3.1068 \(\int \frac{x^3}{(c+a^2 c x^2)^3 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac{4 \sqrt{2 \pi } S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a^4 c^3}-\frac{4 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^4 c^3}+\frac{4 x^4}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}} \]

[Out]

(-2*x^3)/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) - (4*x^2)/(a^2*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (
4*x^4)/(3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (4*Sqrt[2*Pi]*FresnelS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*
a^4*c^3) - (4*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^4*c^3)

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Rubi [A]  time = 0.590159, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4968, 4942, 4970, 4406, 3305, 3351} \[ \frac{4 \sqrt{2 \pi } S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a^4 c^3}-\frac{4 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^4 c^3}+\frac{4 x^4}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{2 x^3}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^3)/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) - (4*x^2)/(a^2*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (
4*x^4)/(3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (4*Sqrt[2*Pi]*FresnelS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*
a^4*c^3) - (4*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^4*c^3)

Rule 4968

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

Rule 4942

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

Rule 4970

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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 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 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]

Rubi steps

\begin{align*} \int \frac{x^3}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{2 \int \frac{x^2}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{3/2}} \, dx}{a}-\frac{1}{3} (2 a) \int \frac{x^4}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16}{3} \int \frac{x^3}{\left (c+a^2 c x^2\right )^3 \sqrt{\tan ^{-1}(a x)}} \, dx-8 \int \frac{x^3}{\left (c+a^2 c x^2\right )^3 \sqrt{\tan ^{-1}(a x)}} \, dx+\frac{8 \int \frac{x}{\left (c+a^2 c x^2\right )^3 \sqrt{\tan ^{-1}(a x)}} \, dx}{a^2}\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^4 c^3}+\frac{8 \operatorname{Subst}\left (\int \frac{\cos ^3(x) \sin (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}-\frac{8 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 \sqrt{x}}-\frac{\sin (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{3 a^4 c^3}-\frac{8 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 \sqrt{x}}-\frac{\sin (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac{8 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 \sqrt{x}}+\frac{\sin (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^4 c^3}+2 \frac{\operatorname{Subst}\left (\int \frac{\sin (4 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}-\frac{4 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^4 c^3}\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 \operatorname{Subst}\left (\int \sin \left (4 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a^4 c^3}+2 \frac{2 \operatorname{Subst}\left (\int \sin \left (4 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^3}-\frac{8 \operatorname{Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a^4 c^3}\\ &=-\frac{2 x^3}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 x^2}{a^2 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 x^4}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}+\frac{4 \sqrt{2 \pi } S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a^4 c^3}-\frac{4 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^4 c^3}\\ \end{align*}

Mathematica [C]  time = 0.442479, size = 227, normalized size = 1.42 \[ \frac{i \sqrt{2} \left (a^2 x^2+1\right )^2 \left (-i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )+\sqrt{2} \left (a^2 x^2+1\right )^2 \sqrt{i \tan ^{-1}(a x)} \tan ^{-1}(a x) \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )-2 \left (i \left (a^2 x^2+1\right )^2 \left (-i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-4 i \tan ^{-1}(a x)\right )+\left (a^2 x^2+1\right )^2 \sqrt{i \tan ^{-1}(a x)} \tan ^{-1}(a x) \text{Gamma}\left (\frac{1}{2},4 i \tan ^{-1}(a x)\right )+a^2 x^2 \left (\left (6-2 a^2 x^2\right ) \tan ^{-1}(a x)+a x\right )\right )}{3 a^4 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*Sqrt[2]*(1 + a^2*x^2)^2*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcTan[a*x]] + Sqrt[2]*(1 + a^2*x^2)^2*S
qrt[I*ArcTan[a*x]]*ArcTan[a*x]*Gamma[1/2, (2*I)*ArcTan[a*x]] - 2*(a^2*x^2*(a*x + (6 - 2*a^2*x^2)*ArcTan[a*x])
+ I*(1 + a^2*x^2)^2*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-4*I)*ArcTan[a*x]] + (1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*
x]]*ArcTan[a*x]*Gamma[1/2, (4*I)*ArcTan[a*x]]))/(3*a^4*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2))

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Maple [A]  time = 0.119, size = 112, normalized size = 0.7 \begin{align*} -{\frac{1}{12\,{c}^{3}{a}^{4}} \left ( -16\,\sqrt{2}\sqrt{\pi }{\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}+16\,\sqrt{\pi }{\it FresnelS} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}+8\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) -8\,\cos \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +2\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) -\sin \left ( 4\,\arctan \left ( ax \right ) \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/a^4/c^3*(-16*2^(1/2)*Pi^(1/2)*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*arctan(a*x)^(3/2)+16*Pi^(1/
2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+8*cos(2*arctan(a*x))*arctan(a*x)-8*cos(4*arctan(a*
x))*arctan(a*x)+2*sin(2*arctan(a*x))-sin(4*arctan(a*x)))/arctan(a*x)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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