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

Optimal. Leaf size=133 \[ -\frac{15 \sqrt{\frac{\pi }{2}} S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4096 a^3 c^3}+\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac{15 \sqrt{\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3} \]

[Out]

ArcTan[a*x]^(7/2)/(28*a^3*c^3) - (5*ArcTan[a*x]^(3/2)*Cos[4*ArcTan[a*x]])/(256*a^3*c^3) - (15*Sqrt[Pi/2]*Fresn
elS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(4096*a^3*c^3) + (15*Sqrt[ArcTan[a*x]]*Sin[4*ArcTan[a*x]])/(2048*a^3*c^3)
- (ArcTan[a*x]^(5/2)*Sin[4*ArcTan[a*x]])/(32*a^3*c^3)

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Rubi [A]  time = 0.175099, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4970, 4406, 3296, 3305, 3351} \[ -\frac{15 \sqrt{\frac{\pi }{2}} S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4096 a^3 c^3}+\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac{15 \sqrt{\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[a*x]^(7/2)/(28*a^3*c^3) - (5*ArcTan[a*x]^(3/2)*Cos[4*ArcTan[a*x]])/(256*a^3*c^3) - (15*Sqrt[Pi/2]*Fresn
elS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(4096*a^3*c^3) + (15*Sqrt[ArcTan[a*x]]*Sin[4*ArcTan[a*x]])/(2048*a^3*c^3)
- (ArcTan[a*x]^(5/2)*Sin[4*ArcTan[a*x]])/(32*a^3*c^3)

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 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 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^2 \tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int x^{5/2} \cos ^2(x) \sin ^2(x) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{x^{5/2}}{8}-\frac{1}{8} x^{5/2} \cos (4 x)\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{\operatorname{Subst}\left (\int x^{5/2} \cos (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{8 a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac{5 \operatorname{Subst}\left (\int x^{3/2} \sin (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{64 a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac{15 \operatorname{Subst}\left (\int \sqrt{x} \cos (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{512 a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}+\frac{15 \sqrt{\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}-\frac{15 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{4096 a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}+\frac{15 \sqrt{\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}-\frac{15 \operatorname{Subst}\left (\int \sin \left (4 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{2048 a^3 c^3}\\ &=\frac{\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac{5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}-\frac{15 \sqrt{\frac{\pi }{2}} S\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4096 a^3 c^3}+\frac{15 \sqrt{\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac{\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}\\ \end{align*}

Mathematica [C]  time = 0.439953, size = 185, normalized size = 1.39 \[ \frac{105 \left (a^2 x^2+1\right )^2 \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-4 i \tan ^{-1}(a x)\right )+105 \left (a^2 x^2+1\right )^2 \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},4 i \tan ^{-1}(a x)\right )+32 \tan ^{-1}(a x) \left (-105 a x \left (a^2 x^2-1\right )+128 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3+448 a x \left (a^2 x^2-1\right ) \tan ^{-1}(a x)^2-70 \left (a^4 x^4-6 a^2 x^2+1\right ) \tan ^{-1}(a x)\right )}{114688 a^3 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(32*ArcTan[a*x]*(-105*a*x*(-1 + a^2*x^2) - 70*(1 - 6*a^2*x^2 + a^4*x^4)*ArcTan[a*x] + 448*a*x*(-1 + a^2*x^2)*A
rcTan[a*x]^2 + 128*(1 + a^2*x^2)^2*ArcTan[a*x]^3) + 105*(1 + a^2*x^2)^2*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*
I)*ArcTan[a*x]] + 105*(1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/(114688*a^3*c^3*(1 +
a^2*x^2)^2*Sqrt[ArcTan[a*x]])

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Maple [A]  time = 0.114, size = 96, normalized size = 0.7 \begin{align*}{\frac{1}{57344\,{c}^{3}{a}^{3}} \left ( 2048\, \left ( \arctan \left ( ax \right ) \right ) ^{4}-1792\, \left ( \arctan \left ( ax \right ) \right ) ^{3}\sin \left ( 4\,\arctan \left ( ax \right ) \right ) -105\,\sqrt{2}\sqrt{\arctan \left ( ax \right ) }\sqrt{\pi }{\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) -1120\, \left ( \arctan \left ( ax \right ) \right ) ^{2}\cos \left ( 4\,\arctan \left ( ax \right ) \right ) +420\,\sin \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) \right ){\frac{1}{\sqrt{\arctan \left ( ax \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/57344/a^3/c^3*(2048*arctan(a*x)^4-1792*arctan(a*x)^3*sin(4*arctan(a*x))-105*2^(1/2)*arctan(a*x)^(1/2)*Pi^(1/
2)*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))-1120*arctan(a*x)^2*cos(4*arctan(a*x))+420*sin(4*arctan(a*x))
*arctan(a*x))/arctan(a*x)^(1/2)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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