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

Optimal. Leaf size=69 \[ \frac {\sqrt {a^2 x^2+1} \text {Ci}\left (\tan ^{-1}(a x)\right )}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {x}{a c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)} \]

[Out]

-x/a/c/arctan(a*x)/(a^2*c*x^2+c)^(1/2)+Ci(arctan(a*x))*(a^2*x^2+1)^(1/2)/a^2/c/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4942, 4905, 4904, 3302} \[ \frac {\sqrt {a^2 x^2+1} \text {CosIntegral}\left (\tan ^{-1}(a x)\right )}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {x}{a c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 55, normalized size = 0.80 \[ \frac {\sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Ci}\left (\tan ^{-1}(a x)\right )-a x}{a^2 c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x}{{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 1.12, size = 210, normalized size = 3.04 \[ -\frac {\left (\arctan \left (a x \right ) \Ei \left (1, -i \arctan \left (a x \right )\right ) x^{2} a^{2}+\Ei \left (1, -i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sqrt {a^{2} x^{2}+1}\, x a -i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right ) c^{2} a^{2}}-\frac {\left (\Ei \left (1, i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{2} a^{2}+\Ei \left (1, i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sqrt {a^{2} x^{2}+1}\, x a +i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right ) c^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \arctan \left (a x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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