Integrand size = 23, antiderivative size = 57 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {1}{a f \sqrt {a+b \sin ^2(e+f x)}} \] Output:
-arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(3/2)/f+1/a/f/(a+b*sin(f*x+e) ^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{a f \sqrt {a+b \sin ^2(e+f x)}} \] Input:
Integrate[Cot[e + f*x]/(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sin[e + f*x]^2)/a]/(a*f*Sqrt[a + b* Sin[e + f*x]^2])
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3673, 61, 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.
\(\displaystyle \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x) \left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\csc ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}}{2 f}\) |
Input:
Int[Cot[e + f*x]/(a + b*Sin[e + f*x]^2)^(3/2),x]
Output:
((-2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {\frac {1}{a \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}}{f}\) | \(62\) |
Input:
int(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
(1/a/(a+b*sin(f*x+e)^2)^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^ 2)^(1/2))/sin(f*x+e)))/f
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (49) = 98\).
Time = 0.15 (sec) , antiderivative size = 241, normalized size of antiderivative = 4.23 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{2 \, {\left (a^{2} b f \cos \left (f x + e\right )^{2} - {\left (a^{3} + a^{2} b\right )} f\right )}}, -\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{a^{2} b f \cos \left (f x + e\right )^{2} - {\left (a^{3} + a^{2} b\right )} f}\right ] \] Input:
integrate(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
[1/2*((b*cos(f*x + e)^2 - a - 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*sq rt(-b*cos(f*x + e)^2 + a + b)*a)/(a^2*b*f*cos(f*x + e)^2 - (a^3 + a^2*b)*f ), -((b*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(b*cos(f*x + e)^2 - a - b)) + sqrt(-b*cos(f*x + e)^2 + a + b)*a)/(a^2*b*f*cos(f*x + e)^2 - (a^3 + a^2*b)*f)]
\[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cot(f*x+e)/(a+b*sin(f*x+e)**2)**(3/2),x)
Output:
Integral(cot(e + f*x)/(a + b*sin(e + f*x)**2)**(3/2), x)
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a}}{f} \] Input:
integrate(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
-(arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) - 1/(sqrt(b*sin(f*x + e )^2 + a)*a))/f
\[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cot \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
integrate(cot(f*x + e)/(b*sin(f*x + e)^2 + a)^(3/2), x)
Timed out. \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\mathrm {cot}\left (e+f\,x\right )}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] 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)
\[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] 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)**4*b**2 + 2*s in(e + f*x)**2*a*b + a**2),x)