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

Optimal. Leaf size=104 \[ -\frac{\sqrt{a^2 x^2+1} \text{Si}\left (\tan ^{-1}(a x)\right )}{2 a^2 c \sqrt{a^2 c x^2+c}}-\frac{x}{2 a c \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}-\frac{1}{2 a^2 c \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.322079, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4942, 4902, 4971, 4970, 3299} \[ -\frac{\sqrt{a^2 x^2+1} \text{Si}\left (\tan ^{-1}(a x)\right )}{2 a^2 c \sqrt{a^2 c x^2+c}}-\frac{x}{2 a c \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}-\frac{1}{2 a^2 c \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4942

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[
((f*x)^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^(p + 1))/(b*c*d*(p + 1)), x] - Dist[(f*m)/(b*c*(p + 1)), Int[
(f*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e
, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

Rule 4902

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

Rule 4971

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q +
1/2)*Sqrt[1 + c^2*x^2])/Sqrt[d + e*x^2], Int[x^m*(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] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

Rule 4970

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(m
+ 1), Subst[Int[((a + b*x)^p*Sin[x]^m)/Cos[x]^(m + 2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d,
e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.129822, size = 63, normalized size = 0.61 \[ -\frac{\sqrt{a^2 x^2+1} \tan ^{-1}(a x)^2 \text{Si}\left (\tan ^{-1}(a x)\right )+a x+\tan ^{-1}(a x)}{2 a^2 c \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x + ArcTan[a*x] + Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*SinIntegral[ArcTan[a*x]])/(2*a^2*c*Sqrt[c + a^2*c*x^2]*A
rcTan[a*x]^2)

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Maple [C]  time = 0.331, size = 294, normalized size = 2.8 \begin{align*}{\frac{-{\frac{i}{4}}}{{c}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}{a}^{2}} \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}{\it Ei} \left ( 1,-i\arctan \left ( ax \right ) \right ){x}^{2}{a}^{2}+\arctan \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}xa+{\it Ei} \left ( 1,-i\arctan \left ( ax \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{2}-i\sqrt{{a}^{2}{x}^{2}+1}xa-i\arctan \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}-\sqrt{{a}^{2}{x}^{2}+1} \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{{\frac{i}{4}}}{{c}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}{a}^{2}} \left ( \left ( \arctan \left ( ax \right ) \right ) ^{2}{\it Ei} \left ( 1,i\arctan \left ( ax \right ) \right ){x}^{2}{a}^{2}+\arctan \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}xa+i\sqrt{{a}^{2}{x}^{2}+1}xa+{\it Ei} \left ( 1,i\arctan \left ( ax \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{2}+i\arctan \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}-\sqrt{{a}^{2}{x}^{2}+1} \right ) \sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \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)^3,x)

[Out]

-1/4*I*(arctan(a*x)^2*Ei(1,-I*arctan(a*x))*x^2*a^2+arctan(a*x)*(a^2*x^2+1)^(1/2)*x*a+Ei(1,-I*arctan(a*x))*arct
an(a*x)^2-I*(a^2*x^2+1)^(1/2)*x*a-I*arctan(a*x)*(a^2*x^2+1)^(1/2)-(a^2*x^2+1)^(1/2))/(a^2*x^2+1)^(3/2)*(c*(a*x
-I)*(a*x+I))^(1/2)/arctan(a*x)^2/c^2/a^2+1/4*I*(arctan(a*x)^2*Ei(1,I*arctan(a*x))*x^2*a^2+arctan(a*x)*(a^2*x^2
+1)^(1/2)*x*a+I*(a^2*x^2+1)^(1/2)*x*a+Ei(1,I*arctan(a*x))*arctan(a*x)^2+I*arctan(a*x)*(a^2*x^2+1)^(1/2)-(a^2*x
^2+1)^(1/2))/(a^2*x^2+1)^(3/2)*(c*(a*x-I)*(a*x+I))^(1/2)/arctan(a*x)^2/c^2/a^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \arctan \left (a x\right )^{3}}\,{d x} \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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} c x^{2} + c} x}{{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )^{3}}, x\right ) \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)^3,x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x/((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)^3), x)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \arctan \left (a x\right )^{3}}\,{d x} \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)^3,x, algorithm="giac")

[Out]

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