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

Optimal. Leaf size=156 \[ -\frac{15 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (a^2 x^2+1\right )}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{15 \sqrt{\tan ^{-1}(a x)}}{64 a^2 c^2} \]

[Out]

(-15*Sqrt[ArcTan[a*x]])/(64*a^2*c^2) + (15*Sqrt[ArcTan[a*x]])/(32*a^2*c^2*(1 + a^2*x^2)) + (5*x*ArcTan[a*x]^(3
/2))/(8*a*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(4*a^2*c^2) - ArcTan[a*x]^(5/2)/(2*a^2*c^2*(1 + a^2*x^2)) - (
15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a^2*c^2)

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Rubi [A]  time = 0.199346, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4930, 4892, 4904, 3312, 3304, 3352} \[ -\frac{15 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (a^2 x^2+1\right )}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{15 \sqrt{\tan ^{-1}(a x)}}{64 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[ArcTan[a*x]])/(64*a^2*c^2) + (15*Sqrt[ArcTan[a*x]])/(32*a^2*c^2*(1 + a^2*x^2)) + (5*x*ArcTan[a*x]^(3
/2))/(8*a*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(4*a^2*c^2) - ArcTan[a*x]^(5/2)/(2*a^2*c^2*(1 + a^2*x^2)) - (
15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a^2*c^2)

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 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 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 3304

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 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[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 \tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 \int \frac{\tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx}{4 a}\\ &=\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15}{16} \int \frac{x \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \int \frac{1}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx}{64 a}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \operatorname{Subst}\left (\int \frac{\cos ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a^2 c^2}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{x}}+\frac{\cos (2 x)}{2 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{64 a^2 c^2}\\ &=-\frac{15 \sqrt{\tan ^{-1}(a x)}}{64 a^2 c^2}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{128 a^2 c^2}\\ &=-\frac{15 \sqrt{\tan ^{-1}(a x)}}{64 a^2 c^2}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{64 a^2 c^2}\\ &=-\frac{15 \sqrt{\tan ^{-1}(a x)}}{64 a^2 c^2}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{15 \sqrt{\pi } C\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2 c^2}\\ \end{align*}

Mathematica [C]  time = 0.169337, size = 234, normalized size = 1.5 \[ \frac{15 i \sqrt{2} \left (a^2 x^2+1\right ) \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )-15 i \sqrt{2} a^2 x^2 \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )-15 i \sqrt{2} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )-60 \sqrt{\pi } \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)} \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )+256 a^2 x^2 \tan ^{-1}(a x)^3-240 a^2 x^2 \tan ^{-1}(a x)-256 \tan ^{-1}(a x)^3+640 a x \tan ^{-1}(a x)^2+240 \tan ^{-1}(a x)}{1024 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(240*ArcTan[a*x] - 240*a^2*x^2*ArcTan[a*x] + 640*a*x*ArcTan[a*x]^2 - 256*ArcTan[a*x]^3 + 256*a^2*x^2*ArcTan[a*
x]^3 - 60*Sqrt[Pi]*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + (15*I)*Sqrt[2]*(
1 + a^2*x^2)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-2*I)*ArcTan[a*x]] - (15*I)*Sqrt[2]*Sqrt[I*ArcTan[a*x]]*Gamma[
1/2, (2*I)*ArcTan[a*x]] - (15*I)*Sqrt[2]*a^2*x^2*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (2*I)*ArcTan[a*x]])/(1024*a^2*
c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]])

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Maple [A]  time = 0.103, size = 88, normalized size = 0.6 \begin{align*} -{\frac{\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{a}^{2}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{5\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{16\,{a}^{2}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{15\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{64\,{a}^{2}{c}^{2}}\sqrt{\arctan \left ( ax \right ) }}-{\frac{15\,\sqrt{\pi }}{128\,{a}^{2}{c}^{2}}{\it FresnelC} \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*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^2,x)

[Out]

-1/4/a^2/c^2*arctan(a*x)^(5/2)*cos(2*arctan(a*x))+5/16/a^2/c^2*arctan(a*x)^(3/2)*sin(2*arctan(a*x))+15/64/a^2/
c^2*arctan(a*x)^(1/2)*cos(2*arctan(a*x))-15/128*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/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*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*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x \operatorname{atan}^{\frac{5}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atan(a*x)**(5/2)/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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