Integrand size = 24, antiderivative size = 160 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=-\frac {2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^{3/2}}-\frac {4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {4 \sqrt {2 \pi } \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a^4 c^3}-\frac {4 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^4 c^3} \]
-2/3*x^3/a/c^3/(a^2*x^2+1)^2/arctan(a*x)^(3/2)-4/3*FresnelS(2*arctan(a*x)^ (1/2)/Pi^(1/2))*Pi^(1/2)/a^4/c^3+4/3*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a* x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4/c^3-4*x^2/a^2/c^3/(a^2*x^2+1)^2/arctan(a*x) ^(1/2)+4/3*x^4/c^3/(a^2*x^2+1)^2/arctan(a*x)^(1/2)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.42 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\frac {i \sqrt {2} \left (1+a^2 x^2\right )^2 (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+\sqrt {2} \left (1+a^2 x^2\right )^2 \sqrt {i \arctan (a x)} \arctan (a x) \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )-2 \left (a^2 x^2 \left (a x+\left (6-2 a^2 x^2\right ) \arctan (a x)\right )+i \left (1+a^2 x^2\right )^2 (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )+\left (1+a^2 x^2\right )^2 \sqrt {i \arctan (a x)} \arctan (a x) \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )\right )}{3 a^4 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^{3/2}} \]
(I*Sqrt[2]*(1 + a^2*x^2)^2*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcT an[a*x]] + Sqrt[2]*(1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]]*ArcTan[a*x]*Gamma[1 /2, (2*I)*ArcTan[a*x]] - 2*(a^2*x^2*(a*x + (6 - 2*a^2*x^2)*ArcTan[a*x]) + I*(1 + a^2*x^2)^2*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-4*I)*ArcTan[a*x]] + (1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]]*ArcTan[a*x]*Gamma[1/2, (4*I)*ArcTan[ a*x]]))/(3*a^4*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2))
Time = 1.13 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5503, 27, 5477, 5503, 5505, 4906, 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.
\(\displaystyle \int \frac {x^3}{\arctan (a x)^{5/2} \left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle \frac {2 \int \frac {x^2}{c^3 \left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx}{a}-\frac {2}{3} a \int \frac {x^4}{c^3 \left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx}{a c^3}-\frac {2 a \int \frac {x^4}{\left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx}{3 c^3}-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5477 |
\(\displaystyle \frac {2 \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx}{a c^3}-\frac {2 a \left (\frac {8 \int \frac {x^3}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx}{a}-\frac {2 x^4}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle \frac {2 \left (\frac {4 \int \frac {x}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx}{a}-4 a \int \frac {x^3}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx-\frac {2 x^2}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{a c^3}-\frac {2 a \left (\frac {8 \int \frac {x^3}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx}{a}-\frac {2 x^4}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle \frac {2 \left (\frac {4 \int \frac {a x}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3}-\frac {4 \int \frac {a^3 x^3}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3}-\frac {2 x^2}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{a c^3}-\frac {2 a \left (\frac {8 \int \frac {a^3 x^3}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^5}-\frac {2 x^4}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {2 a \left (\frac {8 \int \left (\frac {\sin (2 \arctan (a x))}{4 \sqrt {\arctan (a x)}}-\frac {\sin (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^5}-\frac {2 x^4}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}+\frac {2 \left (-\frac {4 \int \left (\frac {\sin (2 \arctan (a x))}{4 \sqrt {\arctan (a x)}}-\frac {\sin (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^3}+\frac {4 \int \left (\frac {\sin (2 \arctan (a x))}{4 \sqrt {\arctan (a x)}}+\frac {\sin (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^3}-\frac {2 x^2}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{a c^3}-\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}-\frac {2 a \left (\frac {8 \left (\frac {1}{4} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )-\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^5}-\frac {2 x^4}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}+\frac {2 \left (-\frac {4 \left (\frac {1}{4} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )-\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^3}+\frac {4 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{4} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )\right )}{a^3}-\frac {2 x^2}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{a c^3}\) |
(-2*x^3)/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) - (2*a*((-2*x^4)/(a*( 1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (8*(-1/8*(Sqrt[Pi/2]*FresnelS[2*Sqrt[2 /Pi]*Sqrt[ArcTan[a*x]]]) + (Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[P i]])/4))/a^5))/(3*c^3) + (2*((-2*x^2)/(a*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]] ) - (4*(-1/8*(Sqrt[Pi/2]*FresnelS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]]) + (Sqrt [Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/4))/a^3 + (4*((Sqrt[Pi/2]*F resnelS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/8 + (Sqrt[Pi]*FresnelS[(2*Sqrt[Ar cTan[a*x]])/Sqrt[Pi]])/4))/a^3))/(a*c^3)
3.11.68.3.1 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[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_)*((f_.)*(x_))^(m_.)*((d_) + (e_. )*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*ArcT an[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[f*(m/(b*c*(p + 1))) Int[(f*x )^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1 ]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[x^m*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] + (-Simp[c*((m + 2*q + 2)/(b*(p + 1))) Int[x^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] - Simp[m/(b*c*(p + 1)) Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x]) /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[e, c^2*d] && IntegerQ[m] && LtQ[q, -1] & & LtQ[p, -1] && NeQ[m + 2*q + 2, 0]
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 = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70
method | result | size |
default | \(-\frac {-16 \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+16 \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+8 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-8 \cos \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+2 \sin \left (2 \arctan \left (a x \right )\right )-\sin \left (4 \arctan \left (a x \right )\right )}{12 a^{4} c^{3} \arctan \left (a x \right )^{\frac {3}{2}}}\) | \(112\) |
-1/12/a^4/c^3*(-16*2^(1/2)*Pi^(1/2)*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x )^(1/2))*arctan(a*x)^(3/2)+16*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/ 2))*arctan(a*x)^(3/2)+8*cos(2*arctan(a*x))*arctan(a*x)-8*cos(4*arctan(a*x) )*arctan(a*x)+2*sin(2*arctan(a*x))-sin(4*arctan(a*x)))/arctan(a*x)^(3/2)
Exception generated. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\frac {\int \frac {x^{3}}{a^{6} x^{6} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{3}} \]
Integral(x**3/(a**6*x**6*atan(a*x)**(5/2) + 3*a**4*x**4*atan(a*x)**(5/2) + 3*a**2*x**2*atan(a*x)**(5/2) + atan(a*x)**(5/2)), x)/c**3
Exception generated. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Timed out. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\int \frac {x^3}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]