3.3.36 \(\int \frac {\text {ArcTan}(a x)}{x (c+a^2 c x^2)^{3/2}} \, dx\) [236]

Optimal. Leaf size=229 \[ -\frac {a x}{c \sqrt {c+a^2 c x^2}}+\frac {\text {ArcTan}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}} \]

[Out]

-a*x/c/(a^2*c*x^2+c)^(1/2)+arctan(a*x)/c/(a^2*c*x^2+c)^(1/2)-2*arctan(a*x)*arctanh((1+I*a*x)^(1/2)/(1-I*a*x)^(
1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)+I*polylog(2,-(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/
c/(a^2*c*x^2+c)^(1/2)-I*polylog(2,(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/c/(a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 229, normalized size of antiderivative = 1.00, 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 = {5086, 5078, 5074, 5050, 197} \begin {gather*} \frac {\text {ArcTan}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {a x}{c \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x)/(c*Sqrt[c + a^2*c*x^2])) + ArcTan[a*x]/(c*Sqrt[c + a^2*c*x^2]) - (2*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcT
anh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(c*Sqrt[c + a^2*c*x^2]) + (I*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a
*x]/Sqrt[1 - I*a*x])])/(c*Sqrt[c + a^2*c*x^2]) - (I*Sqrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*
x]])/(c*Sqrt[c + a^2*c*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 5050

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 5074

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(-2/Sqrt[d])*(a + b
*ArcTan[c*x])*ArcTanh[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]], x] + (Simp[I*(b/Sqrt[d])*PolyLog[2, -Sqrt[1 + I*c*x]/S
qrt[1 - I*c*x]], x] - Simp[I*(b/Sqrt[d])*PolyLog[2, Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]], x]) /; FreeQ[{a, b, c, d
, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 5078

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + c^2*
x^2]/Sqrt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/(x*Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[e, c^2*d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 5086

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\left (a^2 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=\frac {\tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-a \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {\sqrt {1+a^2 x^2} \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=-\frac {a x}{c \sqrt {c+a^2 c x^2}}+\frac {\tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 141, normalized size = 0.62 \begin {gather*} \frac {\sqrt {1+a^2 x^2} \left (-\frac {a x}{\sqrt {1+a^2 x^2}}+\frac {\text {ArcTan}(a x)}{\sqrt {1+a^2 x^2}}+\text {ArcTan}(a x) \log \left (1-e^{i \text {ArcTan}(a x)}\right )-\text {ArcTan}(a x) \log \left (1+e^{i \text {ArcTan}(a x)}\right )+i \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )-i \text {PolyLog}\left (2,e^{i \text {ArcTan}(a x)}\right )\right )}{c \sqrt {c \left (1+a^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*(-((a*x)/Sqrt[1 + a^2*x^2]) + ArcTan[a*x]/Sqrt[1 + a^2*x^2] + ArcTan[a*x]*Log[1 - E^(I*ArcT
an[a*x])] - ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] + I*PolyLog[2, -E^(I*ArcTan[a*x])] - I*PolyLog[2, E^(I*ArcT
an[a*x])]))/(c*Sqrt[c*(1 + a^2*x^2)])

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Maple [A]
time = 0.28, size = 232, normalized size = 1.01

method	result	size
default	\(\frac {\left (\arctan \left (a x \right )+i\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right )}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {i \left (i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2}}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*arctan(a*x)/(a^4*c^2*x^5 + 2*a^2*c^2*x^3 + c^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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