\(\int \frac {x \arctan (a x)^{5/2}}{(c+a^2 c x^2)^3} \, dx\) [875]

Optimal result

Integrand size = 22, antiderivative size = 254 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {225 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4096 a^2 c^3}-\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{256 a^2 c^3} \]

[Out]

5/32*x*arctan(a*x)^(3/2)/a/c^3/(a^2*x^2+1)^2+15/64*x*arctan(a*x)^(3/2)/a/c^3/(a^2*x^2+1)+3/32*arctan(a*x)^(5/2
)/a^2/c^3-1/4*arctan(a*x)^(5/2)/a^2/c^3/(a^2*x^2+1)^2-15/8192*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2
^(1/2)*Pi^(1/2)/a^2/c^3-15/256*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/c^3-225/2048*arctan(a*x)^(1
/2)/a^2/c^3+15/256*arctan(a*x)^(1/2)/a^2/c^3/(a^2*x^2+1)^2+45/256*arctan(a*x)^(1/2)/a^2/c^3/(a^2*x^2+1)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5050, 5020, 5012, 5024, 3393, 3385, 3433} \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4096 a^2 c^3}-\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{256 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (a^2 x^2+1\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (a^2 x^2+1\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (a^2 x^2+1\right )}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {225 \sqrt {\arctan (a x)}}{2048 a^2 c^3} \]

[In]

Int[(x*ArcTan[a*x]^(5/2))/(c + a^2*c*x^2)^3,x]

[Out]

(-225*Sqrt[ArcTan[a*x]])/(2048*a^2*c^3) + (15*Sqrt[ArcTan[a*x]])/(256*a^2*c^3*(1 + a^2*x^2)^2) + (45*Sqrt[ArcT
an[a*x]])/(256*a^2*c^3*(1 + a^2*x^2)) + (5*x*ArcTan[a*x]^(3/2))/(32*a*c^3*(1 + a^2*x^2)^2) + (15*x*ArcTan[a*x]
^(3/2))/(64*a*c^3*(1 + a^2*x^2)) + (3*ArcTan[a*x]^(5/2))/(32*a^2*c^3) - ArcTan[a*x]^(5/2)/(4*a^2*c^3*(1 + a^2*
x^2)^2) - (15*Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(4096*a^2*c^3) - (15*Sqrt[Pi]*FresnelC[(2*S
qrt[ArcTan[a*x]])/Sqrt[Pi]])/(256*a^2*c^3)

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[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]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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])
)

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 5012

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTan[c*x])
^p/(2*d*(d + e*x^2))), x] + (-Dist[b*c*(p/2), Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2), x], x] + Simp
[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p,
0]

Rule 5020

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[b*p*(d + e*x^2)^(q +
1)*((a + b*ArcTan[c*x])^(p - 1)/(4*c*d*(q + 1)^2)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[b^2*p*((p - 1)/(4*(q + 1)^2)), Int[(d + e*x^2)^q*(a + b*ArcTan[c*x])^(
p - 2), x], x] - Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e
}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]

Rule 5024

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(a
+ b*x)^p/Cos[x]^(2*(q + 1)), x], x, ArcTan[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])

Rule 5050

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {5 \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^3} \, dx}{8 a} \\ & = \frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \int \frac {1}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx}{512 a}+\frac {15 \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx}{32 a c} \\ & = \frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \text {Subst}\left (\int \frac {\cos ^4(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{512 a^2 c^3}-\frac {45 \int \frac {x \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{128 c} \\ & = \frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \text {Subst}\left (\int \left (\frac {3}{8 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}+\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{512 a^2 c^3}-\frac {45 \int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\arctan (a x)}} \, dx}{512 a c} \\ & = -\frac {45 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \text {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{4096 a^2 c^3}-\frac {15 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{1024 a^2 c^3}-\frac {45 \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{512 a^2 c^3} \\ & = -\frac {45 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \text {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{2048 a^2 c^3}-\frac {15 \text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{512 a^2 c^3}-\frac {45 \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{512 a^2 c^3} \\ & = -\frac {225 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4096 a^2 c^3}-\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{1024 a^2 c^3}-\frac {45 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{1024 a^2 c^3} \\ & = -\frac {225 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4096 a^2 c^3}-\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{1024 a^2 c^3}-\frac {45 \text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{512 a^2 c^3} \\ & = -\frac {225 \sqrt {\arctan (a x)}}{2048 a^2 c^3}+\frac {15 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {45 \sqrt {\arctan (a x)}}{256 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {5 x \arctan (a x)^{3/2}}{32 a c^3 \left (1+a^2 x^2\right )^2}+\frac {15 x \arctan (a x)^{3/2}}{64 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^{5/2}}{32 a^2 c^3}-\frac {\arctan (a x)^{5/2}}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4096 a^2 c^3}-\frac {15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{256 a^2 c^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.41 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {450 \sqrt {2 \pi } \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {16320 \arctan (a x)-5760 a^2 x^2 \arctan (a x)-14400 a^4 x^4 \arctan (a x)+51200 a x \arctan (a x)^2+30720 a^3 x^3 \arctan (a x)^2-20480 \arctan (a x)^3+24576 a^2 x^2 \arctan (a x)^3+12288 a^4 x^4 \arctan (a x)^3-3600 \sqrt {\pi } \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)} \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+1020 i \sqrt {2} \left (1+a^2 x^2\right )^2 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )-1020 i \sqrt {2} \left (1+a^2 x^2\right )^2 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )+345 i \left (1+a^2 x^2\right )^2 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )-345 i \left (1+a^2 x^2\right )^2 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )}{\left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}}{131072 a^2 c^3} \]

