3.912 \(\int \frac {\tan ^{-1}(a x)^{5/2}}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=337 \[ \frac {45 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{16 a c^2 \sqrt {a^2 c x^2+c}}+\frac {5 \sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{144 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^{5/2}}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{3 a c^2 \sqrt {a^2 c x^2+c}}-\frac {45 x \sqrt {\tan ^{-1}(a x)}}{16 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 \sqrt {a^2 x^2+1} \sqrt {\tan ^{-1}(a x)} \sin \left (3 \tan ^{-1}(a x)\right )}{144 a c^2 \sqrt {a^2 c x^2+c}}+\frac {x \tan ^{-1}(a x)^{5/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{18 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

5/18*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(3/2)+1/3*x*arctan(a*x)^(5/2)/c/(a^2*c*x^2+c)^(3/2)+5/3*arctan(a*x)^(
3/2)/a/c^2/(a^2*c*x^2+c)^(1/2)+2/3*x*arctan(a*x)^(5/2)/c^2/(a^2*c*x^2+c)^(1/2)+5/864*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/c^2/(a^2*c*x^2+c)^(1/2)+45/32*FresnelS(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/c^2/(a^2*c*x^2+c)^(1/2)-45/16*x*arctan(a*x)^(1/2)/c
^2/(a^2*c*x^2+c)^(1/2)-5/144*sin(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*arctan(a*x)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4900, 4898, 4905, 4904, 3296, 3305, 3351, 3312} \[ \frac {45 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{16 a c^2 \sqrt {a^2 c x^2+c}}+\frac {5 \sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{144 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \tan ^{-1}(a x)^{5/2}}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{3 a c^2 \sqrt {a^2 c x^2+c}}-\frac {45 x \sqrt {\tan ^{-1}(a x)}}{16 c^2 \sqrt {a^2 c x^2+c}}-\frac {5 \sqrt {a^2 x^2+1} \sqrt {\tan ^{-1}(a x)} \sin \left (3 \tan ^{-1}(a x)\right )}{144 a c^2 \sqrt {a^2 c x^2+c}}+\frac {x \tan ^{-1}(a x)^{5/2}}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{18 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*x*Sqrt[ArcTan[a*x]])/(16*c^2*Sqrt[c + a^2*c*x^2]) + (5*ArcTan[a*x]^(3/2))/(18*a*c*(c + a^2*c*x^2)^(3/2))
+ (5*ArcTan[a*x]^(3/2))/(3*a*c^2*Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a*x]^(5/2))/(3*c*(c + a^2*c*x^2)^(3/2)) + (2
*x*ArcTan[a*x]^(5/2))/(3*c^2*Sqrt[c + a^2*c*x^2]) + (45*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[
ArcTan[a*x]]])/(16*a*c^2*Sqrt[c + a^2*c*x^2]) + (5*Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTa
n[a*x]]])/(144*a*c^2*Sqrt[c + a^2*c*x^2]) - (5*Sqrt[1 + a^2*x^2]*Sqrt[ArcTan[a*x]]*Sin[3*ArcTan[a*x]])/(144*a*
c^2*Sqrt[c + a^2*c*x^2])

Rule 3296

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 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(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 3312

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 3351

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

Rule 4898

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 4900

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 4904

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 4905

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])

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 176, normalized size = 0.52 \[ \frac {1215 \sqrt {2 \pi } \left (a^2 x^2+1\right )^{3/2} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )+5 \sqrt {6 \pi } \left (a^2 x^2+1\right )^{3/2} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )+24 \sqrt {\tan ^{-1}(a x)} \left (-5 a x \left (20 a^2 x^2+21\right )+12 a x \left (2 a^2 x^2+3\right ) \tan ^{-1}(a x)^2+10 \left (6 a^2 x^2+7\right ) \tan ^{-1}(a x)\right )}{864 c^2 \left (a^3 x^2+a\right ) \sqrt {a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(24*Sqrt[ArcTan[a*x]]*(-5*a*x*(21 + 20*a^2*x^2) + 10*(7 + 6*a^2*x^2)*ArcTan[a*x] + 12*a*x*(3 + 2*a^2*x^2)*ArcT
an[a*x]^2) + 1215*Sqrt[2*Pi]*(1 + a^2*x^2)^(3/2)*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + 5*Sqrt[6*Pi]*(1 + a^
2*x^2)^(3/2)*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(864*c^2*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [F]  time = 1.47, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x \right )^{\frac {5}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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