3.501 \(\int (c+d x^2) \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=57 \[ \frac {\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)+\frac {d x^2}{6 a} \]

[Out]

1/6*d*x^2/a+c*x*arctanh(a*x)+1/3*d*x^3*arctanh(a*x)+1/6*(3*a^2*c+d)*ln(-a^2*x^2+1)/a^3

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Rubi [A]  time = 0.07, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5976, 1593, 444, 43} \[ \frac {\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+c x \tanh ^{-1}(a x)+\frac {d x^2}{6 a}+\frac {1}{3} d x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)*ArcTanh[a*x],x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 5976

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^q, x]}, Dist[a + b*ArcTanh[c*x], u, x] - Dist[b*c, Int[u/(1 - c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x
] && (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rubi steps

\begin {align*} \int \left (c+d x^2\right ) \tanh ^{-1}(a x) \, dx &=c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)-a \int \frac {c x+\frac {d x^3}{3}}{1-a^2 x^2} \, dx\\ &=c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)-a \int \frac {x \left (c+\frac {d x^2}{3}\right )}{1-a^2 x^2} \, dx\\ &=c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)-\frac {1}{2} a \operatorname {Subst}\left (\int \frac {c+\frac {d x}{3}}{1-a^2 x} \, dx,x,x^2\right )\\ &=c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)-\frac {1}{2} a \operatorname {Subst}\left (\int \left (-\frac {d}{3 a^2}+\frac {-3 a^2 c-d}{3 a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {d x^2}{6 a}+c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)+\frac {\left (3 a^2 c+d\right ) \log \left (1-a^2 x^2\right )}{6 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 69, normalized size = 1.21 \[ \frac {c \log \left (1-a^2 x^2\right )}{2 a}+\frac {d \log \left (1-a^2 x^2\right )}{6 a^3}+c x \tanh ^{-1}(a x)+\frac {1}{3} d x^3 \tanh ^{-1}(a x)+\frac {d x^2}{6 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)*ArcTanh[a*x],x]

[Out]

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

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fricas [A]  time = 0.50, size = 65, normalized size = 1.14 \[ \frac {a^{2} d x^{2} + {\left (3 \, a^{2} c + d\right )} \log \left (a^{2} x^{2} - 1\right ) + {\left (a^{3} d x^{3} + 3 \, a^{3} c x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*d*x^2 + (3*a^2*c + d)*log(a^2*x^2 - 1) + (a^3*d*x^3 + 3*a^3*c*x)*log(-(a*x + 1)/(a*x - 1)))/a^3

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giac [B]  time = 0.21, size = 266, normalized size = 4.67 \[ \frac {1}{3} \, a {\left (\frac {{\left (3 \, a^{2} c + d\right )} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {{\left (3 \, a^{2} c + d\right )} \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (a x + 1\right )} d}{{\left (a x - 1\right )} a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{2}} + \frac {{\left (\frac {3 \, {\left (a x + 1\right )}^{2} a^{2} c}{{\left (a x - 1\right )}^{2}} - \frac {6 \, {\left (a x + 1\right )} a^{2} c}{a x - 1} + 3 \, a^{2} c + \frac {3 \, {\left (a x + 1\right )}^{2} d}{{\left (a x - 1\right )}^{2}} + d\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*((3*a^2*c + d)*log(abs(-a*x - 1)/abs(a*x - 1))/a^4 - (3*a^2*c + d)*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^
4 + 2*(a*x + 1)*d/((a*x - 1)*a^4*((a*x + 1)/(a*x - 1) - 1)^2) + (3*(a*x + 1)^2*a^2*c/(a*x - 1)^2 - 6*(a*x + 1)
*a^2*c/(a*x - 1) + 3*a^2*c + 3*(a*x + 1)^2*d/(a*x - 1)^2 + d)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(
a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1))/(a^4*((a*x + 1)/(a*x - 1) -
1)^3))

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maple [A]  time = 0.03, size = 76, normalized size = 1.33 \[ \frac {d \,x^{3} \arctanh \left (a x \right )}{3}+c x \arctanh \left (a x \right )+\frac {d \,x^{2}}{6 a}+\frac {\ln \left (a x -1\right ) c}{2 a}+\frac {\ln \left (a x -1\right ) d}{6 a^{3}}+\frac {\ln \left (a x +1\right ) c}{2 a}+\frac {\ln \left (a x +1\right ) d}{6 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)*arctanh(a*x),x)

[Out]

1/3*d*x^3*arctanh(a*x)+c*x*arctanh(a*x)+1/6*d*x^2/a+1/2/a*ln(a*x-1)*c+1/6/a^3*ln(a*x-1)*d+1/2/a*ln(a*x+1)*c+1/
6/a^3*ln(a*x+1)*d

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maxima [A]  time = 0.32, size = 65, normalized size = 1.14 \[ \frac {1}{6} \, a {\left (\frac {d x^{2}}{a^{2}} + \frac {{\left (3 \, a^{2} c + d\right )} \log \left (a x + 1\right )}{a^{4}} + \frac {{\left (3 \, a^{2} c + d\right )} \log \left (a x - 1\right )}{a^{4}}\right )} + \frac {1}{3} \, {\left (d x^{3} + 3 \, c x\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*(d*x^2/a^2 + (3*a^2*c + d)*log(a*x + 1)/a^4 + (3*a^2*c + d)*log(a*x - 1)/a^4) + 1/3*(d*x^3 + 3*c*x)*arct
anh(a*x)

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mupad [B]  time = 0.94, size = 60, normalized size = 1.05 \[ \frac {\frac {d\,\ln \left (a^2\,x^2-1\right )}{6}+a^2\,\left (\frac {c\,\ln \left (a^2\,x^2-1\right )}{2}+\frac {d\,x^2}{6}\right )}{a^3}+\frac {d\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3}+c\,x\,\mathrm {atanh}\left (a\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)*(c + d*x^2),x)

[Out]

((d*log(a^2*x^2 - 1))/6 + a^2*((c*log(a^2*x^2 - 1))/2 + (d*x^2)/6))/a^3 + (d*x^3*atanh(a*x))/3 + c*x*atanh(a*x
)

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sympy [A]  time = 0.79, size = 73, normalized size = 1.28 \[ \begin {cases} c x \operatorname {atanh}{\left (a x \right )} + \frac {d x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {c \log {\left (x - \frac {1}{a} \right )}}{a} + \frac {c \operatorname {atanh}{\left (a x \right )}}{a} + \frac {d x^{2}}{6 a} + \frac {d \log {\left (x - \frac {1}{a} \right )}}{3 a^{3}} + \frac {d \operatorname {atanh}{\left (a x \right )}}{3 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)*atanh(a*x),x)

[Out]

Piecewise((c*x*atanh(a*x) + d*x**3*atanh(a*x)/3 + c*log(x - 1/a)/a + c*atanh(a*x)/a + d*x**2/(6*a) + d*log(x -
1/a)/(3*a**3) + d*atanh(a*x)/(3*a**3), Ne(a, 0)), (0, True))

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