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

Optimal. Leaf size=163 \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

-Sqrt[ArcTan[a*x]]/(3*a^2*c*(c + a^2*c*x^2)^(3/2)) + (Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[Ar
cTan[a*x]]])/(4*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a
*x]]])/(12*a^2*c^2*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.250406, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4930, 4905, 4904, 3312, 3304, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[ArcTan[a*x]]/(3*a^2*c*(c + a^2*c*x^2)^(3/2)) + (Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[Ar
cTan[a*x]]])/(4*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a
*x]]])/(12*a^2*c^2*Sqrt[c + a^2*c*x^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 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])

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 \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^{5/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{6 a}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{1+a^2 x^2} \int \frac{1}{\left (1+a^2 x^2\right )^{5/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{6 a c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \frac{\cos ^3(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{6 a^2 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \left (\frac{3 \cos (x)}{4 \sqrt{x}}+\frac{\cos (3 x)}{4 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{6 a^2 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \frac{\cos (3 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{24 a^2 c^2 \sqrt{c+a^2 c x^2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt{c+a^2 c x^2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt{c+a^2 c x^2}}+\frac{\sqrt{\frac{\pi }{6}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.416646, size = 167, normalized size = 1.02 \[ \frac{-48 \tan ^{-1}(a x)-i \left (a^2 x^2+1\right )^{3/2} \left (9 \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-i \tan ^{-1}(a x)\right )-9 \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},i \tan ^{-1}(a x)\right )+\sqrt{3} \left (\sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-3 i \tan ^{-1}(a x)\right )-\sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},3 i \tan ^{-1}(a x)\right )\right )\right )}{144 a^2 c \left (a^2 c x^2+c\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-48*ArcTan[a*x] - I*(1 + a^2*x^2)^(3/2)*(9*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] - 9*Sqrt[I*Arc
Tan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] + Sqrt[3]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] - Sqrt[I*
ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])))/(144*a^2*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]])

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Maple [F]  time = 0.823, size = 0, normalized size = 0. \begin{align*} \int{x\sqrt{\arctan \left ( ax \right ) } \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

[Out]

int(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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