(b*tan(e + f*x))^(3/2),x)
 

Output:

int((a*sin(e + f*x))^(11/2)/(b*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a \sin (e+f x))^{11/2}}{(b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}\, \sin \left (f x +e \right )^{5}}{\tan \left (f x +e \right )^{2}}d x \right ) a^{5}}{b^{2}} \] Input:

int((a*sin(f*x+e))^(11/2)/(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)** 
5)/tan(e + f*x)**2,x)*a**5)/b**2