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

Optimal. Leaf size=157 \[ -\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^3 c^2}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2} \]

[Out]

(15*x*Sqrt[ArcTan[a*x]])/(32*a^2*c^2*(1 + a^2*x^2)) + (5*ArcTan[a*x]^(3/2))/(16*a^3*c^2) - (5*ArcTan[a*x]^(3/2
))/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(5/2))/(2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a^3*c^2)
- (15*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a^3*c^2)

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Rubi [A]  time = 0.224478, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4936, 4930, 4892, 4970, 4406, 12, 3305, 3351} \[ -\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^3 c^2}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*x*Sqrt[ArcTan[a*x]])/(32*a^2*c^2*(1 + a^2*x^2)) + (5*ArcTan[a*x]^(3/2))/(16*a^3*c^2) - (5*ArcTan[a*x]^(3/2
))/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(5/2))/(2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a^3*c^2)
- (15*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a^3*c^2)

Rule 4936

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[(b*p)/(2*c), Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^
2, x], x] - Simp[(x*(a + b*ArcTan[c*x])^p)/(2*c^2*d*(d + e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

Rule 4930

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 4892

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 )^2} \, dx &=-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}+\frac{5 \int \frac{x \tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx}{4 a}\\ &=-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}+\frac{15 \int \frac{\sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{16 a^2}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \int \frac{x}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx}{64 a}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a^3 c^2}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 \sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a^3 c^2}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{128 a^3 c^2}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \operatorname{Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{64 a^3 c^2}\\ &=\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a^3 c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a^3 c^2}-\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^3 c^2}\\ \end{align*}

Mathematica [A]  time = 0.200296, size = 111, normalized size = 0.71 \[ \frac{4 \sqrt{\tan ^{-1}(a x)} \left (32 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3+70 \left (a^2 x^2-1\right ) \tan ^{-1}(a x)+105 a x-112 a x \tan ^{-1}(a x)^2\right )-105 \sqrt{\pi } \left (a^2 x^2+1\right ) S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{896 a^3 c^2 \left (a^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[ArcTan[a*x]]*(105*a*x + 70*(-1 + a^2*x^2)*ArcTan[a*x] - 112*a*x*ArcTan[a*x]^2 + 32*(1 + a^2*x^2)*ArcTa
n[a*x]^3) - 105*Sqrt[Pi]*(1 + a^2*x^2)*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(896*a^3*c^2*(1 + a^2*x^2))

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Maple [A]  time = 0.106, size = 102, normalized size = 0.7 \begin{align*}{\frac{1}{7\,{a}^{3}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{a}^{3}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{5\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{16\,{a}^{3}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{15\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{64\,{a}^{3}{c}^{2}}\sqrt{\arctan \left ( ax \right ) }}-{\frac{15\,\sqrt{\pi }}{128\,{a}^{3}{c}^{2}}{\it FresnelS} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \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)^2,x)

[Out]

1/7*arctan(a*x)^(7/2)/a^3/c^2-1/4/a^3/c^2*arctan(a*x)^(5/2)*sin(2*arctan(a*x))-5/16/a^3/c^2*arctan(a*x)^(3/2)*
cos(2*arctan(a*x))+15/64/a^3/c^2*arctan(a*x)^(1/2)*sin(2*arctan(a*x))-15/128*FresnelS(2*arctan(a*x)^(1/2)/Pi^(
1/2))*Pi^(1/2)/a^3/c^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)^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^2*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^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**2*atan(a*x)**(5/2)/(a**2*c*x**2+c)**2,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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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