3.3.6 \(\int \frac {\sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{x^3} \, dx\) [206]

Optimal. Leaf size=240 \[ -\frac {a \sqrt {c+a^2 c x^2}}{2 x}-\frac {\sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{2 x^2}-\frac {a^2 c \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^2 c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {i a^2 c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5066, 5082, 270, 5078, 5074} \begin {gather*} -\frac {\text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{2 x^2}-\frac {a^2 c \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 )}{\sqrt {a^2 c x^2+c}}+\frac {i a^2 c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {i a^2 c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {a \sqrt {a^2 c x^2+c}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

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

Rule 5066

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

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 5082

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

Rubi steps

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

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Mathematica [A]
time = 0.83, size = 165, normalized size = 0.69 \begin {gather*} \frac {a^2 \sqrt {c \left (1+a^2 x^2\right )} \left (-2 \cot \left (\frac {1}{2} \text {ArcTan}(a x)\right )-\text {ArcTan}(a x) \csc ^2\left (\frac {1}{2} \text {ArcTan}(a x)\right )+4 \text {ArcTan}(a x) \log \left (1-e^{i \text {ArcTan}(a x)}\right )-4 \text {ArcTan}(a x) \log \left (1+e^{i \text {ArcTan}(a x)}\right )+4 i \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )-4 i \text {PolyLog}\left (2,e^{i \text {ArcTan}(a x)}\right )+\text {ArcTan}(a x) \sec ^2\left (\frac {1}{2} \text {ArcTan}(a x)\right )-2 \tan \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )}{8 \sqrt {1+a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[c*(1 + a^2*x^2)]*(-2*Cot[ArcTan[a*x]/2] - ArcTan[a*x]*Csc[ArcTan[a*x]/2]^2 + 4*ArcTan[a*x]*Log[1 - E
^(I*ArcTan[a*x])] - 4*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] + (4*I)*PolyLog[2, -E^(I*ArcTan[a*x])] - (4*I)*Po
lyLog[2, E^(I*ArcTan[a*x])] + ArcTan[a*x]*Sec[ArcTan[a*x]/2]^2 - 2*Tan[ArcTan[a*x]/2]))/(8*Sqrt[1 + a^2*x^2])

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Maple [A]
time = 0.37, size = 169, normalized size = 0.70

method	result	size
default	\(-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +\arctan \left (a x \right )\right )}{2 x^{2}}+\frac {i a^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 \sqrt {a^{2} x^{2}+1}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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 )\,\sqrt {c\,a^2\,x^2+c}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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