\(\int \frac {x^3 \arctan (a x)^{5/2}}{(c+a^2 c x^2)^{5/2}} \, dx\) [909]

Optimal result

Integrand size = 26, antiderivative size = 350 \[ \int \frac {x^3 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {45 \sqrt {\arctan (a x)}}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{144 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {45 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 \sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{144 a^4 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

5/18*x^3*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(3/2)-1/3*x^2*arctan(a*x)^(5/2)/a^2/c/(a^2*c*x^2+c)^(3/2)+5/3*x*a
rctan(a*x)^(3/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)-2/3*arctan(a*x)^(5/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)+5/864*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^4/c^2/(a^2*c*x^2+c)^(1/2)-45/32*Fresn
elC(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^4/c^2/(a^2*c*x^2+c)^(1/2)+45/16*a
rctan(a*x)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)-5/144*cos(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*arctan(a*x)^(1/2)/a^4/
c^2/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5060, 5050, 5018, 5025, 5024, 3385, 3433, 5091, 5090, 3393, 3377} \[ \int \frac {x^3 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {45 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{16 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {5 \sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{144 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {45 \sqrt {\arctan (a x)}}{16 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{144 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(45*Sqrt[ArcTan[a*x]])/(16*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (5*x^3*ArcTan[a*x]^(3/2))/(18*a*c*(c + a^2*c*x^2)^(3
/2)) + (5*x*ArcTan[a*x]^(3/2))/(3*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x]^(5/2))/(3*a^2*c*(c + a^2*c*x
^2)^(3/2)) - (2*ArcTan[a*x]^(5/2))/(3*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (5*Sqrt[1 + a^2*x^2]*Sqrt[ArcTan[a*x]]*Co
s[3*ArcTan[a*x]])/(144*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (45*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqr
t[ArcTan[a*x]]])/(16*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (5*Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[A
rcTan[a*x]]])/(144*a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 5018

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

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 5025

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[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])

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]

Rule 5060

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

Rule 5090

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[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])

Rule 5091

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[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] &&  !(IntegerQ[q] || GtQ[d, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5}{12} \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx+\frac {2 \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^2 c} \\ & = \frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^3 c}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (1+a^2 x^2\right )^{5/2}} \, dx}{12 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {5 \sqrt {\arctan (a x)}}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx}{4 a^3 c}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \sqrt {x} \sin ^3(x) \, dx,x,\arctan (a x)\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {5 \sqrt {\arctan (a x)}}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {3}{4} \sqrt {x} \sin (x)-\frac {1}{4} \sqrt {x} \sin (3 x)\right ) \, dx,x,\arctan (a x)\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \int \frac {1}{\left (1+a^2 x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx}{4 a^3 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {5 \sqrt {\arctan (a x)}}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \sqrt {x} \sin (3 x) \, dx,x,\arctan (a x)\right )}{48 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \sqrt {x} \sin (x) \, dx,x,\arctan (a x)\right )}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {45 \sqrt {\arctan (a x)}}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{144 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{288 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{32 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {45 \sqrt {\arctan (a x)}}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{144 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{144 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (5 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{16 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {45 \sqrt {\arctan (a x)}}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 x^3 \arctan (a x)^{3/2}}{18 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {5 x \arctan (a x)^{3/2}}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \arctan (a x)^{5/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \arctan (a x)^{5/2}}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {5 \sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{144 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {45 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{16 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {5 \sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{144 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {4800 \arctan (a x)+5040 a^2 x^2 \arctan (a x)+2880 a x \arctan (a x)^2+3360 a^3 x^3 \arctan (a x)^2-1152 \arctan (a x)^3-1728 a^2 x^2 \arctan (a x)^3+1215 i \left (1+a^2 x^2\right )^{3/2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-1215 i \left (1+a^2 x^2\right )^{3/2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )-5 i \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )-5 i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+5 i \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )+5 i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )}{1728 a^4 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

(4800*ArcTan[a*x] + 5040*a^2*x^2*ArcTan[a*x] + 2880*a*x*ArcTan[a*x]^2 + 3360*a^3*x^3*ArcTan[a*x]^2 - 1152*ArcT
an[a*x]^3 - 1728*a^2*x^2*ArcTan[a*x]^3 + (1215*I)*(1 + a^2*x^2)^(3/2)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*A
rcTan[a*x]] - (1215*I)*(1 + a^2*x^2)^(3/2)*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] - (5*I)*Sqrt[3 + 3*a^
2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] - (5*I)*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[(-I)*Arc
Tan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + (5*I)*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcT
an[a*x]] + (5*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]])/(1728*a^4*c^2*
(1 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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