Integrand size = 19, antiderivative size = 416 \[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\frac {c x}{2 b d}-\frac {x^2}{4 b}-\frac {c^2 \log (c+d x)}{2 b d^2}+\frac {x^2 \log (c+d x)}{2 b}+\frac {a^{2/3} \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{5/3}}-\frac {\sqrt [3]{-1} a^{2/3} \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{5/3}}+\frac {(-1)^{2/3} a^{2/3} \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{5/3}}+\frac {a^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}+\frac {(-1)^{2/3} a^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}-\frac {\sqrt [3]{-1} a^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 b^{5/3}} \] Output:
1/2*c*x/b/d-1/4*x^2/b-1/2*c^2*ln(d*x+c)/b/d^2+1/2*x^2*ln(d*x+c)/b+1/3*a^(2 /3)*ln(-d*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*c-a^(1/3)*d))*ln(d*x+c)/b^(5/3)-1/3 *(-1)^(1/3)*a^(2/3)*ln(d*(a^(1/3)-(-1)^(1/3)*b^(1/3)*x)/((-1)^(1/3)*b^(1/3 )*c+a^(1/3)*d))*ln(d*x+c)/b^(5/3)+1/3*(-1)^(2/3)*a^(2/3)*ln(-d*(a^(1/3)+(- 1)^(2/3)*b^(1/3)*x)/((-1)^(2/3)*b^(1/3)*c-a^(1/3)*d))*ln(d*x+c)/b^(5/3)+1/ 3*a^(2/3)*polylog(2,b^(1/3)*(d*x+c)/(b^(1/3)*c-a^(1/3)*d))/b^(5/3)+1/3*(-1 )^(2/3)*a^(2/3)*polylog(2,(-1)^(2/3)*b^(1/3)*(d*x+c)/((-1)^(2/3)*b^(1/3)*c -a^(1/3)*d))/b^(5/3)-1/3*(-1)^(1/3)*a^(2/3)*polylog(2,(-1)^(1/3)*b^(1/3)*( d*x+c)/((-1)^(1/3)*b^(1/3)*c+a^(1/3)*d))/b^(5/3)
Time = 0.39 (sec) , antiderivative size = 403, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\frac {6 b^{2/3} c d x-3 b^{2/3} d^2 x^2-6 b^{2/3} c^2 \log (c+d x)+6 b^{2/3} d^2 x^2 \log (c+d x)+4 a^{2/3} d^2 \log \left (\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)-4 \sqrt [3]{-1} a^{2/3} d^2 \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)+4 (-1)^{2/3} a^{2/3} d^2 \log \left (\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{-(-1)^{2/3} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)+4 a^{2/3} d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )+4 (-1)^{2/3} a^{2/3} d^2 \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )-4 \sqrt [3]{-1} a^{2/3} d^2 \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{12 b^{5/3} d^2} \] Input:
Integrate[(x^4*Log[c + d*x])/(a + b*x^3),x]
Output:
(6*b^(2/3)*c*d*x - 3*b^(2/3)*d^2*x^2 - 6*b^(2/3)*c^2*Log[c + d*x] + 6*b^(2 /3)*d^2*x^2*Log[c + d*x] + 4*a^(2/3)*d^2*Log[(d*(a^(1/3) + b^(1/3)*x))/(-( b^(1/3)*c) + a^(1/3)*d)]*Log[c + d*x] - 4*(-1)^(1/3)*a^(2/3)*d^2*Log[(d*(a ^(1/3) - (-1)^(1/3)*b^(1/3)*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)]*Log[c + d*x] + 4*(-1)^(2/3)*a^(2/3)*d^2*Log[(d*(a^(1/3) + (-1)^(2/3)*b^(1/3)*x)) /(-((-1)^(2/3)*b^(1/3)*c) + a^(1/3)*d)]*Log[c + d*x] + 4*a^(2/3)*d^2*PolyL og[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a^(1/3)*d)] + 4*(-1)^(2/3)*a^(2/3)* d^2*PolyLog[2, ((-1)^(2/3)*b^(1/3)*(c + d*x))/((-1)^(2/3)*b^(1/3)*c - a^(1 /3)*d)] - 4*(-1)^(1/3)*a^(2/3)*d^2*PolyLog[2, ((-1)^(1/3)*b^(1/3)*(c + d*x ))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)])/(12*b^(5/3)*d^2)
Time = 1.52 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2863, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {x \log (c+d x)}{b}-\frac {a x \log (c+d x)}{b \left (a+b x^3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}+\frac {(-1)^{2/3} a^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}-\frac {\sqrt [3]{-1} a^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 b^{5/3}}+\frac {a^{2/3} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}-\frac {\sqrt [3]{-1} a^{2/3} \log (c+d x) \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} c}\right )}{3 b^{5/3}}+\frac {(-1)^{2/3} a^{2/3} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{5/3}}-\frac {c^2 \log (c+d x)}{2 b d^2}+\frac {x^2 \log (c+d x)}{2 b}+\frac {c x}{2 b d}-\frac {x^2}{4 b}\) |
