3.2.14 \(\int \frac {a+b \tanh ^{-1}(c x^3)}{x^2} \, dx\) [114]

Optimal. Leaf size=104 \[ \frac {1}{2} \sqrt {3} b \sqrt [3]{c} \text {ArcTan}\left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}-\frac {1}{2} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right ) \]

[Out]

(-a-b*arctanh(c*x^3))/x-1/2*b*c^(1/3)*ln(1-c^(2/3)*x^2)+1/4*b*c^(1/3)*ln(1+c^(2/3)*x^2+c^(4/3)*x^4)+1/2*b*c^(1
/3)*arctan(1/3*(1+2*c^(2/3)*x^2)*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6037, 281, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}+\frac {1}{2} \sqrt {3} b \sqrt [3]{c} \text {ArcTan}\left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )-\frac {1}{2} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^2,x]

[Out]

(Sqrt[3]*b*c^(1/3)*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/2 - (a + b*ArcTanh[c*x^3])/x - (b*c^(1/3)*Log[1 - c^(2
/3)*x^2])/2 + (b*c^(1/3)*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 210

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c x^3\right )}{x^2} \, dx &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}+(3 b c) \int \frac {x}{1-c^2 x^6} \, dx\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}+\frac {1}{2} (3 b c) \text {Subst}\left (\int \frac {1}{1-c^2 x^3} \, dx,x,x^2\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}+\frac {1}{2} (b c) \text {Subst}\left (\int \frac {1}{1-c^{2/3} x} \, dx,x,x^2\right )+\frac {1}{2} (b c) \text {Subst}\left (\int \frac {2+c^{2/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}-\frac {1}{2} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2\right )+\frac {1}{4} \left (b \sqrt [3]{c}\right ) \text {Subst}\left (\int \frac {c^{2/3}+2 c^{4/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )+\frac {1}{4} (3 b c) \text {Subst}\left (\int \frac {1}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}-\frac {1}{2} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )-\frac {1}{2} \left (3 b \sqrt [3]{c}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 c^{2/3} x^2\right )\\ &=\frac {1}{2} \sqrt {3} b \sqrt [3]{c} \tan ^{-1}\left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{x}-\frac {1}{2} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 183, normalized size = 1.76 \begin {gather*} -\frac {a}{x}+\frac {1}{2} \sqrt {3} b \sqrt [3]{c} \text {ArcTan}\left (\frac {-1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {1}{2} \sqrt {3} b \sqrt [3]{c} \text {ArcTan}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {b \tanh ^{-1}\left (c x^3\right )}{x}-\frac {1}{2} b \sqrt [3]{c} \log \left (1-\sqrt [3]{c} x\right )-\frac {1}{2} b \sqrt [3]{c} \log \left (1+\sqrt [3]{c} x\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )+\frac {1}{4} b \sqrt [3]{c} \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/x) + (Sqrt[3]*b*c^(1/3)*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/2 - (Sqrt[3]*b*c^(1/3)*ArcTan[(1 + 2*c^(1/3)*x
)/Sqrt[3]])/2 - (b*ArcTanh[c*x^3])/x - (b*c^(1/3)*Log[1 - c^(1/3)*x])/2 - (b*c^(1/3)*Log[1 + c^(1/3)*x])/2 + (
b*c^(1/3)*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/4 + (b*c^(1/3)*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/4

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Maple [A]
time = 0.04, size = 105, normalized size = 1.01

method	result	size
default	\(-\frac {a}{x}-\frac {b \arctanh \left (c \,x^{3}\right )}{x}-\frac {b \ln \left (x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}}+\frac {b \ln \left (x^{4}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}} x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}}\)	\(105\)
risch	\(-\frac {b \ln \left (c \,x^{3}+1\right )}{2 x}-\frac {a}{x}+\frac {b \ln \left (-c \,x^{3}+1\right )}{2 x}-\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 \left (\frac {1}{c}\right )^{\frac {1}{3}}}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/x-b/x*arctanh(c*x^3)-1/2*b/c/(1/c^2)^(2/3)*ln(x^2-(1/c^2)^(1/3))+1/4*b/c/(1/c^2)^(2/3)*ln(x^4+(1/c^2)^(1/3)
*x^2+(1/c^2)^(2/3))+1/2*b/c/(1/c^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c^2)^(1/3)*x^2+1))

