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

Optimal. Leaf size=107 \[ \frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2}-\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a^4 c^{3/2}}-\frac{x}{a^3 c \sqrt{a^2 c x^2+c}}+\frac{\tan ^{-1}(a x)}{a^4 c \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.202065, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4964, 4930, 217, 206, 191} \[ \frac{\sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2}-\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a^4 c^{3/2}}-\frac{x}{a^3 c \sqrt{a^2 c x^2+c}}+\frac{\tan ^{-1}(a x)}{a^4 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4964

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.126293, size = 107, normalized size = 1. \[ \frac{-a x \sqrt{a^2 c x^2+c}-\sqrt{c} \left (a^2 x^2+1\right ) \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+a c x\right )+\left (a^2 x^2+2\right ) \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2 \left (a^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 1.029, size = 242, normalized size = 2.3 \begin{align*}{\frac{ \left ( \arctan \left ( ax \right ) +i \right ) \left ( 1+iax \right ) }{ \left ( 2\,{a}^{2}{x}^{2}+2 \right ){a}^{4}{c}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{ \left ( -1+iax \right ) \left ( \arctan \left ( ax \right ) -i \right ) }{ \left ( 2\,{a}^{2}{x}^{2}+2 \right ){a}^{4}{c}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{\arctan \left ( ax \right ) }{{a}^{4}{c}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{1}{{a}^{4}{c}^{2}}\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-i \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{1}{{a}^{4}{c}^{2}}\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+i \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(arctan(a*x)+I)*(1+I*a*x)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^2*x^2+1)/a^4/c^2-1/2*(c*(a*x-I)*(a*x+I))^(1/2)*(-1+
I*a*x)*(arctan(a*x)-I)/(a^2*x^2+1)/a^4/c^2+arctan(a*x)*(c*(a*x-I)*(a*x+I))^(1/2)/a^4/c^2+ln((1+I*a*x)/(a^2*x^2
+1)^(1/2)-I)/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/a^4/c^2-ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+I)/(a^2*x^2+1)
^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/a^4/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.41378, size = 228, normalized size = 2.13 \begin{align*} \frac{{\left (a^{2} x^{2} + 1\right )} \sqrt{c} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt{a^{2} c x^{2} + c} a \sqrt{c} x - c\right ) - 2 \, \sqrt{a^{2} c x^{2} + c}{\left (a x -{\left (a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )}}{2 \,{\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*x^2 + 1)*sqrt(c)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*sqrt(a^2*c*x^2 + c)*(
a*x - (a^2*x^2 + 2)*arctan(a*x)))/(a^6*c^2*x^2 + a^4*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.35594, size = 130, normalized size = 1.21 \begin{align*} -\frac{x}{\sqrt{a^{2} c x^{2} + c} a^{3} c} + \frac{{\left (\sqrt{a^{2} c x^{2} + c} + \frac{c}{\sqrt{a^{2} c x^{2} + c}}\right )} \arctan \left (a x\right )}{a^{4} c^{2}} + \frac{\log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{a^{3} c^{\frac{3}{2}}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(sqrt(a^2*c*x^2 + c)*a^3*c) + (sqrt(a^2*c*x^2 + c) + c/sqrt(a^2*c*x^2 + c))*arctan(a*x)/(a^4*c^2) + log(abs
(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/(a^3*c^(3/2)*abs(a))