3.283 \(\int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx\)

Optimal. Leaf size=371 \[ -\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{a} d+\sqrt [3]{b} c}\right )}{3 b^2}+\frac {c^3 \log (c+d x)}{3 b d^3}-\frac {c^2 x}{3 b d^2}+\frac {x^3 \log (c+d x)}{3 b}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b} \]

[Out]

-1/3*c^2*x/b/d^2+1/6*c*x^2/b/d-1/9*x^3/b+1/3*c^3*ln(d*x+c)/b/d^3+1/3*x^3*ln(d*x+c)/b-1/3*a*ln(-d*(a^(1/3)+b^(1
/3)*x)/(b^(1/3)*c-a^(1/3)*d))*ln(d*x+c)/b^2-1/3*a*ln(-d*((-1)^(2/3)*a^(1/3)+b^(1/3)*x)/(b^(1/3)*c-(-1)^(2/3)*a
^(1/3)*d))*ln(d*x+c)/b^2-1/3*a*ln((-1)^(1/3)*d*(a^(1/3)+(-1)^(2/3)*b^(1/3)*x)/(b^(1/3)*c+(-1)^(1/3)*a^(1/3)*d)
)*ln(d*x+c)/b^2-1/3*a*polylog(2,b^(1/3)*(d*x+c)/(b^(1/3)*c-a^(1/3)*d))/b^2-1/3*a*polylog(2,b^(1/3)*(d*x+c)/(b^
(1/3)*c+(-1)^(1/3)*a^(1/3)*d))/b^2-1/3*a*polylog(2,b^(1/3)*(d*x+c)/(b^(1/3)*c-(-1)^(2/3)*a^(1/3)*d))/b^2

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Rubi [A]  time = 0.60, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {266, 43, 2416, 2395, 260, 2394, 2393, 2391} \[ -\frac {a \text {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \text {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{a} d+\sqrt [3]{b} c}\right )}{3 b^2}-\frac {a \text {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{a} d+\sqrt [3]{b} c}\right )}{3 b^2}-\frac {c^2 x}{3 b d^2}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {c x^2}{6 b d}+\frac {x^3 \log (c+d x)}{3 b}-\frac {x^3}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*Log[c + d*x])/(a + b*x^3),x]

[Out]

-(c^2*x)/(3*b*d^2) + (c*x^2)/(6*b*d) - x^3/(9*b) + (c^3*Log[c + d*x])/(3*b*d^3) + (x^3*Log[c + d*x])/(3*b) - (
a*Log[-((d*(a^(1/3) + b^(1/3)*x))/(b^(1/3)*c - a^(1/3)*d))]*Log[c + d*x])/(3*b^2) - (a*Log[-((d*((-1)^(2/3)*a^
(1/3) + b^(1/3)*x))/(b^(1/3)*c - (-1)^(2/3)*a^(1/3)*d))]*Log[c + d*x])/(3*b^2) - (a*Log[((-1)^(1/3)*d*(a^(1/3)
+ (-1)^(2/3)*b^(1/3)*x))/(b^(1/3)*c + (-1)^(1/3)*a^(1/3)*d)]*Log[c + d*x])/(3*b^2) - (a*PolyLog[2, (b^(1/3)*(
c + d*x))/(b^(1/3)*c - a^(1/3)*d)])/(3*b^2) - (a*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c + (-1)^(1/3)*a^(1/3
)*d)])/(3*b^2) - (a*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - (-1)^(2/3)*a^(1/3)*d)])/(3*b^2)

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 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 266

