3.62 \(\int \frac{a+b \tanh ^{-1}(c x^2)}{x^4} \, dx\)

Optimal. Leaf size=63 \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}-\frac{1}{3} b c^{3/2} \tan ^{-1}\left (\sqrt{c} x\right )+\frac{1}{3} b c^{3/2} \tanh ^{-1}\left (\sqrt{c} x\right )-\frac{2 b c}{3 x} \]

[Out]

(-2*b*c)/(3*x) - (b*c^(3/2)*ArcTan[Sqrt[c]*x])/3 + (b*c^(3/2)*ArcTanh[Sqrt[c]*x])/3 - (a + b*ArcTanh[c*x^2])/(
3*x^3)

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Rubi [A]  time = 0.0334463, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6097, 325, 298, 203, 206} \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}-\frac{1}{3} b c^{3/2} \tan ^{-1}\left (\sqrt{c} x\right )+\frac{1}{3} b c^{3/2} \tanh ^{-1}\left (\sqrt{c} x\right )-\frac{2 b c}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x^4,x]

[Out]

(-2*b*c)/(3*x) - (b*c^(3/2)*ArcTan[Sqrt[c]*x])/3 + (b*c^(3/2)*ArcTanh[Sqrt[c]*x])/3 - (a + b*ArcTanh[c*x^2])/(
3*x^3)

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (c x^2\right )}{x^4} \, dx &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} (2 b c) \int \frac{1}{x^2 \left (1-c^2 x^4\right )} \, dx\\ &=-\frac{2 b c}{3 x}-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} \left (2 b c^3\right ) \int \frac{x^2}{1-c^2 x^4} \, dx\\ &=-\frac{2 b c}{3 x}-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} \left (b c^2\right ) \int \frac{1}{1-c x^2} \, dx-\frac{1}{3} \left (b c^2\right ) \int \frac{1}{1+c x^2} \, dx\\ &=-\frac{2 b c}{3 x}-\frac{1}{3} b c^{3/2} \tan ^{-1}\left (\sqrt{c} x\right )+\frac{1}{3} b c^{3/2} \tanh ^{-1}\left (\sqrt{c} x\right )-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0273203, size = 91, normalized size = 1.44 \[ -\frac{a}{3 x^3}-\frac{1}{6} b c^{3/2} \log \left (1-\sqrt{c} x\right )+\frac{1}{6} b c^{3/2} \log \left (\sqrt{c} x+1\right )-\frac{1}{3} b c^{3/2} \tan ^{-1}\left (\sqrt{c} x\right )-\frac{b \tanh ^{-1}\left (c x^2\right )}{3 x^3}-\frac{2 b c}{3 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])/x^4,x]

[Out]

-a/(3*x^3) - (2*b*c)/(3*x) - (b*c^(3/2)*ArcTan[Sqrt[c]*x])/3 - (b*ArcTanh[c*x^2])/(3*x^3) - (b*c^(3/2)*Log[1 -
Sqrt[c]*x])/6 + (b*c^(3/2)*Log[1 + Sqrt[c]*x])/6

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Maple [A]  time = 0.014, size = 51, normalized size = 0.8 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b{\it Artanh} \left ( c{x}^{2} \right ) }{3\,{x}^{3}}}-{\frac{b}{3}{c}^{{\frac{3}{2}}}\arctan \left ( x\sqrt{c} \right ) }-{\frac{2\,bc}{3\,x}}+{\frac{b}{3}{c}^{{\frac{3}{2}}}{\it Artanh} \left ( x\sqrt{c} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))/x^4,x)

[Out]

-1/3*a/x^3-1/3*b/x^3*arctanh(c*x^2)-1/3*b*c^(3/2)*arctan(x*c^(1/2))-2/3*b*c/x+1/3*b*c^(3/2)*arctanh(x*c^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13394, size = 440, normalized size = 6.98 \begin{align*} \left [-\frac{2 \, b c^{\frac{3}{2}} x^{3} \arctan \left (\sqrt{c} x\right ) - b c^{\frac{3}{2}} x^{3} \log \left (\frac{c x^{2} + 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right ) + 4 \, b c x^{2} + b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{6 \, x^{3}}, -\frac{2 \, b \sqrt{-c} c x^{3} \arctan \left (\sqrt{-c} x\right ) - b \sqrt{-c} c x^{3} \log \left (\frac{c x^{2} - 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right ) + 4 \, b c x^{2} + b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{6 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^4,x, algorithm="fricas")

