Integrand size = 22, antiderivative size = 94 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {1}{4} \text {arctanh}\left (\frac {2 \tan (x)}{\sqrt {1+5 \tan ^2(x)}}\right )-\frac {\cos (x)}{4 \sqrt {1+5 \tan ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}-\frac {1}{8} \cos (x) \sqrt {1+5 \tan ^2(x)}+\frac {9}{2} \cot (x) \sqrt {1+5 \tan ^2(x)} \]
-1/4*arctanh(2*tan(x)/(1+5*tan(x)^2)^(1/2))-1/4*cos(x)/(1+5*tan(x)^2)^(1/2 )-5/2*cot(x)/(1+5*tan(x)^2)^(1/2)-1/8*cos(x)*(1+5*tan(x)^2)^(1/2)+9/2*cot( x)*(1+5*tan(x)^2)^(1/2)
Time = 0.60 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {(-3+2 \cos (2 x))^{3/2} \left (-1+2 \cot ^2(x) \csc (x)\right ) \sin ^2(x) \left (-2 \text {arcsinh}(2 \sin (x)) \left (4+\csc ^2(x)\right )+\left (-2+164 \csc (x)-3 \csc ^2(x)+16 \csc ^3(x)\right ) \sqrt {1+4 \sin ^2(x)}\right ) \tan (x)}{2 \sqrt {-(3-2 \cos (2 x))^2} \left (5+\cot ^2(x)\right ) (4+4 \cos (2 x)-3 \sin (x)+\sin (3 x)) \sqrt {1+5 \tan ^2(x)}} \]
-1/2*((-3 + 2*Cos[2*x])^(3/2)*(-1 + 2*Cot[x]^2*Csc[x])*Sin[x]^2*(-2*ArcSin h[2*Sin[x]]*(4 + Csc[x]^2) + (-2 + 164*Csc[x] - 3*Csc[x]^2 + 16*Csc[x]^3)* Sqrt[1 + 4*Sin[x]^2])*Tan[x])/(Sqrt[-(3 - 2*Cos[2*x])^2]*(5 + Cot[x]^2)*(4 + 4*Cos[2*x] - 3*Sin[x] + Sin[3*x])*Sqrt[1 + 5*Tan[x]^2])
Time = 0.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4877, 27, 3042, 4147, 245, 208, 4153, 374, 25, 445, 25, 291, 219}
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.
\(\displaystyle \int \frac {\sin (x)-2 \cot ^2(x)}{\left (5 \tan ^2(x)+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)-2 \cot (x)^2}{\left (5 \tan (x)^2+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4877 |
\(\displaystyle \int \frac {\sin (x)}{\left (5 \tan ^2(x)+1\right )^{3/2}}dx+\int -\frac {2 \cot ^2(x)}{\left (5 \tan ^2(x)+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\sin (x)}{\left (5 \tan ^2(x)+1\right )^{3/2}}dx-2 \int \frac {\cot ^2(x)}{\left (5 \tan ^2(x)+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\left (5 \tan (x)^2+1\right )^{3/2}}dx-2 \int \frac {1}{\tan (x)^2 \left (5 \tan (x)^2+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \int \frac {\cos ^2(x)}{\left (5 \sec ^2(x)-4\right )^{3/2}}d\sec (x)-2 \int \frac {1}{\tan (x)^2 \left (5 \tan (x)^2+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -2 \int \frac {1}{\tan (x)^2 \left (5 \tan (x)^2+1\right )^{3/2}}dx+\frac {5}{2} \int \frac {1}{\left (5 \sec ^2(x)-4\right )^{3/2}}d\sec (x)+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle -2 \int \frac {1}{\tan (x)^2 \left (5 \tan (x)^2+1\right )^{3/2}}dx-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -2 \int \frac {\cot ^2(x)}{\left (\tan ^2(x)+1\right ) \left (5 \tan ^2(x)+1\right )^{3/2}}d\tan (x)-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 374 |
\(\displaystyle -2 \left (\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}-\frac {1}{4} \int -\frac {\cot ^2(x) \left (10 \tan ^2(x)+9\right )}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan (x)\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \left (\frac {1}{4} \int \frac {\cot ^2(x) \left (10 \tan ^2(x)+9\right )}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan (x)+\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 445 |
\(\displaystyle -2 \left (\frac {1}{4} \left (-\int -\frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan (x)-9 \sqrt {5 \tan ^2(x)+1} \cot (x)\right )+\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \left (\frac {1}{4} \left (\int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan (x)-9 \sqrt {5 \tan ^2(x)+1} \cot (x)\right )+\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -2 \left (\frac {1}{4} \left (\int \frac {1}{1-\frac {4 \tan ^2(x)}{5 \tan ^2(x)+1}}d\frac {\tan (x)}{\sqrt {5 \tan ^2(x)+1}}-9 \sqrt {5 \tan ^2(x)+1} \cot (x)\right )+\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \left (\frac {1}{4} \left (\frac {1}{2} \text {arctanh}\left (\frac {2 \tan (x)}{\sqrt {5 \tan ^2(x)+1}}\right )-9 \sqrt {5 \tan ^2(x)+1} \cot (x)\right )+\frac {5 \cot (x)}{4 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}\) |
Cos[x]/(4*Sqrt[-4 + 5*Sec[x]^2]) - (5*Sec[x])/(8*Sqrt[-4 + 5*Sec[x]^2]) - 2*((5*Cot[x])/(4*Sqrt[1 + 5*Tan[x]^2]) + (ArcTanh[(2*Tan[x])/Sqrt[1 + 5*Ta n[x]^2]]/2 - 9*Cot[x]*Sqrt[1 + 5*Tan[x]^2])/4)
3.5.36.3.1 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_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] : > With[{e = FreeFactors[Cos[c*(a + b*x)], x]}, Int[ActivateTrig[u*v], x] + Simp[d Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[Cos[ c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] && !FreeQ[v, x] && Intege rQ[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Sin] || EqQ[F, sin])
Leaf count of result is larger than twice the leaf count of optimal. \(150\) vs. \(2(74)=148\).
