Integrand size = 25, antiderivative size = 126 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{5/2} f}+\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 a f} \]
-1/8*(8*a^2+8*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2) /f+1/8*(8*a+3*b)*csc(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)/a^2/f-1/4*csc(f*x+e )^4*(a+b*sin(f*x+e)^2)^(1/2)/a/f
Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-\left (\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )\right )+\sqrt {a} \csc ^2(e+f x) \left (8 a+3 b-2 a \csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^{5/2} f} \]
(-((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]]) + Sqrt[a]*Csc[e + f*x]^2*(8*a + 3*b - 2*a*Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*a^(5/2)*f)
Time = 0.31 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 100, 27, 87, 73, 221}
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.
\(\displaystyle \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^5 \sqrt {a+b \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\csc ^6(e+f x) \left (1-\sin ^2(e+f x)\right )^2}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\frac {\int -\frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+3 b\right )}{2 \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\int \frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+3 b\right )}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{2 a}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{4 a}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a}}{2 f}\) |
(-1/2*(Csc[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/a - (((8*a^2 + 8*a*b + 3 *b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/a^(3/2) - ((8*a + 3*b)* Csc[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/a)/(4*a))/(2*f)
3.6.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Time = 1.12 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.63
method | result | size |
default | \(\frac {-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{\sqrt {a}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 a \sin \left (f x +e \right )^{4}}+\frac {3 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 a^{2} \sin \left (f x +e \right )^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 a^{\frac {5}{2}}}+\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{a \sin \left (f x +e \right )^{2}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}}{f}\) | \(205\) |
(-1/a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-1/4/a/ sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)+3/8/a^2*b/sin(f*x+e)^2*(a+b*sin(f*x+ e)^2)^(1/2)-3/8/a^(5/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/si n(f*x+e))+1/a/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)-b/a^(3/2)*ln((2*a+2*a^ (1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)))/f
Time = 0.39 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.08 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \]
[1/16*(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2) *cos(f*x + e)^2 + 8*a^2 + 8*a*b + 3*b^2)*sqrt(a)*log(2*(b*cos(f*x + e)^2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1) ) - 2*((8*a^2 + 3*a*b)*cos(f*x + e)^2 - 6*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e )^2 + a + b))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/8 *(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2)*cos( f*x + e)^2 + 8*a^2 + 8*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((8*a^2 + 3*a*b)*cos(f*x + e)^2 - 6*a^2 - 3*a*b)*s qrt(-b*cos(f*x + e)^2 + a + b))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)]
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\cot ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.25 \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {8 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} + \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}} - \frac {3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \]
-1/8*(8*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/sqrt(a) + 8*b*arcsinh(a/( sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 3*b^2*arcsinh(a/(sqrt(a*b)*abs(sin (f*x + e))))/a^(5/2) - 8*sqrt(b*sin(f*x + e)^2 + a)/(a*sin(f*x + e)^2) - 3 *sqrt(b*sin(f*x + e)^2 + a)*b/(a^2*sin(f*x + e)^2) + 2*sqrt(b*sin(f*x + e) ^2 + a)/(a*sin(f*x + e)^4))/f
\[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{5}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^5}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]