[In]

Integrate[(x*ArcTan[a*x]^(5/2))/(c + a^2*c*x^2)^3,x]

[Out]

(450*Sqrt[2*Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + (16320*ArcTan[a*x] - 5760*a^2*x^2*ArcTan[a*x] - 144
00*a^4*x^4*ArcTan[a*x] + 51200*a*x*ArcTan[a*x]^2 + 30720*a^3*x^3*ArcTan[a*x]^2 - 20480*ArcTan[a*x]^3 + 24576*a
^2*x^2*ArcTan[a*x]^3 + 12288*a^4*x^4*ArcTan[a*x]^3 - 3600*Sqrt[Pi]*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]*FresnelC[
(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + (1020*I)*Sqrt[2]*(1 + a^2*x^2)^2*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-2*I)*Ar
cTan[a*x]] - (1020*I)*Sqrt[2]*(1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (2*I)*ArcTan[a*x]] + (345*I)*(1 +
a^2*x^2)^2*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan[a*x]] - (345*I)*(1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]
]*Gamma[1/2, (4*I)*ArcTan[a*x]])/((1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]))/(131072*a^2*c^3)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.62

\[-\frac {1024 \arctan \left (a x \right )^{\frac {5}{2}} \cos \left (2 \arctan \left (a x \right )\right ) \sqrt {\pi }+256 \arctan \left (a x \right )^{\frac {5}{2}} \cos \left (4 \arctan \left (a x \right )\right ) \sqrt {\pi }-1280 \arctan \left (a x \right )^{\frac {3}{2}} \sin \left (2 \arctan \left (a x \right )\right ) \sqrt {\pi }+15 \pi \sqrt {2}\, \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-160 \arctan \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sin \left (4 \arctan \left (a x \right )\right )-960 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \cos \left (2 \arctan \left (a x \right )\right )+480 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-60 \cos \left (4 \arctan \left (a x \right )\right ) \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }}{8192 c^{3} a^{2} \sqrt {\pi }}\]

[In]

int(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,x)

[Out]

-1/8192/c^3/a^2/Pi^(1/2)*(1024*arctan(a*x)^(5/2)*cos(2*arctan(a*x))*Pi^(1/2)+256*arctan(a*x)^(5/2)*cos(4*arcta
n(a*x))*Pi^(1/2)-1280*arctan(a*x)^(3/2)*sin(2*arctan(a*x))*Pi^(1/2)+15*Pi*2^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*
arctan(a*x)^(1/2))-160*arctan(a*x)^(3/2)*Pi^(1/2)*sin(4*arctan(a*x))-960*arctan(a*x)^(1/2)*Pi^(1/2)*cos(2*arct
an(a*x))+480*Pi*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))-60*cos(4*arctan(a*x))*arctan(a*x)^(1/2)*Pi^(1/2))

Fricas [F(-2)]

Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]

[In]

integrate(x*atan(a*x)**(5/2)/(a**2*c*x**2+c)**3,x)

[Out]

Integral(x*atan(a*x)**(5/2)/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {x \arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx=\int \frac {x\,{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

[In]

int((x*atan(a*x)^(5/2))/(c + a^2*c*x^2)^3,x)

[Out]

int((x*atan(a*x)^(5/2))/(c + a^2*c*x^2)^3, x)