Integrand size = 25, antiderivative size = 106 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {2 b}{9 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{9/2}}-\frac {16 b}{45 a^3 f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2}}-\frac {64 b}{45 a^5 f \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)}} \] Output:
-2/9*b/a/f/(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(9/2)-16/45*b/a^3/f/(b*sec( f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2)-64/45*b/a^5/f/(b*sec(f*x+e))^(1/2)/(a*s in(f*x+e))^(1/2)
Time = 1.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\frac {2 b (-21+20 \cos (2 (e+f x))-4 \cos (4 (e+f x))) \csc ^5(e+f x) \sqrt {a \sin (e+f x)}}{45 a^6 f \sqrt {b \sec (e+f x)}} \] Input:
Integrate[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(11/2),x]
Output:
(2*b*(-21 + 20*Cos[2*(e + f*x)] - 4*Cos[4*(e + f*x)])*Csc[e + f*x]^5*Sqrt[ a*Sin[e + f*x]])/(45*a^6*f*Sqrt[b*Sec[e + f*x]])
Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3064, 3042, 3064, 3042, 3058}
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.
\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}}dx\) |
\(\Big \downarrow \) 3064 |
\(\displaystyle \frac {8 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{7/2}}dx}{9 a^2}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{7/2}}dx}{9 a^2}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3064 |
\(\displaystyle \frac {8 \left (\frac {4 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{5 a^2}-\frac {2 b}{5 a f (a \sin (e+f x))^{5/2} \sqrt {b \sec (e+f x)}}\right )}{9 a^2}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {4 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}}dx}{5 a^2}-\frac {2 b}{5 a f (a \sin (e+f x))^{5/2} \sqrt {b \sec (e+f x)}}\right )}{9 a^2}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3058 |
\(\displaystyle \frac {8 \left (-\frac {8 b}{5 a^3 f \sqrt {a \sin (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b}{5 a f (a \sin (e+f x))^{5/2} \sqrt {b \sec (e+f x)}}\right )}{9 a^2}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}}\) |
Input:
Int[Sqrt[b*Sec[e + f*x]]/(a*Sin[e + f*x])^(11/2),x]
Output:
(-2*b)/(9*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(9/2)) + (8*((-2*b)/(5 *a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(5/2)) - (8*b)/(5*a^3*f*Sqrt[b* Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])))/(9*a^2)
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] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2, 0] & & NeQ[m, -1]
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]
Time = 0.55 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61
method | result | size |
default | \(-\frac {2 \left (32 \cos \left (f x +e \right )^{4}-72 \cos \left (f x +e \right )^{2}+45\right ) \sqrt {b \sec \left (f x +e \right )}\, \cot \left (f x +e \right ) \csc \left (f x +e \right )^{3}}{45 f \sqrt {a \sin \left (f x +e \right )}\, a^{5}}\) | \(65\) |
Input:
int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)
Output:
-2/45/f*(32*cos(f*x+e)^4-72*cos(f*x+e)^2+45)*(b*sec(f*x+e))^(1/2)/(a*sin(f *x+e))^(1/2)/a^5*cot(f*x+e)*csc(f*x+e)^3
Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {2 \, {\left (32 \, \cos \left (f x + e\right )^{5} - 72 \, \cos \left (f x + e\right )^{3} + 45 \, \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, {\left (a^{6} f \cos \left (f x + e\right )^{4} - 2 \, a^{6} f \cos \left (f x + e\right )^{2} + a^{6} f\right )} \sin \left (f x + e\right )} \] Input:
integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="fricas" )
Output:
-2/45*(32*cos(f*x + e)^5 - 72*cos(f*x + e)^3 + 45*cos(f*x + e))*sqrt(a*sin (f*x + e))*sqrt(b/cos(f*x + e))/((a^6*f*cos(f*x + e)^4 - 2*a^6*f*cos(f*x + e)^2 + a^6*f)*sin(f*x + e))
Timed out. \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(11/2),x)
Output:
Timed out
\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\int { \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {11}{2}}} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="maxima" )
Output:
integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(11/2), x)
\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\int { \frac {\sqrt {b \sec \left (f x + e\right )}}{\left (a \sin \left (f x + e\right )\right )^{\frac {11}{2}}} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="giac")
Output:
integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(11/2), x)
Time = 32.71 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {{\mathrm {e}}^{-e\,5{}\mathrm {i}-f\,x\,5{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {352\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}}{45\,a^5\,f}-\frac {256\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )}{45\,a^5\,f}+\frac {64\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )}{45\,a^5\,f}\right )}{16\,{\sin \left (e+f\,x\right )}^4\,\sqrt {a\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int((b/cos(e + f*x))^(1/2)/(a*sin(e + f*x))^(11/2),x)
Output:
-(exp(- e*5i - f*x*5i)*(b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)) ^(1/2)*((352*cos(e + f*x)*exp(e*5i + f*x*5i))/(45*a^5*f) - (256*exp(e*5i + f*x*5i)*cos(3*e + 3*f*x))/(45*a^5*f) + (64*exp(e*5i + f*x*5i)*cos(5*e + 5 *f*x))/(45*a^5*f)))/(16*sin(e + f*x)^4*(a*((exp(- e*1i - f*x*1i)*1i)/2 - ( exp(e*1i + f*x*1i)*1i)/2))^(1/2))
\[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sin \left (f x +e \right )^{6}}d x \right )}{a^{6}} \] Input:
int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x)
Output:
(sqrt(b)*sqrt(a)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/sin(e + f*x)* *6,x))/a**6