Integrand size = 25, antiderivative size = 140 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \]
-1/6*(2*a-3*b)*(a+b*sin(f*x+e)^2)^(3/2)/a/f-1/2*csc(f*x+e)^2*(a+b*sin(f*x+ e)^2)^(5/2)/a/f+1/2*(2*a-3*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))*a^ (1/2)/f-1/2*(2*a-3*b)*(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\left (-8 a+5 b+b \cos (2 (e+f x))-3 a \csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{6 f} \]
(3*Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + (-8*a + 5*b + b*Cos[2*(e + f*x)] - 3*a*Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^ 2])/(6*f)
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 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.
| \(\displaystyle \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\tan (e+f x)^3}dx\) |
| \(\Big \downarrow \) 3673 |
| \(\displaystyle \frac {\int \csc ^4(e+f x) \left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {-\frac {(2 a-3 b) \int \csc ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {-\frac {(2 a-3 b) \left (a \int \csc ^2(e+f x) \sqrt {b \sin ^2(e+f x)+a}d\sin ^2(e+f x)+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {-\frac {(2 a-3 b) \left (a \left (a \int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {-\frac {(2 a-3 b) \left (a \left (\frac {2 a \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {(2 a-3 b) \left (a \left (2 \sqrt {a+b \sin ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
(-((Csc[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2))/a) - ((2*a - 3*b)*((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))/(2*f)
3.6.3.3.1 Defintions of rubi rules used
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)^(-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 = 1.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.18
| method | result | size |
| default | \(\frac {-\frac {b \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}-\frac {4 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}+b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\frac {a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 \sin \left (f x +e \right )^{2}}-\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2}+a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\) | \(165\) |
(-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)+b*(a+b*sin(f*x+e)^2)^(1/2)-1/2*a/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)- 3/2*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 = 0.97 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.01 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}, -\frac {3 \, {\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \]
[-1/12*(3*((2*a - 3*b)*cos(f*x + e)^2 - 2*a + 3*b)*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*(2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e)^2 + 11*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - f), -1/6*(3*((2* a - 3*b)*cos(f*x + e)^2 - 2*a + 3*b)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^ 2 + a + b)*sqrt(-a)/a) - (2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e)^2 + 11*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - f)]
\[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{3}{\left (e + f x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {6 \, a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - 9 \, \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a + 9 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b + \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a} - \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \sin \left (f x + e\right )^{2}}}{6 \, f} \]
1/6*(6*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - 9*sqrt(a)*b*arcs inh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - 2*(b*sin(f*x + e)^2 + a)^(3/2) - 6* sqrt(b*sin(f*x + e)^2 + a)*a + 9*sqrt(b*sin(f*x + e)^2 + a)*b + 3*(b*sin(f *x + e)^2 + a)^(3/2)*b/a - 3*(b*sin(f*x + e)^2 + a)^(5/2)/(a*sin(f*x + e)^ 2))/f
Timed out. \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]