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx &=\int \left (\frac {x^2 \log (c+d x)}{b}-\frac {a x^2 \log (c+d x)}{b \left (a+b x^3\right )}\right ) \, dx\\ &=\frac {\int x^2 \log (c+d x) \, dx}{b}-\frac {a \int \frac {x^2 \log (c+d x)}{a+b x^3} \, dx}{b}\\ &=\frac {x^3 \log (c+d x)}{3 b}-\frac {a \int \left (\frac {\log (c+d x)}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac {\log (c+d x)}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac {\log (c+d x)}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}\right ) \, dx}{b}-\frac {d \int \frac {x^3}{c+d x} \, dx}{3 b}\\ &=\frac {x^3 \log (c+d x)}{3 b}-\frac {a \int \frac {\log (c+d x)}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}-\frac {a \int \frac {\log (c+d x)}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}-\frac {a \int \frac {\log (c+d x)}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{5/3}}-\frac {d \int \left (\frac {c^2}{d^3}-\frac {c x}{d^2}+\frac {x^2}{d}-\frac {c^3}{d^3 (c+d x)}\right ) \, dx}{3 b}\\ &=-\frac {c^2 x}{3 b d^2}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {x^3 \log (c+d x)}{3 b}-\frac {a \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}+\frac {(a d) \int \frac {\log \left (\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{c+d x} \, dx}{3 b^2}+\frac {(a d) \int \frac {\log \left (\frac {d \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c-\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{c+d x} \, dx}{3 b^2}+\frac {(a d) \int \frac {\log \left (\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c+(-1)^{2/3} \sqrt [3]{a} d}\right )}{c+d x} \, dx}{3 b^2}\\ &=-\frac {c^2 x}{3 b d^2}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {x^3 \log (c+d x)}{3 b}-\frac {a \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}+\frac {a \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt [3]{b} x}{-\sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{x} \, dx,x,c+d x\right )}{3 b^2}+\frac {a \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt [3]{b} x}{-\sqrt [3]{b} c-\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{x} \, dx,x,c+d x\right )}{3 b^2}+\frac {a \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt [3]{b} x}{-\sqrt [3]{b} c+(-1)^{2/3} \sqrt [3]{a} d}\right )}{x} \, dx,x,c+d x\right )}{3 b^2}\\ &=-\frac {c^2 x}{3 b d^2}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {x^3 \log (c+d x)}{3 b}-\frac {a \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 345, normalized size = 0.93 \[ -\frac {6 a d^3 \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )+6 a d^3 \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )+6 a d^3 \text {Li}_2\left (\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )+6 a d^3 \log (c+d x) \log \left (\frac {d \left (\sqrt [3]{-1} \sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{a} d+\sqrt [3]{b} c}\right )+6 a d^3 \log (c+d x) \log \left (\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} c}\right )+6 a d^3 \log (c+d x) \log \left (\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{a} d-\sqrt [3]{b} c}\right )-6 b c^3 \log (c+d x)+6 b c^2 d x-6 b d^3 x^3 \log (c+d x)-3 b c d^2 x^2+2 b d^3 x^3}{18 b^2 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Log[c + d*x])/(a + b*x^3),x]

[Out]

-1/18*(6*b*c^2*d*x - 3*b*c*d^2*x^2 + 2*b*d^3*x^3 - 6*b*c^3*Log[c + d*x] - 6*b*d^3*x^3*Log[c + d*x] + 6*a*d^3*L
og[(d*((-1)^(1/3)*a^(1/3) - b^(1/3)*x))/(b^(1/3)*c + (-1)^(1/3)*a^(1/3)*d)]*Log[c + d*x] + 6*a*d^3*Log[(d*(a^(
1/3) + b^(1/3)*x))/(-(b^(1/3)*c) + a^(1/3)*d)]*Log[c + d*x] + 6*a*d^3*Log[(d*((-1)^(2/3)*a^(1/3) + b^(1/3)*x))
/(-(b^(1/3)*c) + (-1)^(2/3)*a^(1/3)*d)]*Log[c + d*x] + 6*a*d^3*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a^(
1/3)*d)] + 6*a*d^3*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c + (-1)^(1/3)*a^(1/3)*d)] + 6*a*d^3*PolyLog[2, (b^
(1/3)*(c + d*x))/(b^(1/3)*c - (-1)^(2/3)*a^(1/3)*d)])/(b^2*d^3)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="fricas")

[Out]

integral(x^5*log(d*x + c)/(b*x^3 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="giac")

[Out]

integrate(x^5*log(d*x + c)/(b*x^3 + a), x)

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maple [C]  time = 0.26, size = 153, normalized size = 0.41 \[ \frac {x^{3} \ln \left (d x +c \right )}{3 b}-\frac {x^{3}}{9 b}+\frac {c \,x^{2}}{6 b d}-\frac {a \left (\ln \left (\frac {-d x +\RootOf \left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )-c}{\RootOf \left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}\right ) \ln \left (d x +c \right )+\dilog \left (\frac {-d x +\RootOf \left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )-c}{\RootOf \left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}\right )\right )}{3 b^{2}}+\frac {c^{3} \ln \left (d x +c \right )}{3 b \,d^{3}}-\frac {c^{2} x}{3 b \,d^{2}}-\frac {11 c^{3}}{18 b \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*ln(d*x+c)/(b*x^3+a),x)

[Out]

1/3*x^3*ln(d*x+c)/b+1/3*c^3*ln(d*x+c)/b/d^3-1/9*x^3/b+1/6*c*x^2/b/d-1/3*c^2*x/b/d^2-11/18/d^3/b*c^3-1/3/b^2*su
m(ln(d*x+c)*ln((-d*x+_R1-c)/_R1)+dilog((-d*x+_R1-c)/_R1),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*_Z*b*c^2+a*d^3-b*c^3))
*a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="maxima")

[Out]

integrate(x^5*log(d*x + c)/(b*x^3 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\ln \left (c+d\,x\right )}{b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*log(c + d*x))/(a + b*x^3),x)

[Out]

int((x^5*log(c + d*x))/(a + b*x^3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*ln(d*x+c)/(b*x**3+a),x)

[Out]

Timed out

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