Integrand size = 25, antiderivative size = 110 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}} \]
1/2*(2*a+3*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+1/2*(-2* a-3*b)/a^2/f/(a+b*sin(f*x+e)^2)^(1/2)-1/2*csc(f*x+e)^2/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 = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-a \csc ^2(e+f x)-(2 a+3 b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
(-(a*Csc[e + f*x]^2) - (2*a + 3*b)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b* Sin[e + f*x]^2)/a])/(2*a^2*f*Sqrt[a + b*Sin[e + f*x]^2])
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3673, 87, 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.
\(\displaystyle \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\csc ^4(e+f x) \left (1-\sin ^2(e+f x)\right )}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {-\frac {(2 a+3 b) \int \frac {\csc ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {-\frac {(2 a+3 b) \left (\frac {\int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {-\frac {(2 a+3 b) \left (\frac {2 \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {(2 a+3 b) \left (\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\) |
(-(Csc[e + f*x]^2/(a*Sqrt[a + b*Sin[e + f*x]^2])) - ((2*a + 3*b)*((-2*ArcT anh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sin[e + f*x]^2])))/(2*a))/(2*f)
3.6.25.3.1 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 + 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_))*((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 = 1.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {-\frac {1}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}-\frac {1}{2 a \sin \left (f x +e \right )^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {3 b}{2 a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2 a^{\frac {5}{2}}}}{f}\) | \(148\) |
(-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))-1/2/a/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)-3/2/a^2 *b/(a+b*sin(f*x+e)^2)^(1/2)+3/2/a^(5/2)*b*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. 190 vs. \(2 (94) = 188\).
Time = 0.38 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.69 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + a^{3} b\right )} f\right )}}, -\frac {{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a^{3} b f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + a^{3} b\right )} f\right )}}\right ] \]
[1/4*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2)*cos(f*x + e)^2 + 2*a^2 + 5*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*((2*a ^2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b ))/(a^3*b*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b)*f*cos(f*x + e)^2 + (a^4 + a^3 *b)*f), -1/2*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2)*co s(f*x + e)^2 + 2*a^2 + 5*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e) ^2 + a + b)*sqrt(-a)/a) - ((2*a^2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b) *sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b )*f*cos(f*x + e)^2 + (a^4 + a^3*b)*f)]
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {2 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {3 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}}}{2 \, f} \]
1/2*(2*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 3*b*arcsinh(a/(s qrt(a*b)*abs(sin(f*x + e))))/a^(5/2) - 2/(sqrt(b*sin(f*x + e)^2 + a)*a) - 3*b/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - 1/(sqrt(b*sin(f*x + e)^2 + a)*a*sin (f*x + e)^2))/f
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]