Integrand size = 22, antiderivative size = 101 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^2 c^2} \]
-2/3*x/a/c^2/(a^2*x^2+1)/arctan(a*x)^(3/2)-8/3*FresnelS(2*arctan(a*x)^(1/2 )/Pi^(1/2))*Pi^(1/2)/a^2/c^2-4/3*(-a^2*x^2+1)/a^2/c^2/(a^2*x^2+1)/arctan(a *x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2 \left (a x+\left (2-2 a^2 x^2\right ) \arctan (a x)+4 \sqrt {\pi } \left (1+a^2 x^2\right ) \arctan (a x)^{3/2} \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}} \]
(-2*(a*x + (2 - 2*a^2*x^2)*ArcTan[a*x] + 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a *x]^(3/2)*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]))/(3*a^2*c^2*(1 + a^2*x ^2)*ArcTan[a*x]^(3/2))
Time = 0.52 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5467, 27, 5505, 4906, 27, 3042, 3786, 3832}
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.
\(\displaystyle \int \frac {x}{\arctan (a x)^{5/2} \left (a^2 c x^2+c\right )^2} \, dx\) |
\(\Big \downarrow \) 5467 |
\(\displaystyle -\frac {16}{3} \int \frac {x}{c^2 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {16 \int \frac {x}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{3 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle -\frac {16 \int \frac {a x}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {16 \int \frac {\sin (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}d\arctan (a x)}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8 \int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {8 \int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle -\frac {16 \int \sin (2 \arctan (a x))d\sqrt {\arctan (a x)}}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
(-2*x)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (4*(1 - a^2*x^2))/(3*a^ 2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) - (8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcT an[a*x]])/Sqrt[Pi]])/(3*a^2*c^2)
3.11.61.3.1 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[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f }, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1) Subst[Int[(a + b*x)^p*(Sin[x]^m/ Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p }, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q ] || GtQ[d, 0])
Time = 1.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58
method | result | size |
default | \(-\frac {8 \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+4 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sin \left (2 \arctan \left (a x \right )\right )}{3 c^{2} a^{2} \arctan \left (a x \right )^{\frac {3}{2}}}\) | \(59\) |
-1/3/c^2/a^2*(8*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x )^(3/2)+4*cos(2*arctan(a*x))*arctan(a*x)+sin(2*arctan(a*x)))/arctan(a*x)^( 3/2)
Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\frac {\int \frac {x}{a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{2}} \]
Integral(x/(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. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Timed out. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]