3.35 \(\int \frac{a+b \tanh ^{-1}(c x^3)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=414 \[ -\frac{a+b \tanh ^{-1}\left (c x^3\right )}{e (d+e x)}-\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{4 \left (c d^3-e^3\right )}-\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d-e\right ) \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{4 \left (c d^3+e^3\right )}-\frac{3 b c d^2 e^2 \log (d+e x)}{c^2 d^6-e^6}-\frac{\sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{2 \left (c^{2/3} d^2+\sqrt [3]{c} d e+e^2\right )}-\frac{b c d^2 \log \left (1-c x^3\right )}{2 e \left (c d^3+e^3\right )}+\frac{b c d^2 \log \left (c x^3+1\right )}{2 e \left (c d^3-e^3\right )}+\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d-e\right ) \log \left (1-\sqrt [3]{c} x\right )}{2 \left (c d^3+e^3\right )}+\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \log \left (\sqrt [3]{c} x+1\right )}{2 \left (c d^3-e^3\right )}-\frac{\sqrt{3} b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )}{2 \left (c d^3+e^3\right )} \]

[Out]

-(Sqrt[3]*b*c^(1/3)*ArcTan[(1 - 2*c^(1/3)*x)/Sqrt[3]])/(2*(c^(2/3)*d^2 + c^(1/3)*d*e + e^2)) - (Sqrt[3]*b*c^(1
/3)*(c^(1/3)*d + e)*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(2*(c*d^3 + e^3)) - (a + b*ArcTanh[c*x^3])/(e*(d + e*x)
) + (b*c^(1/3)*(c^(1/3)*d - e)*Log[1 - c^(1/3)*x])/(2*(c*d^3 + e^3)) + (b*c^(1/3)*(c^(1/3)*d + e)*Log[1 + c^(1
/3)*x])/(2*(c*d^3 - e^3)) - (3*b*c*d^2*e^2*Log[d + e*x])/(c^2*d^6 - e^6) - (b*c^(1/3)*(c^(1/3)*d + e)*Log[1 -
c^(1/3)*x + c^(2/3)*x^2])/(4*(c*d^3 - e^3)) - (b*c^(1/3)*(c^(1/3)*d - e)*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(4*
(c*d^3 + e^3)) - (b*c*d^2*Log[1 - c*x^3])/(2*e*(c*d^3 + e^3)) + (b*c*d^2*Log[1 + c*x^3])/(2*e*(c*d^3 - e^3))

________________________________________________________________________________________

Rubi [A]  time = 0.772685, antiderivative size = 414, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 12, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6273, 12, 6725, 1871, 1861, 31, 634, 617, 204, 628, 260, 1860} \[ -\frac{a+b \tanh ^{-1}\left (c x^3\right )}{e (d+e x)}-\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{4 \left (c d^3-e^3\right )}-\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d-e\right ) \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{4 \left (c d^3+e^3\right )}-\frac{3 b c d^2 e^2 \log (d+e x)}{c^2 d^6-e^6}-\frac{\sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{1-2 \sqrt [3]{c} x}{\sqrt{3}}\right )}{2 \left (c^{2/3} d^2+\sqrt [3]{c} d e+e^2\right )}-\frac{b c d^2 \log \left (1-c x^3\right )}{2 e \left (c d^3+e^3\right )}+\frac{b c d^2 \log \left (c x^3+1\right )}{2 e \left (c d^3-e^3\right )}+\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d-e\right ) \log \left (1-\sqrt [3]{c} x\right )}{2 \left (c d^3+e^3\right )}+\frac{b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \log \left (\sqrt [3]{c} x+1\right )}{2 \left (c d^3-e^3\right )}-\frac{\sqrt{3} b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )}{2 \left (c d^3+e^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3]*b*c^(1/3)*ArcTan[(1 - 2*c^(1/3)*x)/Sqrt[3]])/(2*(c^(2/3)*d^2 + c^(1/3)*d*e + e^2)) - (Sqrt[3]*b*c^(1
/3)*(c^(1/3)*d + e)*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(2*(c*d^3 + e^3)) - (a + b*ArcTanh[c*x^3])/(e*(d + e*x)
) + (b*c^(1/3)*(c^(1/3)*d - e)*Log[1 - c^(1/3)*x])/(2*(c*d^3 + e^3)) + (b*c^(1/3)*(c^(1/3)*d + e)*Log[1 + c^(1
/3)*x])/(2*(c*d^3 - e^3)) - (3*b*c*d^2*e^2*Log[d + e*x])/(c^2*d^6 - e^6) - (b*c^(1/3)*(c^(1/3)*d + e)*Log[1 -
c^(1/3)*x + c^(2/3)*x^2])/(4*(c*d^3 - e^3)) - (b*c^(1/3)*(c^(1/3)*d - e)*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(4*
(c*d^3 + e^3)) - (b*c*d^2*Log[1 - c*x^3])/(2*e*(c*d^3 + e^3)) + (b*c*d^2*Log[1 + c*x^3])/(2*e*(c*d^3 - e^3))

