Integrand size = 19, antiderivative size = 383 \[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=-\frac {x}{b}+\frac {(c+d x) \log (c+d x)}{b d}-\frac {\sqrt [3]{a} \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{a} \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{a} \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^{4/3}}-\frac {\sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 b^{4/3}} \] Output:
-x/b+(d*x+c)*ln(d*x+c)/b/d-1/3*a^(1/3)*ln(-d*(a^(1/3)+b^(1/3)*x)/(b^(1/3)* c-a^(1/3)*d))*ln(d*x+c)/b^(4/3)-1/3*(-1)^(2/3)*a^(1/3)*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^(4/3)+1/3*( -1)^(1/3)*a^(1/3)*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^(4/3)-1/3*a^(1/3)*polylog(2,b^(1/3)*(d*x+c)/(b^ (1/3)*c-a^(1/3)*d))/b^(4/3)+1/3*(-1)^(1/3)*a^(1/3)*polylog(2,(-1)^(2/3)*b^ (1/3)*(d*x+c)/((-1)^(2/3)*b^(1/3)*c-a^(1/3)*d))/b^(4/3)-1/3*(-1)^(2/3)*a^( 1/3)*polylog(2,(-1)^(1/3)*b^(1/3)*(d*x+c)/((-1)^(1/3)*b^(1/3)*c+a^(1/3)*d) )/b^(4/3)
Time = 0.19 (sec) , antiderivative size = 369, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\frac {-3 \sqrt [3]{b} d x+3 \sqrt [3]{b} c \log (c+d x)+3 \sqrt [3]{b} d x \log (c+d x)-\sqrt [3]{a} d \log \left (\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)-(-1)^{2/3} \sqrt [3]{a} d \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)+\sqrt [3]{-1} \sqrt [3]{a} d \log \left (\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{-(-1)^{2/3} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)-\sqrt [3]{a} d \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )+\sqrt [3]{-1} \sqrt [3]{a} d \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )-(-1)^{2/3} \sqrt [3]{a} d \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 b^{4/3} d} \] Input:
Integrate[(x^3*Log[c + d*x])/(a + b*x^3),x]
Output:
(-3*b^(1/3)*d*x + 3*b^(1/3)*c*Log[c + d*x] + 3*b^(1/3)*d*x*Log[c + d*x] - a^(1/3)*d*Log[(d*(a^(1/3) + b^(1/3)*x))/(-(b^(1/3)*c) + a^(1/3)*d)]*Log[c + d*x] - (-1)^(2/3)*a^(1/3)*d*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] + (-1)^(1/3)*a^(1/3)*d*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] - a^(1/3)*d*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a^(1 /3)*d)] + (-1)^(1/3)*a^(1/3)*d*PolyLog[2, ((-1)^(2/3)*b^(1/3)*(c + d*x))/( (-1)^(2/3)*b^(1/3)*c - a^(1/3)*d)] - (-1)^(2/3)*a^(1/3)*d*PolyLog[2, ((-1) ^(1/3)*b^(1/3)*(c + d*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)])/(3*b^(4/3)* d)
Time = 1.17 (sec) , antiderivative size = 383, 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.
| \(\displaystyle \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {\log (c+d x)}{b}-\frac {a \log (c+d x)}{b \left (a+b x^3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{a} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 b^{4/3}}-\frac {\sqrt [3]{a} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{a} \log (c+d x) \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} c}\right )}{3 b^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{a} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^{4/3}}+\frac {(c+d x) \log (c+d x)}{b d}-\frac {x}{b}\) |
Input:
Int[(x^3*Log[c + d*x])/(a + b*x^3),x]
Output:
-(x/b) + ((c + d*x)*Log[c + d*x])/(b*d) - (a^(1/3)*Log[-((d*(a^(1/3) + b^( 1/3)*x))/(b^(1/3)*c - a^(1/3)*d))]*Log[c + d*x])/(3*b^(4/3)) - ((-1)^(2/3) *a^(1/3)*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^(4/3)) + ((-1)^(1/3)*a^(1/3)*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^(4/3)) - (a^(1/3)*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a ^(1/3)*d)])/(3*b^(4/3)) + ((-1)^(1/3)*a^(1/3)*PolyLog[2, ((-1)^(2/3)*b^(1/ 3)*(c + d*x))/((-1)^(2/3)*b^(1/3)*c - a^(1/3)*d)])/(3*b^(4/3)) - ((-1)^(2/ 3)*a^(1/3)*PolyLog[2, ((-1)^(1/3)*b^(1/3)*(c + d*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/3)*d)])/(3*b^(4/3))
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 = 1.44 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.33
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right ) d^{3}}{b}-\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right ) a \,d^{6}}{3 b^{2}}}{d^{4}}\) | \(127\) |
| default | \(\frac {\frac {\left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right ) d^{3}}{b}-\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right ) a \,d^{6}}{3 b^{2}}}{d^{4}}\) | \(127\) |
| risch | \(\frac {x \ln \left (d x +c \right )}{b}+\frac {\ln \left (d x +c \right ) c}{d b}-\frac {x}{b}-\frac {c}{b d}-\frac {d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right ) a}{3 b^{2}}\) | \(136\) |
Input:
int(x^3*ln(d*x+c)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
1/d^4*(((d*x+c)*ln(d*x+c)-d*x-c)*d^3/b-1/3/b^2*sum(1/(_R1^2-2*_R1*c+c^2)*( 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*d^6)
\[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{3} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^3*log(d*x+c)/(b*x^3+a),x, algorithm="fricas")
Output:
integral(x^3*log(d*x + c)/(b*x^3 + a), x)
Timed out. \[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\text {Timed out} \] Input:
integrate(x**3*ln(d*x+c)/(b*x**3+a),x)
Output:
Timed out
\[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{3} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^3*log(d*x+c)/(b*x^3+a),x, algorithm="maxima")
Output:
integrate(x^3*log(d*x + c)/(b*x^3 + a), x)
\[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{3} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \] Input:
integrate(x^3*log(d*x+c)/(b*x^3+a),x, algorithm="giac")
Output:
integrate(x^3*log(d*x + c)/(b*x^3 + a), x)
Timed out. \[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\int \frac {x^3\,\ln \left (c+d\,x\right )}{b\,x^3+a} \,d x \] Input:
int((x^3*log(c + d*x))/(a + b*x^3),x)
Output:
int((x^3*log(c + d*x))/(a + b*x^3), x)
\[ \int \frac {x^3 \log (c+d x)}{a+b x^3} \, dx=\frac {-\left (\int \frac {\mathrm {log}\left (d x +c \right )}{b d \,x^{4}+b c \,x^{3}+a d x +a c}d x \right ) a c d -\left (\int \frac {\mathrm {log}\left (d x +c \right ) x}{b d \,x^{4}+b c \,x^{3}+a d x +a c}d x \right ) a \,d^{2}+\mathrm {log}\left (d x +c \right ) c +\mathrm {log}\left (d x +c \right ) d x -d x}{b d} \] Input:
int(x^3*log(d*x+c)/(b*x^3+a),x)
Output:
( - int(log(c + d*x)/(a*c + a*d*x + b*c*x**3 + b*d*x**4),x)*a*c*d - int((l og(c + d*x)*x)/(a*c + a*d*x + b*c*x**3 + b*d*x**4),x)*a*d**2 + log(c + d*x )*c + log(c + d*x)*d*x - d*x)/(b*d)