\(\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx\) [455]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {2 b}{9 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{9/2}}-\frac {16 b}{45 a^3 f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2}}-\frac {64 b}{45 a^5 f \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)}} \]

[Out]

-2/9*b/a/f/(a*sin(f*x+e))^(9/2)/(b*sec(f*x+e))^(1/2)-16/45*b/a^3/f/(a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(1/2)-6
4/45*b/a^5/f/(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2664, 2658} \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {64 b}{45 a^5 f \sqrt {a \sin (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {16 b}{45 a^3 f (a \sin (e+f x))^{5/2} \sqrt {b \sec (e+f x)}}-\frac {2 b}{9 a f (a \sin (e+f x))^{9/2} \sqrt {b \sec (e+f x)}} \]

[In]

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

[Out]

(-2*b)/(9*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(9/2)) - (16*b)/(45*a^3*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e +
f*x])^(5/2)) - (64*b)/(45*a^5*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])

Rule 2658

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]

Rule 2664

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b}{9 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{9/2}}+\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 \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 {32 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx}{45 a^4} \\ & = -\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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=\frac {2 b (-21+20 \cos (2 (e+f x))-4 \cos (4 (e+f x))) \csc ^5(e+f x) \sqrt {a \sin (e+f x)}}{45 a^6 f \sqrt {b \sec (e+f x)}} \]

[In]

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

[Out]

(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*Se
c[e + f*x]])

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {2 \sqrt {b \sec \left (f x +e \right )}\, \left (32 \left (\cos ^{4}\left (f x +e \right )\right )-72 \left (\cos ^{2}\left (f x +e \right )\right )+45\right ) \cot \left (f x +e \right ) \left (\csc ^{3}\left (f x +e \right )\right )}{45 f \sqrt {a \sin \left (f x +e \right )}\, a^{5}}\)	\(65\)

[In]

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/45/f*(b*sec(f*x+e))^(1/2)*(32*cos(f*x+e)^4-72*cos(f*x+e)^2+45)/(a*sin(f*x+e))^(1/2)/a^5*cot(f*x+e)*csc(f*x+
e)^3

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {2 \, {\left (32 \, \cos \left (f x + e\right )^{5} - 72 \, \cos \left (f x + e\right )^{3} + 45 \, \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, {\left (a^{6} f \cos \left (f x + e\right )^{4} - 2 \, a^{6} f \cos \left (f x + e\right )^{2} + a^{6} f\right )} \sin \left (f x + e\right )} \]

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-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))

Sympy [F(-1)]

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

[In]

integrate((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{11/2}} \, dx=-\frac {{\mathrm {e}}^{-e\,5{}\mathrm {i}-f\,x\,5{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {352\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}}{45\,a^5\,f}-\frac {256\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )}{45\,a^5\,f}+\frac {64\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )}{45\,a^5\,f}\right )}{16\,{\sin \left (e+f\,x\right )}^4\,\sqrt {a\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int((b/cos(e + f*x))^(1/2)/(a*sin(e + f*x))^(11/2),x)

[Out]

-(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))