Integrand size = 23, antiderivative size = 78 \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {a \sqrt {a+b \sin ^2(e+f x)}}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f} \] Output:
-a^(3/2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/f+a*(a+b*sin(f*x+e)^2)^ (1/2)/f+1/3*(a+b*sin(f*x+e)^2)^(3/2)/f
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)\right )}{3 f} \] Input:
Integrate[Cot[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
(-3*a^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + Sqrt[a + b*Sin[e + f*x]^2]*(4*a + b*Sin[e + f*x]^2))/(3*f)
Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3673, 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.
\(\displaystyle \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^{3/2}}{\tan (e+f x)}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \csc ^2(e+f x) \left (b \sin ^2(e+f x)+a\right )^{3/2}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \int \csc ^2(e+f x) \sqrt {b \sin ^2(e+f x)+a}d\sin ^2(e+f x)+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (a \int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a \left (\frac {2 a \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sin ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {a \left (2 \sqrt {a+b \sin ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 f}\) |
Input:
Int[Cot[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
((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*f)
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_)^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 = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10
method | result | size |
default | \(\frac {-a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )+\frac {b \sin \left (f x +e \right )^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}{3}+\frac {4 a \sqrt {a +b \sin \left (f x +e \right )^{2}}}{3}}{f}\) | \(86\) |
Input:
int(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
(-a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))+1/3*b*si n(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))/f
Time = 0.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.44 \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, a^{\frac {3}{2}} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{2} - 4 \, a - b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f}, -\frac {3 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + {\left (b \cos \left (f x + e\right )^{2} - 4 \, a - b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, f}\right ] \] Input:
integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
[1/6*(3*a^(3/2)*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*(b*cos(f*x + e)^2 - 4*a - b )*sqrt(-b*cos(f*x + e)^2 + a + b))/f, -1/3*(3*sqrt(-a)*a*arctan(sqrt(-b*co s(f*x + e)^2 + a + b)*sqrt(-a)/(b*cos(f*x + e)^2 - a - b)) + (b*cos(f*x + e)^2 - 4*a - b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f]
\[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot {\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)*(a+b*sin(f*x+e)**2)**(3/2),x)
Output:
Integral((a + b*sin(e + f*x)**2)**(3/2)*cot(e + f*x), x)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 \, a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a}{3 \, f} \] Input:
integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
-1/3*(3*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - (b*sin(f*x + e) ^2 + a)^(3/2) - 3*sqrt(b*sin(f*x + e)^2 + a)*a)/f
Timed out. \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \mathrm {cot}\left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \] Input:
int(cot(e + f*x)*(a + b*sin(e + f*x)^2)^(3/2),x)
Output:
int(cot(e + f*x)*(a + b*sin(e + f*x)^2)^(3/2), x)
\[ \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right ) \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )d x \right ) a \] Input:
int(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x)
Output:
int(sqrt(sin(e + f*x)**2*b + a)*cot(e + f*x)*sin(e + f*x)**2,x)*b + int(sq rt(sin(e + f*x)**2*b + a)*cot(e + f*x),x)*a