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Maxima [A]
time = 0.47, size = 94, normalized size = 0.90 \begin {gather*} \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {2}{3}}} + \frac {\log \left (c^{\frac {4}{3}} x^{4} + c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {2}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} - 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {2}{3}}}\right )} - \frac {4 \, \operatorname {artanh}\left (c x^{3}\right )}{x}\right )} b - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(4/3)*x^2 + c^(2/3))/c^(2/3))/c^(2/3) + log(c^(4/3)*x^4 + c^(2/3)*x^
2 + 1)/c^(2/3) - 2*log((c^(2/3)*x^2 - 1)/c^(2/3))/c^(2/3)) - 4*arctanh(c*x^3)/x)*b - a/x

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Fricas [A]
time = 0.36, size = 117, normalized size = 1.12 \begin {gather*} -\frac {2 \, \sqrt {3} b \left (-c\right )^{\frac {1}{3}} x \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-c\right )^{\frac {2}{3}} x^{2} + \frac {1}{3} \, \sqrt {3}\right ) + b \left (-c\right )^{\frac {1}{3}} x \log \left (c^{2} x^{4} - \left (-c\right )^{\frac {1}{3}} c x^{2} + \left (-c\right )^{\frac {2}{3}}\right ) - 2 \, b \left (-c\right )^{\frac {1}{3}} x \log \left (c x^{2} + \left (-c\right )^{\frac {1}{3}}\right ) + 2 \, b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.42, size = 106, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, b c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )} {\left | c \right |}^{\frac {2}{3}}\right )}{{\left | c \right |}^{\frac {2}{3}}} + \frac {\log \left (x^{4} + \frac {x^{2}}{{\left | c \right |}^{\frac {2}{3}}} + \frac {1}{{\left | c \right |}^{\frac {4}{3}}}\right )}{{\left | c \right |}^{\frac {2}{3}}} - \frac {2 \, \log \left ({\left | x^{2} - \frac {1}{{\left | c \right |}^{\frac {2}{3}}} \right |}\right )}{{\left | c \right |}^{\frac {2}{3}}}\right )} - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{2 \, x} - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1/abs(c)^(2/3))*abs(c)^(2/3))/abs(c)^(2/3) + log(x^4 + x^2/abs(
c)^(2/3) + 1/abs(c)^(4/3))/abs(c)^(2/3) - 2*log(abs(x^2 - 1/abs(c)^(2/3)))/abs(c)^(2/3)) - 1/2*b*log(-(c*x^3 +
1)/(c*x^3 - 1))/x - a/x

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Mupad [B]
time = 2.39, size = 117, normalized size = 1.12 \begin {gather*} \frac {b\,\ln \left (1-c\,x^3\right )}{2\,x}-\frac {b\,c^{1/3}\,\ln \left (1-c^{2/3}\,x^2\right )}{2}-\frac {b\,\ln \left (c\,x^3+1\right )}{2\,x}-\frac {a}{x}-\frac {b\,c^{1/3}\,\ln \left (-\sqrt {3}-c^{2/3}\,x^2\,2{}\mathrm {i}-\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}+\frac {b\,c^{1/3}\,\ln \left (-\sqrt {3}+c^{2/3}\,x^2\,2{}\mathrm {i}+1{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))/x^2,x)

[Out]

(b*log(1 - c*x^3))/(2*x) - (b*c^(1/3)*log(1 - c^(2/3)*x^2))/2 - (b*log(c*x^3 + 1))/(2*x) - a/x - (b*c^(1/3)*lo
g(- 3^(1/2) - c^(2/3)*x^2*2i - 1i)*(3^(1/2)*1i - 1))/4 + (b*c^(1/3)*log(c^(2/3)*x^2*2i - 3^(1/2) + 1i)*(3^(1/2
)*1i + 1))/4

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