Integrand size = 24, antiderivative size = 127 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{32 a^3 c^2} \]
-1/2*x*arctan(a*x)^(3/2)/a^2/c^2/(a^2*x^2+1)+1/5*arctan(a*x)^(5/2)/a^3/c^2 +3/32*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^3/c^2+3/16*arctan( a*x)^(1/2)/a^3/c^2-3/8*arctan(a*x)^(1/2)/a^3/c^2/(a^2*x^2+1)
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.47 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\frac {16 \sqrt {\arctan (a x)} \left (15 \left (-1+a^2 x^2\right )-40 a x \arctan (a x)+16 \left (1+a^2 x^2\right ) \arctan (a x)^2\right )}{1+a^2 x^2}+60 \left (-2 \sqrt {\arctan (a x)}+\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )\right )+\frac {15 \left (8 \arctan (a x)-i \sqrt {2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+i \sqrt {2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )\right )}{\sqrt {\arctan (a x)}}}{1280 a^3 c^2} \]
((16*Sqrt[ArcTan[a*x]]*(15*(-1 + a^2*x^2) - 40*a*x*ArcTan[a*x] + 16*(1 + a ^2*x^2)*ArcTan[a*x]^2))/(1 + a^2*x^2) + 60*(-2*Sqrt[ArcTan[a*x]] + Sqrt[Pi ]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]) + (15*(8*ArcTan[a*x] - I*Sqrt[ 2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + I*Sqrt[2]*Sqrt[ I*ArcTan[a*x]]*Gamma[1/2, (2*I)*ArcTan[a*x]]))/Sqrt[ArcTan[a*x]])/(1280*a^ 3*c^2)
Time = 0.55 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5471, 27, 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.
\(\displaystyle \int \frac {x^2 \arctan (a x)^{3/2}}{\left (a^2 c x^2+c\right )^2} \, dx\) |
\(\Big \downarrow \) 5471 |
\(\displaystyle \frac {3 \int \frac {x \sqrt {\arctan (a x)}}{c^2 \left (a^2 x^2+1\right )^2}dx}{4 a}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \frac {x \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx}{4 a c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 5465 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{4 a}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^2}{\sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {3 \left (\frac {\int \left (\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {1}{2 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \left (\frac {\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {\arctan (a x)}}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\) |
-1/2*(x*ArcTan[a*x]^(3/2))/(a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5* a^3*c^2) + (3*(-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)))/(4 *a*c^2)
3.8.85.3.1 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_ 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_)^2)/((d_) + (e_.)*(x_)^2) ^2, x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (-Simp[x*((a + b*ArcTan[c*x])^p/(2*c^2*d*(d + e*x^2))), x] + Simp[b*(p/( 2*c)) Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ [{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
Time = 7.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {32 \arctan \left (a x \right )^{\frac {5}{2}} \sqrt {\pi }-40 \arctan \left (a x \right )^{\frac {3}{2}} \sin \left (2 \arctan \left (a x \right )\right ) \sqrt {\pi }+15 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-30 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \cos \left (2 \arctan \left (a x \right )\right )}{160 c^{2} a^{3} \sqrt {\pi }}\) | \(75\) |
1/160/c^2/a^3/Pi^(1/2)*(32*arctan(a*x)^(5/2)*Pi^(1/2)-40*arctan(a*x)^(3/2) *sin(2*arctan(a*x))*Pi^(1/2)+15*Pi*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))- 30*arctan(a*x)^(1/2)*Pi^(1/2)*cos(2*arctan(a*x)))
Exception generated. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]
Exception generated. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{\frac {3}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^{3/2}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]