Input:
Int[(x^4*Log[c + d*x])/(a + b*x^3),x]
Output:
(c*x)/(2*b*d) - x^2/(4*b) - (c^2*Log[c + d*x])/(2*b*d^2) + (x^2*Log[c + d* x])/(2*b) + (a^(2/3)*Log[-((d*(a^(1/3) + b^(1/3)*x))/(b^(1/3)*c - a^(1/3)* d))]*Log[c + d*x])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*Log[(d*(a^(1/3) - (-1 )^(1/3)*b^(1/3)*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)]*Log[c + d*x])/(3*b ^(5/3)) + ((-1)^(2/3)*a^(2/3)*Log[-((d*(a^(1/3) + (-1)^(2/3)*b^(1/3)*x))/( (-1)^(2/3)*b^(1/3)*c - a^(1/3)*d))]*Log[c + d*x])/(3*b^(5/3)) + (a^(2/3)*P olyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a^(1/3)*d)])/(3*b^(5/3)) + ((-1 )^(2/3)*a^(2/3)*PolyLog[2, ((-1)^(2/3)*b^(1/3)*(c + d*x))/((-1)^(2/3)*b^(1 /3)*c - a^(1/3)*d)])/(3*b^(5/3)) - ((-1)^(1/3)*a^(2/3)*PolyLog[2, ((-1)^(1 /3)*b^(1/3)*(c + d*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)])/(3*b^(5/3))
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.49 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.36
| method | result | size |
| risch | \(\frac {x^{2} \ln \left (d x +c \right )}{2 b}-\frac {c^{2} \ln \left (d x +c \right )}{2 b \,d^{2}}-\frac {x^{2}}{4 b}+\frac {c x}{2 d b}+\frac {3 c^{2}}{4 d^{2} b}+\frac {d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{-\textit {\_R1} +c}\right ) a}{3 b^{2}}\) | \(148\) |
| derivativedivides | \(\frac {-\frac {\left (-\frac {\left (d x +c \right )^{2} \ln \left (d x +c \right )}{2}+\frac {\left (d x +c \right )^{2}}{4}+c \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )\right ) d^{3}}{b}+\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{-\textit {\_R1} +c}\right ) a \,d^{6}}{3 b^{2}}}{d^{5}}\) | \(149\) |
| default | \(\frac {-\frac {\left (-\frac {\left (d x +c \right )^{2} \ln \left (d x +c \right )}{2}+\frac {\left (d x +c \right )^{2}}{4}+c \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )\right ) d^{3}}{b}+\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{-\textit {\_R1} +c}\right ) a \,d^{6}}{3 b^{2}}}{d^{5}}\) | \(149\) |
Input:
int(x^4*ln(d*x+c)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*ln(d*x+c)/b-1/2*c^2*ln(d*x+c)/b/d^2-1/4*x^2/b+1/2/d/b*c*x+3/4/d^2/ b*c^2+1/3*d/b^2*sum(1/(-_R1+c)*(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
\[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{4} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^4*log(d*x+c)/(b*x^3+a),x, algorithm="fricas")
Output:
integral(x^4*log(d*x + c)/(b*x^3 + a), x)
Timed out. \[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\text {Timed out} \] Input:
integrate(x**4*ln(d*x+c)/(b*x**3+a),x)
Output:
Timed out
\[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{4} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^4*log(d*x+c)/(b*x^3+a),x, algorithm="maxima")
Output:
integrate(x^4*log(d*x + c)/(b*x^3 + a), x)
\[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{4} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^4*log(d*x+c)/(b*x^3+a),x, algorithm="giac")
Output:
integrate(x^4*log(d*x + c)/(b*x^3 + a), x)
Timed out. \[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\int \frac {x^4\,\ln \left (c+d\,x\right )}{b\,x^3+a} \,d x \] Input:
int((x^4*log(c + d*x))/(a + b*x^3),x)
Output:
int((x^4*log(c + d*x))/(a + b*x^3), x)
\[ \int \frac {x^4 \log (c+d x)}{a+b x^3} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (d x +c \right ) x}{b \,x^{3}+a}d x \right ) a \,d^{2}-2 \,\mathrm {log}\left (d x +c \right ) c^{2}+2 \,\mathrm {log}\left (d x +c \right ) d^{2} x^{2}+2 c d x -d^{2} x^{2}}{4 b \,d^{2}} \] Input:
int(x^4*log(d*x+c)/(b*x^3+a),x)
Output:
( - 4*int((log(c + d*x)*x)/(a + b*x**3),x)*a*d**2 - 2*log(c + d*x)*c**2 + 2*log(c + d*x)*d**2*x**2 + 2*c*d*x - d**2*x**2)/(4*b*d**2)