Integrand size = 26, antiderivative size = 247 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {x^3 \arctan (a x)^{3/2}}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{24 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{24 a^3 c^2 \sqrt {c+a^2 c x^2}} \]
1/3*x^3*arctan(a*x)^(3/2)/c/(a^2*c*x^2+c)^(3/2)+1/144*FresnelC(6^(1/2)/Pi^ (1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^3/c^2/(a^2*c *x^2+c)^(1/2)-3/16*FresnelC(2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi ^(1/2)*(a^2*x^2+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)+3/8*arctan(a*x)^(1/2) /a^3/c^2/(a^2*c*x^2+c)^(1/2)-1/24*cos(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*arc tan(a*x)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.37 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {96 \arctan (a x)+144 a^2 x^2 \arctan (a x)+96 a^3 x^3 \arctan (a x)^2+27 i \left (1+a^2 x^2\right )^{3/2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-27 i \left (1+a^2 x^2\right )^{3/2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )-i \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )-i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+i \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )+i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )}{288 a^3 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]
(96*ArcTan[a*x] + 144*a^2*x^2*ArcTan[a*x] + 96*a^3*x^3*ArcTan[a*x]^2 + (27 *I)*(1 + a^2*x^2)^(3/2)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x] ] - (27*I)*(1 + a^2*x^2)^(3/2)*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x ]] - I*Sqrt[3 + 3*a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan [a*x]] - I*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, ( -3*I)*ArcTan[a*x]] + I*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]] + I*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gam ma[1/2, (3*I)*ArcTan[a*x]])/(288*a^3*c^2*(1 + a^2*x^2)*Sqrt[c + a^2*c*x^2] *Sqrt[ArcTan[a*x]])
Time = 0.77 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5479, 5506, 5505, 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.
\(\displaystyle \int \frac {x^2 \arctan (a x)^{3/2}}{\left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5479 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{2} a \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (a^2 c x^2+c\right )^{5/2}}dx\) |
\(\Big \downarrow \) 5506 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a \sqrt {a^2 x^2+1} \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^{5/2}}dx}{2 c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \frac {a^3 x^3 \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^{3/2}}d\arctan (a x)}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \sqrt {\arctan (a x)} \sin (\arctan (a x))^3d\arctan (a x)}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \left (\frac {3 a x \sqrt {\arctan (a x)}}{4 \sqrt {a^2 x^2+1}}-\frac {1}{4} \sqrt {\arctan (a x)} \sin (3 \arctan (a x))\right )d\arctan (a x)}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^3 \arctan (a x)^{3/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \left (-\frac {3 \sqrt {\arctan (a x)}}{4 \sqrt {a^2 x^2+1}}+\frac {3}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{12} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{12} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))\right )}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}\) |
(x^3*ArcTan[a*x]^(3/2))/(3*c*(c + a^2*c*x^2)^(3/2)) - (Sqrt[1 + a^2*x^2]*( (-3*Sqrt[ArcTan[a*x]])/(4*Sqrt[1 + a^2*x^2]) + (Sqrt[ArcTan[a*x]]*Cos[3*Ar cTan[a*x]])/12 + (3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/4 - (Sqrt[Pi/6]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/12))/(2*a^3*c^2*Sqrt[ c + a^2*c*x^2])
3.9.32.3.1 Defintions of rubi rules used
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_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 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 + 3, 0] && GtQ[p, 0] && NeQ[m, -1]
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])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]) Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && !(I ntegerQ[q] || GtQ[d, 0])
\[\int \frac {x^{2} \arctan \left (a x \right )^{\frac {3}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{\frac {3}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^{3/2}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]