Time = 4.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.61
method | result | size |
default | \(-\frac {\left (\sec ^{3}\left (x \right )\right ) \csc \left (x \right ) \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (-2 \cos \left (x \right ) \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \left (\cos ^{2}\left (x \right )\right ) \sin \left (x \right )-2 \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}-164 \left (\cos ^{2}\left (x \right )\right )-5 \sin \left (x \right )+180\right )}{8 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}}}\) | \(151\) |
parts | \(-\frac {8 \cos \left (x \right )-30 \sec \left (x \right )+25 \left (\sec ^{3}\left (x \right )\right )}{8 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}} \left (-2+\sqrt {5}\right )^{2} \left (2+\sqrt {5}\right )^{2}}+\frac {\left (\sec ^{3}\left (x \right )\right ) \csc \left (x \right ) \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (\cos \left (x \right ) \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+\sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+82 \left (\cos ^{2}\left (x \right )\right )-90\right )}{4 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}}}\) | \(179\) |
-1/8*sec(x)^3*csc(x)*(4*cos(x)^2-5)*(-2*cos(x)*sin(x)*arctanh(2*sin(x)/(co s(x)+1)/(-(4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2))*(-(4*cos(x)^2-5)/(cos(x)+1)^ 2)^(1/2)+2*cos(x)^2*sin(x)-2*sin(x)*arctanh(2*sin(x)/(cos(x)+1)/(-(4*cos(x )^2-5)/(cos(x)+1)^2)^(1/2))*(-(4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2)-164*cos(x )^2-5*sin(x)+180)/(5*sec(x)^2-4)^(3/2)
Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\frac {2 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \log \left (\sqrt {-\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right ) - 2 \, \sin \left (x\right )\right ) \sin \left (x\right ) + {\left (164 \, \cos \left (x\right )^{3} - {\left (2 \, \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )\right )} \sin \left (x\right ) - 180 \, \cos \left (x\right )\right )} \sqrt {-\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}}}{8 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \sin \left (x\right )} \]
1/8*(2*(4*cos(x)^2 - 5)*log(sqrt(-(4*cos(x)^2 - 5)/cos(x)^2)*cos(x) - 2*si n(x))*sin(x) + (164*cos(x)^3 - (2*cos(x)^3 - 5*cos(x))*sin(x) - 180*cos(x) )*sqrt(-(4*cos(x)^2 - 5)/cos(x)^2))/((4*cos(x)^2 - 5)*sin(x))
\[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=- \int \left (- \frac {\sin {\left (x \right )}}{5 \sqrt {5 \tan ^{2}{\left (x \right )} + 1} \tan ^{2}{\left (x \right )} + \sqrt {5 \tan ^{2}{\left (x \right )} + 1}}\right )\, dx - \int \frac {2 \cot ^{2}{\left (x \right )}}{5 \sqrt {5 \tan ^{2}{\left (x \right )} + 1} \tan ^{2}{\left (x \right )} + \sqrt {5 \tan ^{2}{\left (x \right )} + 1}}\, dx \]
-Integral(-sin(x)/(5*sqrt(5*tan(x)**2 + 1)*tan(x)**2 + sqrt(5*tan(x)**2 + 1)), x) - Integral(2*cot(x)**2/(5*sqrt(5*tan(x)**2 + 1)*tan(x)**2 + sqrt(5 *tan(x)**2 + 1)), x)
Timed out. \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\int { -\frac {2 \, \cot \left (x\right )^{2} - \sin \left (x\right )}{{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\int \frac {\sin \left (x\right )-2\,{\mathrm {cot}\left (x\right )}^2}{{\left (5\,{\mathrm {tan}\left (x\right )}^2+1\right )}^{3/2}} \,d x \]