3.202 \(\int x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=86 \[ -\frac{x \sqrt{a^2 c x^2+c}}{6 a}+\frac{\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{6 a^2} \]

[Out]

-(x*Sqrt[c + a^2*c*x^2])/(6*a) + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*a^2*c) - (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x
)/Sqrt[c + a^2*c*x^2]])/(6*a^2)

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Rubi [A]  time = 0.0605705, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4930, 195, 217, 206} \[ -\frac{x \sqrt{a^2 c x^2+c}}{6 a}+\frac{\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x],x]

[Out]

-(x*Sqrt[c + a^2*c*x^2])/(6*a) + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*a^2*c) - (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x
)/Sqrt[c + a^2*c*x^2]])/(6*a^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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

Rubi steps

\begin{align*} \int x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x) \, dx &=\frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{\int \sqrt{c+a^2 c x^2} \, dx}{3 a}\\ &=-\frac{x \sqrt{c+a^2 c x^2}}{6 a}+\frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{c \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{6 a}\\ &=-\frac{x \sqrt{c+a^2 c x^2}}{6 a}+\frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{c \operatorname{Subst}\left (\int \frac{1}{1-a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c+a^2 c x^2}}\right )}{6 a}\\ &=-\frac{x \sqrt{c+a^2 c x^2}}{6 a}+\frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c+a^2 c x^2}}\right )}{6 a^2}\\ \end{align*}

Mathematica [A]  time = 0.110979, size = 86, normalized size = 1. \[ -\frac{a x \sqrt{a^2 c x^2+c}+\sqrt{c} \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+a c x\right )-2 \left (a^2 x^2+1\right ) \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x],x]

[Out]

-(a*x*Sqrt[c + a^2*c*x^2] - 2*(1 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + Sqrt[c]*Log[a*c*x + Sqrt[c]*Sqrt
[c + a^2*c*x^2]])/(6*a^2)

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Maple [C]  time = 0.408, size = 156, normalized size = 1.8 \begin{align*}{\frac{2\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-ax+2\,\arctan \left ( ax \right ) }{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{1}{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-i \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{1}{6\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+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*arctan(a*x)*(a^2*c*x^2+c)^(1/2),x)

[Out]

1/6/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*(2*arctan(a*x)*a^2*x^2-a*x+2*arctan(a*x))+1/6/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*
ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)/(a^2*x^2+1)^(1/2)-1/6/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)
^(1/2)+I)/(a^2*x^2+1)^(1/2)

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Maxima [B]  time = 1.85979, size = 351, normalized size = 4.08 \begin{align*} \frac{4 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \sqrt{c} \arctan \left (a x\right ) - 2 \,{\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}}{\left (a x \cos \left (\frac{1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, \sin \left (\frac{1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt{c} + \sqrt{c}{\left (\arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}} \sin \left (\frac{1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x +{\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) + \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}} \sin \left (\frac{1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x +{\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right )\right )}}{12 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(a^2*x^2 + 1)^(3/2)*sqrt(c)*arctan(a*x) - 2*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*x*cos(1/2*arctan2(4*a*
x, -a^2*x^2 + 3)) + 2*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) + sqrt(c)*(arctan2((a^4*x^4 + 10*a^2*x^2
+ 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*
a*x, a^2*x^2 - 3))) + arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) - 2, -a*x
+ (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3)))))/a^2

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Fricas [A]  time = 1.73939, size = 188, normalized size = 2.19 \begin{align*} -\frac{2 \, \sqrt{a^{2} c x^{2} + c}{\left (a x - 2 \,{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\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 )}{12 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*sqrt(a^2*c*x^2 + c)*(a*x - 2*(a^2*x^2 + 1)*arctan(a*x)) - sqrt(c)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2
+ c)*a*sqrt(c)*x - c))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{c \left (a^{2} x^{2} + 1\right )} \operatorname{atan}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)*(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x*sqrt(c*(a**2*x**2 + 1))*atan(a*x), x)

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Giac [A]  time = 1.15356, size = 107, normalized size = 1.24 \begin{align*} -\frac{\sqrt{a^{2} c x^{2} + c} x - \frac{\sqrt{c} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}}}{6 \, a} + \frac{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \arctan \left (a x\right )}{3 \, a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(sqrt(a^2*c*x^2 + c)*x - sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/abs(a))/a + 1/3*(a^2*c*x^
2 + c)^(3/2)*arctan(a*x)/(a^2*c)