Integrand size = 24, antiderivative size = 248 \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {3 x \sqrt {\arctan (a x)}}{8 a c^2 \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{24 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \sin (3 \arctan (a x))}{24 a^2 c^2 \sqrt {c+a^2 c x^2}} \]
-1/3*arctan(a*x)^(3/2)/a^2/c/(a^2*c*x^2+c)^(3/2)-1/144*FresnelS(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^2/c^2/(a^2* c*x^2+c)^(1/2)-3/16*FresnelS(2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*P i^(1/2)*(a^2*x^2+1)^(1/2)/a^2/c^2/(a^2*c*x^2+c)^(1/2)+3/8*x*arctan(a*x)^(1 /2)/a/c^2/(a^2*c*x^2+c)^(1/2)+1/24*sin(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*ar ctan(a*x)^(1/2)/a^2/c^2/(a^2*c*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 0.93 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.05 \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {48 \left (3 a x+2 a^3 x^3-2 \arctan (a x)\right ) \arctan (a x)-4 \sqrt {6 \pi } \left (1+a^2 x^2\right )^{3/2} \sqrt {\arctan (a x)} \left (3 \sqrt {3} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )+3 \left (1+a^2 x^2\right )^{3/2} \left (3 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+3 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3} \left (\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )\right )}{288 a^2 c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \]
(48*(3*a*x + 2*a^3*x^3 - 2*ArcTan[a*x])*ArcTan[a*x] - 4*Sqrt[6*Pi]*(1 + a^ 2*x^2)^(3/2)*Sqrt[ArcTan[a*x]]*(3*Sqrt[3]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[ a*x]]] - FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]]) + 3*(1 + a^2*x^2)^(3/2)*( 3*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] + 3*Sqrt[I*ArcTan[a* x]]*Gamma[1/2, I*ArcTan[a*x]] + Sqrt[3]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])) )/(288*a^2*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]])
Time = 0.60 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5465, 5440, 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.
\(\displaystyle \int \frac {x \arctan (a x)^{3/2}}{\left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5465 |
\(\displaystyle \frac {\int \frac {\sqrt {\arctan (a x)}}{\left (a^2 c x^2+c\right )^{5/2}}dx}{2 a}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5440 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {\sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^{5/2}}dx}{2 a c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {\sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^{3/2}}d\arctan (a x)}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \sqrt {\arctan (a x)} \sin \left (\arctan (a x)+\frac {\pi }{2}\right )^3d\arctan (a x)}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \left (\frac {1}{4} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))+\frac {3 \sqrt {\arctan (a x)}}{4 \sqrt {a^2 x^2+1}}\right )d\arctan (a x)}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {3 a x \sqrt {\arctan (a x)}}{4 \sqrt {a^2 x^2+1}}-\frac {3}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{12} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{12} \sqrt {\arctan (a x)} \sin (3 \arctan (a x))\right )}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\) |
-1/3*ArcTan[a*x]^(3/2)/(a^2*c*(c + a^2*c*x^2)^(3/2)) + (Sqrt[1 + a^2*x^2]* ((3*a*x*Sqrt[ArcTan[a*x]])/(4*Sqrt[1 + a^2*x^2]) - (3*Sqrt[Pi/2]*FresnelS[ Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/4 - (Sqrt[Pi/6]*FresnelS[Sqrt[6/Pi]*Sqrt[Ar cTan[a*x]]])/12 + (Sqrt[ArcTan[a*x]]*Sin[3*ArcTan[a*x]])/12))/(2*a^2*c^2*S qrt[c + a^2*c*x^2])
3.9.33.3.1 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_.)*((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_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]) Int[(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] && 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]
\[\int \frac {x \arctan \left (a x \right )^{\frac {3}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x \arctan \left (a x\right )^{\frac {3}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x\,{\mathrm {atan}\left (a\,x\right )}^{3/2}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]