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

Optimal. Leaf size=109 \[ -\frac{3 c^{3/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{40 a^2}-\frac{x \left (a^2 c x^2+c\right )^{3/2}}{20 a}-\frac{3 c x \sqrt{a^2 c x^2+c}}{40 a}+\frac{\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c} \]

[Out]

(-3*c*x*Sqrt[c + a^2*c*x^2])/(40*a) - (x*(c + a^2*c*x^2)^(3/2))/(20*a) + ((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/(
5*a^2*c) - (3*c^(3/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(40*a^2)

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Rubi [A]  time = 0.0746509, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4930, 195, 217, 206} \[ -\frac{3 c^{3/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{40 a^2}-\frac{x \left (a^2 c x^2+c\right )^{3/2}}{20 a}-\frac{3 c x \sqrt{a^2 c x^2+c}}{40 a}+\frac{\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c*x*Sqrt[c + a^2*c*x^2])/(40*a) - (x*(c + a^2*c*x^2)^(3/2))/(20*a) + ((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/(
5*a^2*c) - (3*c^(3/2)*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(40*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 \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx &=\frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac{\int \left (c+a^2 c x^2\right )^{3/2} \, dx}{5 a}\\ &=-\frac{x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac{(3 c) \int \sqrt{c+a^2 c x^2} \, dx}{20 a}\\ &=-\frac{3 c x \sqrt{c+a^2 c x^2}}{40 a}-\frac{x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac{\left (3 c^2\right ) \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{40 a}\\ &=-\frac{3 c x \sqrt{c+a^2 c x^2}}{40 a}-\frac{x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c+a^2 c x^2}}\right )}{40 a}\\ &=-\frac{3 c x \sqrt{c+a^2 c x^2}}{40 a}-\frac{x \left (c+a^2 c x^2\right )^{3/2}}{20 a}+\frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{5 a^2 c}-\frac{3 c^{3/2} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c+a^2 c x^2}}\right )}{40 a^2}\\ \end{align*}

Mathematica [A]  time = 0.162349, size = 101, normalized size = 0.93 \[ -\frac{3 c^{3/2} \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+a c x\right )+a c x \left (2 a^2 x^2+5\right ) \sqrt{a^2 c x^2+c}-8 c \left (a^2 x^2+1\right )^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{40 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*x*(5 + 2*a^2*x^2)*Sqrt[c + a^2*c*x^2] - 8*c*(1 + a^2*x^2)^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + 3*c^(3/2)*
Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(40*a^2)

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Maple [C]  time = 0.3, size = 179, normalized size = 1.6 \begin{align*}{\frac{c \left ( 8\,\arctan \left ( ax \right ){x}^{4}{a}^{4}-2\,{a}^{3}{x}^{3}+16\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-5\,ax+8\,\arctan \left ( ax \right ) \right ) }{40\,{a}^{2}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{\frac{3\,c}{40\,{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{3\,c}{40\,{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*(a^2*c*x^2+c)^(3/2)*arctan(a*x),x)

[Out]

1/40*c/a^2*(c*(a*x-I)*(a*x+I))^(1/2)*(8*arctan(a*x)*x^4*a^4-2*a^3*x^3+16*arctan(a*x)*a^2*x^2-5*a*x+8*arctan(a*
x))+3/40*c/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)-3/40*c/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 = 2.05227, size = 576, normalized size = 5.28 \begin{align*} \frac{40 \,{\left (a^{2} c x^{2} + c\right )} \sqrt{a^{2} x^{2} + 1} \sqrt{c} \arctan \left (a x\right ) - 20 \,{\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac{1}{4}}{\left (a c x \cos \left (\frac{1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, c \sin \left (\frac{1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt{c} -{\left ({\left (a{\left (\frac{3 \,{\left (\frac{2 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{a^{2}} - \frac{\sqrt{a^{2} x^{2} + 1} x}{a^{2}} - \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{2}}\right )}}{a^{2}} - \frac{8 \,{\left (\sqrt{a^{2} x^{2} + 1} x + \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}}\right )}}{a^{4}}\right )} - 8 \,{\left (\frac{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{2}}{a^{2}} - \frac{2 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4}}\right )} \arctan \left (a x\right )\right )} a^{4} c - 10 \, c \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 ) - 10 \, c \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 )} \sqrt{c}}{120 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(40*(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 20*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*c*x*c
os(1/2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) - ((a*(3*(2*(a^2*x^2
+ 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2))/a^2 - 8*(sqrt(a^2*x^2
+ 1)*x + arcsinh(a^2*x/sqrt(a^2))/sqrt(a^2))/a^4) - 8*(3*(a^2*x^2 + 1)^(3/2)*x^2/a^2 - 2*(a^2*x^2 + 1)^(3/2)/a
^4)*arctan(a*x))*a^4*c - 10*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))) - 10*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*ar
ctan2(4*a*x, a^2*x^2 - 3))))*sqrt(c))/a^2

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Fricas [A]  time = 1.76499, size = 235, normalized size = 2.16 \begin{align*} \frac{3 \, c^{\frac{3}{2}} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt{a^{2} c x^{2} + c} a \sqrt{c} x - c\right ) - 2 \,{\left (2 \, a^{3} c x^{3} + 5 \, a c x - 8 \,{\left (a^{4} c x^{4} + 2 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c}}{80 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(3*c^(3/2)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c) - 2*(2*a^3*c*x^3 + 5*a*c*x - 8*(a^4*
c*x^4 + 2*a^2*c*x^2 + c)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{3}{2}} \operatorname{atan}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(c*(a**2*x**2 + 1))**(3/2)*atan(a*x), x)

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Giac [A]  time = 1.28191, size = 170, normalized size = 1.56 \begin{align*} -\frac{{\left (2 \, a^{2} c x^{2} + 5 \, c\right )} \sqrt{a^{2} c x^{2} + c} x - \frac{3 \, c^{\frac{3}{2}} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}}}{40 \, a} + \frac{{\left (5 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} + \frac{3 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} - 5 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c}{c}\right )} \arctan \left (a x\right )}{15 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*((2*a^2*c*x^2 + 5*c)*sqrt(a^2*c*x^2 + c)*x - 3*c^(3/2)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/ab
s(a))/a + 1/15*(5*(a^2*c*x^2 + c)^(3/2) + (3*(a^2*c*x^2 + c)^(5/2) - 5*(a^2*c*x^2 + c)^(3/2)*c)/c)*arctan(a*x)
/a^2