Integrand size = 25, antiderivative size = 208 \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\left (8 a^2-24 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 \sqrt {a} f}+\frac {\left (8 a^2-24 a b+3 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a f}+\frac {\left (8 a^2-24 a b+3 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 a^2 f}+\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 a f} \]
1/24*(8*a^2-24*a*b+3*b^2)*(a+b*sin(f*x+e)^2)^(3/2)/a^2/f+1/8*(8*a-b)*csc(f *x+e)^2*(a+b*sin(f*x+e)^2)^(5/2)/a^2/f-1/4*csc(f*x+e)^4*(a+b*sin(f*x+e)^2) ^(5/2)/a/f-1/8*(8*a^2-24*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/ 2))/f/a^(1/2)+1/8*(8*a^2-24*a*b+3*b^2)*(a+b*sin(f*x+e)^2)^(1/2)/a/f
Time = 0.84 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.59 \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-3 \left (8 a^2-24 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b \sin ^2(e+f x)} \left (3 (8 a-5 b) \csc ^2(e+f x)-6 a \csc ^4(e+f x)+8 \left (4 a-6 b+b \sin ^2(e+f x)\right )\right )}{24 \sqrt {a} f} \]
(-3*(8*a^2 - 24*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + Sqrt[a]*Sqrt[a + b*Sin[e + f*x]^2]*(3*(8*a - 5*b)*Csc[e + f*x]^2 - 6*a*Cs c[e + f*x]^4 + 8*(4*a - 6*b + b*Sin[e + f*x]^2)))/(24*Sqrt[a]*f)
Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3673, 100, 27, 87, 60, 60, 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.
| \(\displaystyle \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\tan (e+f x)^5}dx\) |
| \(\Big \downarrow \) 3673 |
| \(\displaystyle \frac {\int \csc ^6(e+f x) \left (1-\sin ^2(e+f x)\right )^2 \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {\frac {\int -\frac {1}{2} \csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a-b\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a-b\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {-\frac {-\frac {\left (8 a^2-24 a b+3 b^2\right ) \int \csc ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 a}-\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {-\frac {-\frac {\left (8 a^2-24 a b+3 b^2\right ) \left (a \int \csc ^2(e+f x) \sqrt {b \sin ^2(e+f x)+a}d\sin ^2(e+f x)+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {-\frac {-\frac {\left (8 a^2-24 a b+3 b^2\right ) \left (a \left (a \int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {-\frac {-\frac {\left (8 a^2-24 a b+3 b^2\right ) \left (a \left (\frac {2 a \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {-\frac {\left (8 a^2-24 a b+3 b^2\right ) \left (a \left (2 \sqrt {a+b \sin ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {(8 a-b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{4 a}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a}}{2 f}\) |
(-1/2*(Csc[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(5/2))/a - (-(((8*a - b)*Csc[ e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2))/a) - ((8*a^2 - 24*a*b + 3*b^2)*(( 2*(a + b*Sin[e + f*x]^2)^(3/2))/3 + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + b*Sin[e + f*x]^2])))/(2*a))/(4*a))/(2*f)
3.6.4.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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 = 1.28 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.24
| method | result | size |
| default | \(\frac {\frac {b \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}+\frac {4 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}-\frac {a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 \sin \left (f x +e \right )^{4}}-\frac {5 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 \sin \left (f x +e \right )^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 \sqrt {a}}-2 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}+\frac {a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}+3 \sqrt {a}\, b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )-a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\) | \(257\) |
(1/3*b*sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)+4/3*a*(a+b*sin(f*x+e)^2)^(1/2 )-1/4*a/sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)-5/8*b/sin(f*x+e)^2*(a+b*sin( f*x+e)^2)^(1/2)-3/8/a^(1/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2) )/sin(f*x+e))-2*b*(a+b*sin(f*x+e)^2)^(1/2)+a/sin(f*x+e)^2*(a+b*sin(f*x+e)^ 2)^(1/2)+3*a^(1/2)*b*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e ))-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 = 1.35 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.12 \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left ({\left (8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (8 \, a b \cos \left (f x + e\right )^{6} - 8 \, {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (88 \, a^{2} - 87 \, a b\right )} \cos \left (f x + e\right )^{2} - 50 \, a^{2} + 55 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{48 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}, \frac {3 \, {\left ({\left (8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 24 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (8 \, a b \cos \left (f x + e\right )^{6} - 8 \, {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{4} + {\left (88 \, a^{2} - 87 \, a b\right )} \cos \left (f x + e\right )^{2} - 50 \, a^{2} + 55 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{24 \, {\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f\right )}}\right ] \]
[1/48*(3*((8*a^2 - 24*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 24*a*b + 3* b^2)*cos(f*x + e)^2 + 8*a^2 - 24*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*b*cos(f*x + e)^6 - 8*(4*a^2 - 3*a*b)*cos(f*x + e)^4 + (88* a^2 - 87*a*b)*cos(f*x + e)^2 - 50*a^2 + 55*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^2 + a*f), 1/24*(3*((8*a^2 - 24*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 24*a*b + 3*b^2)*cos(f*x + e) ^2 + 8*a^2 - 24*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (8*a*b*cos(f*x + e)^6 - 8*(4*a^2 - 3*a*b)*cos(f*x + e)^4 + (88*a^2 - 87*a*b)*cos(f*x + e)^2 - 50*a^2 + 55*a*b)*sqrt(-b*cos(f*x + e) ^2 + a + b))/(a*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^2 + a*f)]
Timed out. \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.31 \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {24 \, a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - 72 \, \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) + \frac {9 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - 8 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - 24 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a + 72 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b + \frac {24 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a} - \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2}}{a^{2}} - \frac {9 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}}{a} - \frac {24 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \sin \left (f x + e\right )^{2}} + \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {6 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \sin \left (f x + e\right )^{4}}}{24 \, f} \]
-1/24*(24*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - 72*sqrt(a)*b* arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) + 9*b^2*arcsinh(a/(sqrt(a*b)*abs( sin(f*x + e))))/sqrt(a) - 8*(b*sin(f*x + e)^2 + a)^(3/2) - 24*sqrt(b*sin(f *x + e)^2 + a)*a + 72*sqrt(b*sin(f*x + e)^2 + a)*b + 24*(b*sin(f*x + e)^2 + a)^(3/2)*b/a - 3*(b*sin(f*x + e)^2 + a)^(3/2)*b^2/a^2 - 9*sqrt(b*sin(f*x + e)^2 + a)*b^2/a - 24*(b*sin(f*x + e)^2 + a)^(5/2)/(a*sin(f*x + e)^2) + 3*(b*sin(f*x + e)^2 + a)^(5/2)*b/(a^2*sin(f*x + e)^2) + 6*(b*sin(f*x + e)^ 2 + a)^(5/2)/(a*sin(f*x + e)^4))/f
Timed out. \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]