Integrand size = 25, antiderivative size = 174 \[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=-\frac {2}{11 a b f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac {2}{77 a^3 b f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac {4}{77 a^5 b f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac {4 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{77 a^6 b^2 f \sqrt {a \sin (e+f x)}} \] Output:
-2/11/a/b/f/(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2)+2/77/a^3/b/f/(b*sec (f*x+e))^(1/2)/(a*sin(f*x+e))^(7/2)+4/77/a^5/b/f/(b*sec(f*x+e))^(1/2)/(a*s in(f*x+e))^(3/2)-4/77*InverseJacobiAM(e-1/4*Pi+f*x,2^(1/2))*(b*sec(f*x+e)) ^(1/2)*sin(2*f*x+2*e)^(1/2)/a^6/b^2/f/(a*sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.47 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\frac {2 \cot (2 (e+f x)) \csc (2 (e+f x)) \sqrt {a \sin (e+f x)} \left ((23+6 \cos (2 (e+f x))-\cos (4 (e+f x))) \csc ^4(e+f x)+8 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{77 a^7 b f \sqrt {b \sec (e+f x)} \left (-2+\sec ^2(e+f x)\right )} \] Input:
Integrate[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(13/2)),x]
Output:
(2*Cot[2*(e + f*x)]*Csc[2*(e + f*x)]*Sqrt[a*Sin[e + f*x]]*((23 + 6*Cos[2*( e + f*x)] - Cos[4*(e + f*x)])*Csc[e + f*x]^4 + 8*Hypergeometric2F1[1/2, 3/ 4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(77*a^7*b*f*Sqrt[b*Sec[e + f*x]]*(-2 + Sec[e + f*x]^2))
Time = 1.45 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3061, 3042, 3064, 3042, 3064, 3042, 3065, 3042, 3053, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x))^{13/2} (b \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x))^{13/2} (b \sec (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3061 |
| \(\displaystyle -\frac {\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{9/2}}dx}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{9/2}}dx}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3064 |
| \(\displaystyle -\frac {\frac {6 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}}dx}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3064 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3065 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{3 a^2}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3053 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{3 a^2 \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{3 a^2 \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {6 \left (\frac {2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}}\right )}{7 a^2}-\frac {2 b}{7 a f (a \sin (e+f x))^{7/2} \sqrt {b \sec (e+f x)}}}{11 a^2 b^2}-\frac {2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt {b \sec (e+f x)}}\) |
Input:
Int[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(13/2)),x]
Output:
-2/(11*a*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(11/2)) - ((-2*b)/(7*a* f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(7/2)) + (6*((-2*b)/(3*a*f*Sqrt[b* Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2)) + (2*EllipticF[e - Pi/4 + f*x, 2]*Sq rt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(3*a^2*f*Sqrt[a*Sin[e + f*x]])) )/(7*a^2))/(11*a^2*b^2)
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n + 1)/(a *b*f*(m + 1))), x] - Simp[(n + 1)/(a^2*b^2*(m + 1)) Int[(a*Sin[e + f*x])^ (m + 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n , -1] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/ (a*f*(m + 1))), x] + Simp[(m - n + 2)/(a^2*(m + 1)) Int[(a*Sin[e + f*x])^ (m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, - 1] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int egerQ[m - 1/2] && IntegerQ[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]
Time = 1.15 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89
| method | result | size |
| default | \(-\frac {2 \left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (2+2 \sec \left (f x +e \right )\right )+\left (-2 \cos \left (f x +e \right )^{4}+5 \cos \left (f x +e \right )^{2}+4\right ) \csc \left (f x +e \right )^{5}\right )}{77 f \sqrt {a \sin \left (f x +e \right )}\, \sqrt {b \sec \left (f x +e \right )}\, a^{6} b}\) | \(154\) |
Input:
int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x,method=_RETURNVERBOSE)
Output:
-2/77/f/(a*sin(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/2)/a^6/b*((csc(f*x+e)-cot(f *x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e ))^(1/2)*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*(2+2*sec(f *x+e))+(-2*cos(f*x+e)^4+5*cos(f*x+e)^2+4)*csc(f*x+e)^5)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\frac {2 \, {\left (2 \, {\left (\cos \left (f x + e\right )^{6} - 3 \, \cos \left (f x + e\right )^{4} + 3 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt {i \, a b} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + 2 \, {\left (\cos \left (f x + e\right )^{6} - 3 \, \cos \left (f x + e\right )^{4} + 3 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-i \, a b} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - {\left (2 \, \cos \left (f x + e\right )^{5} - 5 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{77 \, {\left (a^{7} b^{2} f \cos \left (f x + e\right )^{6} - 3 \, a^{7} b^{2} f \cos \left (f x + e\right )^{4} + 3 \, a^{7} b^{2} f \cos \left (f x + e\right )^{2} - a^{7} b^{2} f\right )}} \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="frica s")
Output:
2/77*(2*(cos(f*x + e)^6 - 3*cos(f*x + e)^4 + 3*cos(f*x + e)^2 - 1)*sqrt(I* a*b)*elliptic_f(arcsin(cos(f*x + e) + I*sin(f*x + e)), -1) + 2*(cos(f*x + e)^6 - 3*cos(f*x + e)^4 + 3*cos(f*x + e)^2 - 1)*sqrt(-I*a*b)*elliptic_f(ar csin(cos(f*x + e) - I*sin(f*x + e)), -1) - (2*cos(f*x + e)^5 - 5*cos(f*x + e)^3 - 4*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b/cos(f*x + e)))/(a^7*b^ 2*f*cos(f*x + e)^6 - 3*a^7*b^2*f*cos(f*x + e)^4 + 3*a^7*b^2*f*cos(f*x + e) ^2 - a^7*b^2*f)
Timed out. \[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*sec(f*x+e))**(3/2)/(a*sin(f*x+e))**(13/2),x)
Output:
Timed out
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \left (a \sin \left (f x + e\right )\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="maxim a")
Output:
integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(13/2)), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \left (a \sin \left (f x + e\right )\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="giac" )
Output:
integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(13/2)), x)
Timed out. \[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{13/2}\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/((a*sin(e + f*x))^(13/2)*(b/cos(e + f*x))^(3/2)),x)
Output:
int(1/((a*sin(e + f*x))^(13/2)*(b/cos(e + f*x))^(3/2)), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{7}}d x \right )}{a^{7} b^{2}} \] Input:
int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x)
Output:
(sqrt(b)*sqrt(a)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/(sec(e + f*x) **2*sin(e + f*x)**7),x))/(a**7*b**2)