Rule 6273

Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1861

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 3]], s = Denominato
r[Rt[-(a/b), 3]]}, Dist[(r*(B*r + A*s))/(3*a*s), Int[1/(r - s*x), x], x] - Dist[r/(3*a*s), Int[(r*(B*r - 2*A*s
) - s*(B*r + A*s)*x)/(r^2 + r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && Ne
gQ[a/b]

Rule 31

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

Rule 634

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 617

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 204

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

Rule 628

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 260

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

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rubi steps

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

Mathematica [A]  time = 0.558207, size = 534, normalized size = 1.29 \[ \frac{1}{4} \left (-\frac{4 a}{e (d+e x)}+\frac{2 b c d^2 e^2 \log \left (1-c^2 x^6\right )}{c^2 d^6-e^6}+\frac{b \sqrt [3]{c} \left (-c^{4/3} d^4 e+2 c^{5/3} d^5-c d^3 e^2-\sqrt [3]{c} d e^4-e^5\right ) \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{c^2 d^6 e-e^7}+\frac{b \sqrt [3]{c} \left (c^{4/3} d^4 e+2 c^{5/3} d^5-c d^3 e^2-\sqrt [3]{c} d e^4+e^5\right ) \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{e^7-c^2 d^6 e}-\frac{12 b c d^2 e^2 \log (d+e x)}{c^2 d^6-e^6}+\frac{2 b \sqrt [3]{c} \left (-c^{4/3} d^4 e+c^{5/3} d^5+c d^3 e^2+\sqrt [3]{c} d e^4-e^5\right ) \log \left (1-\sqrt [3]{c} x\right )}{e^7-c^2 d^6 e}-\frac{2 b \sqrt [3]{c} \left (c^{4/3} d^4 e+c^{5/3} d^5+c d^3 e^2+\sqrt [3]{c} d e^4+e^5\right ) \log \left (\sqrt [3]{c} x+1\right )}{e^7-c^2 d^6 e}+\frac{2 \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x-1}{\sqrt{3}}\right )}{c^{2/3} d^2+\sqrt [3]{c} d e+e^2}-\frac{2 \sqrt{3} b \sqrt [3]{c} \left (\sqrt [3]{c} d+e\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )}{c d^3+e^3}-\frac{4 b \tanh ^{-1}\left (c x^3\right )}{e (d+e x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a)/(e*(d + e*x)) + (2*Sqrt[3]*b*c^(1/3)*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(c^(2/3)*d^2 + c^(1/3)*d*e +
e^2) - (2*Sqrt[3]*b*c^(1/3)*(c^(1/3)*d + e)*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(c*d^3 + e^3) - (4*b*ArcTanh[c*
x^3])/(e*(d + e*x)) + (2*b*c^(1/3)*(c^(5/3)*d^5 - c^(4/3)*d^4*e + c*d^3*e^2 + c^(1/3)*d*e^4 - e^5)*Log[1 - c^(
1/3)*x])/(-(c^2*d^6*e) + e^7) - (2*b*c^(1/3)*(c^(5/3)*d^5 + c^(4/3)*d^4*e + c*d^3*e^2 + c^(1/3)*d*e^4 + e^5)*L
og[1 + c^(1/3)*x])/(-(c^2*d^6*e) + e^7) - (12*b*c*d^2*e^2*Log[d + e*x])/(c^2*d^6 - e^6) + (b*c^(1/3)*(2*c^(5/3
)*d^5 - c^(4/3)*d^4*e - c*d^3*e^2 - c^(1/3)*d*e^4 - e^5)*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(c^2*d^6*e - e^7) +
(b*c^(1/3)*(2*c^(5/3)*d^5 + c^(4/3)*d^4*e - c*d^3*e^2 - c^(1/3)*d*e^4 + e^5)*Log[1 + c^(1/3)*x + c^(2/3)*x^2]
)/(-(c^2*d^6*e) + e^7) + (2*b*c*d^2*e^2*Log[1 - c^2*x^6])/(c^2*d^6 - e^6))/4

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Maple [A]  time = 0.039, size = 591, normalized size = 1.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/(e*x+d)^2,x)

[Out]

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

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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^3))/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out