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

Optimal. Leaf size=39 \[ \frac {\sqrt {a^2 x^2+1} \text {Ci}\left (\tan ^{-1}(a x)\right )}{a c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4905, 4904, 3302} \[ \frac {\sqrt {a^2 x^2+1} \text {CosIntegral}\left (\tan ^{-1}(a x)\right )}{a c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 37, normalized size = 0.95 \[ \frac {\left (a^2 x^2+1\right )^{3/2} \text {Ci}\left (\tan ^{-1}(a x)\right )}{a \left (c \left (a^2 x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]),x]

[Out]

((1 + a^2*x^2)^(3/2)*CosIntegral[ArcTan[a*x]])/(a*(c*(1 + a^2*x^2))^(3/2))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c}}{{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.63, size = 136, normalized size = 3.49 \[ -\frac {i \mathrm {csgn}\left (\arctan \left (a x \right )\right ) \mathrm {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{2} a}+\frac {i \mathrm {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{2} a}+\frac {\Ci \left (\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \arctan \left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x),x, algorithm="maxima")

[Out]

integrate(1/((a^2*c*x^2 + c)^(3/2)*arctan(a*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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