Integrand size = 21, antiderivative size = 174 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {16 \sqrt {\arctan (a x)}}{3 a c^2}+\frac {32 \sqrt {\arctan (a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {8 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a c^2} \]
-2/3/a/c^2/(a^2*x^2+1)/arctan(a*x)^(3/2)-8/3*FresnelC(2*arctan(a*x)^(1/2)/ Pi^(1/2))*Pi^(1/2)/a/c^2+8/3*x/c^2/(a^2*x^2+1)/arctan(a*x)^(1/2)-16/3*arct an(a*x)^(1/2)/a/c^2+32/3*arctan(a*x)^(1/2)/a/c^2/(a^2*x^2+1)-16/3*(-a^2*x^ 2+1)*arctan(a*x)^(1/2)/a/c^2/(a^2*x^2+1)
Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\frac {-2+8 a x \arctan (a x)-4 \sqrt {\pi } \left (1+a^2 x^2\right ) \arctan (a x)^{3/2} \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {2} \left (1+a^2 x^2\right ) \sqrt {i \arctan (a x)} \sqrt {\arctan (a x)^2} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+\frac {\sqrt {2} \left (1+a^2 x^2\right ) \arctan (a x)^2 \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )}{\sqrt {i \arctan (a x)}}}{3 c^2 \left (a+a^3 x^2\right ) \arctan (a x)^{3/2}} \]
(-2 + 8*a*x*ArcTan[a*x] - 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)*Fresn elC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + Sqrt[2]*(1 + a^2*x^2)*Sqrt[I*ArcTan[ a*x]]*Sqrt[ArcTan[a*x]^2]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + (Sqrt[2]*(1 + a ^2*x^2)*ArcTan[a*x]^2*Gamma[1/2, (2*I)*ArcTan[a*x]])/Sqrt[I*ArcTan[a*x]])/ (3*c^2*(a + a^3*x^2)*ArcTan[a*x]^(3/2))
Time = 0.68 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5437, 27, 5467, 5465, 5439, 3042, 3793, 2009}
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.
\(\displaystyle \int \frac {1}{\arctan (a x)^{5/2} \left (a^2 c x^2+c\right )^2} \, dx\) |
\(\Big \downarrow \) 5437 |
\(\displaystyle -\frac {4}{3} a \int \frac {x}{c^2 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}dx-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 a \int \frac {x}{\left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}dx}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5467 |
\(\displaystyle -\frac {4 a \left (16 \int \frac {x \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5465 |
\(\displaystyle -\frac {4 a \left (16 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{4 a}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle -\frac {4 a \left (16 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {4 a \left (16 \left (\frac {\int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^2}{\sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {4 a \left (16 \left (\frac {\int \left (\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {1}{2 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 a \left (16 \left (\frac {\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {\arctan (a x)}}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 c^2}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\) |
-2/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (4*a*((-2*x)/(a*(1 + a^2*x^ 2)*Sqrt[ArcTan[a*x]]) + (4*(1 - a^2*x^2)*Sqrt[ArcTan[a*x]])/(a^2*(1 + a^2* x^2)) + 16*(-1/2*Sqrt[ArcTan[a*x]]/(a^2*(1 + a^2*x^2)) + (Sqrt[ArcTan[a*x] ] + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/2)/(4*a^2))))/(3*c ^2)
3.11.62.3.1 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[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_S ymbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*Arc Tan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cos[x]^(2*(q + 1)), x], x, Ar cTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ[2*( q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_ .), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*e*(q + 1))), x] - Simp[b*(p/(2*c*(q + 1))) Int[(d + e*x^2)^q*(a + b*ArcTan[c*x]) ^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2 ))), x] + (-Simp[(1 - c^2*x^2)*((a + b*ArcTan[c*x])^(p + 2)/(b^2*e*(p + 1)* (p + 2)*(d + e*x^2))), x] - Simp[4/(b^2*(p + 1)*(p + 2)) Int[x*((a + b*Ar cTan[c*x])^(p + 2)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]
Time = 0.94 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.36
method | result | size |
default | \(\frac {-8 \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {\pi }\, \arctan \left (a x \right )^{\frac {3}{2}}+4 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\cos \left (2 \arctan \left (a x \right )\right )-1}{3 c^{2} a \arctan \left (a x \right )^{\frac {3}{2}}}\) | \(62\) |
1/3/c^2/a*(-8*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)*arctan(a*x)^ (3/2)+4*sin(2*arctan(a*x))*arctan(a*x)-cos(2*arctan(a*x))-1)/arctan(a*x)^( 3/2)
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\frac {\int \frac {1}{a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{2}} \]
Integral(1/(a**4*x**4*atan(a*x)**(5/2) + 2*a**2*x**2*atan(a*x)**(5/2) + at an(a*x)**(5/2)), x)/c**2
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]