[Out]

[-1/6*(2*b*c^(3/2)*x^3*arctan(sqrt(c)*x) - b*c^(3/2)*x^3*log((c*x^2 + 2*sqrt(c)*x + 1)/(c*x^2 - 1)) + 4*b*c*x^
2 + b*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a)/x^3, -1/6*(2*b*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x) - b*sqrt(-c)*c*x^3
*log((c*x^2 - 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 4*b*c*x^2 + b*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a)/x^3]

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Sympy [A]  time = 21.2908, size = 813, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))/x**4,x)

[Out]

Piecewise((-a/(3*x**3), Eq(c, 0)), (-(a - oo*b)/(3*x**3), Eq(c, -1/x**2)), (-(a + oo*b)/(3*x**3), Eq(c, x**(-2
))), (-4*a*x**4/(12*x**7 - 12*x**3/c**2) + 4*a/(12*c**2*x**7 - 12*x**3) - b*c**3*x**7*(1/c)**(3/2)*log(x + I*s
qrt(1/c))/(12*x**7 - 12*x**3/c**2) + I*b*c**3*x**7*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2)
+ 2*b*c**2*x**7*sqrt(1/c)*log(x - I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) + 2*I*b*c**2*x**7*sqrt(1/c)*log(x - I*
sqrt(1/c))/(12*x**7 - 12*x**3/c**2) + 3*b*c**2*x**7*sqrt(1/c)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) -
3*I*b*c**2*x**7*sqrt(1/c)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) - 4*b*c**2*x**7*sqrt(1/c)*log(x - sqrt
(1/c))/(12*x**7 - 12*x**3/c**2) - 4*b*c**2*x**7*sqrt(1/c)*atanh(c*x**2)/(12*x**7 - 12*x**3/c**2) - 8*b*c*x**6/
(12*x**7 - 12*x**3/c**2) + b*c*x**3*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) - I*b*c*x**3*(1
/c)**(3/2)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) - 4*b*x**4*atanh(c*x**2)/(12*x**7 - 12*x**3/c**2) - 2
*b*x**3*sqrt(1/c)*log(x - I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) - 2*I*b*x**3*sqrt(1/c)*log(x - I*sqrt(1/c))/(1
2*x**7 - 12*x**3/c**2) - 3*b*x**3*sqrt(1/c)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) + 3*I*b*x**3*sqrt(1/
c)*log(x + I*sqrt(1/c))/(12*x**7 - 12*x**3/c**2) + 4*b*x**3*sqrt(1/c)*log(x - sqrt(1/c))/(12*x**7 - 12*x**3/c*
*2) + 4*b*x**3*sqrt(1/c)*atanh(c*x**2)/(12*x**7 - 12*x**3/c**2) + 8*b*x**2/(12*c*x**7 - 12*x**3/c) + 4*b*atanh
(c*x**2)/(12*c**2*x**7 - 12*x**3), True))

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Giac [B]  time = 1.35212, size = 132, normalized size = 2.1 \begin{align*} -\frac{1}{6} \, b c^{3}{\left (\frac{2 \, \sqrt{{\left | c \right |}} \arctan \left (x \sqrt{{\left | c \right |}}\right )}{c^{2}} - \frac{\sqrt{{\left | c \right |}} \log \left ({\left | x + \frac{1}{\sqrt{{\left | c \right |}}} \right |}\right )}{c^{2}} + \frac{\sqrt{{\left | c \right |}} \log \left ({\left | x - \frac{1}{\sqrt{{\left | c \right |}}} \right |}\right )}{c^{2}}\right )} - \frac{b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{6 \, x^{3}} - \frac{2 \, b c x^{2} + a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^4,x, algorithm="giac")

[Out]

-1/6*b*c^3*(2*sqrt(abs(c))*arctan(x*sqrt(abs(c)))/c^2 - sqrt(abs(c))*log(abs(x + 1/sqrt(abs(c))))/c^2 + sqrt(a
bs(c))*log(abs(x - 1/sqrt(abs(c))))/c^2) - 1/6*b*log(-(c*x^2 + 1)/(c*x^2 - 1))/x^3 - 1/3*(2*b*c*x^2 + a)/x^3