Integrand size = 19, antiderivative size = 423 \[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=-\frac {d}{2 a c x}-\frac {d^2 \log (x)}{2 a c^2}+\frac {d^2 \log (c+d x)}{2 a c^2}-\frac {\log (c+d x)}{2 a x^2}-\frac {b^{2/3} \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 a^{5/3}}-\frac {(-1)^{2/3} b^{2/3} \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)}{3 a^{5/3}}+\frac {\sqrt [3]{-1} b^{2/3} \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 a^{5/3}}-\frac {b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}+\frac {\sqrt [3]{-1} b^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}-\frac {(-1)^{2/3} b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 a^{5/3}} \] Output:
-1/2*d/a/c/x-1/2*d^2*ln(x)/a/c^2+1/2*d^2*ln(d*x+c)/a/c^2-1/2*ln(d*x+c)/a/x ^2-1/3*b^(2/3)*ln(-d*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*c-a^(1/3)*d))*ln(d*x+c)/ a^(5/3)-1/3*(-1)^(2/3)*b^(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)/a^(5/3)+1/3*(-1)^(1/3)*b^(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) /a^(5/3)-1/3*b^(2/3)*polylog(2,b^(1/3)*(d*x+c)/(b^(1/3)*c-a^(1/3)*d))/a^(5 /3)+1/3*(-1)^(1/3)*b^(2/3)*polylog(2,(-1)^(2/3)*b^(1/3)*(d*x+c)/((-1)^(2/3 )*b^(1/3)*c-a^(1/3)*d))/a^(5/3)-1/3*(-1)^(2/3)*b^(2/3)*polylog(2,(-1)^(1/3 )*b^(1/3)*(d*x+c)/((-1)^(1/3)*b^(1/3)*c+a^(1/3)*d))/a^(5/3)
Time = 0.29 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.88 \[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\frac {-\frac {3 a^{2/3} \log (c+d x)}{x^2}-2 b^{2/3} \log \left (\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)-2 (-1)^{2/3} b^{2/3} \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)+2 \sqrt [3]{-1} b^{2/3} \log \left (\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{-(-1)^{2/3} \sqrt [3]{b} c+\sqrt [3]{a} d}\right ) \log (c+d x)-\frac {3 a^{2/3} d (c+d x \log (x)-d x \log (c+d x))}{c^2 x}-2 b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )+2 \sqrt [3]{-1} b^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )-2 (-1)^{2/3} b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{6 a^{5/3}} \] Input:
Integrate[Log[c + d*x]/(x^3*(a + b*x^3)),x]
Output:
((-3*a^(2/3)*Log[c + d*x])/x^2 - 2*b^(2/3)*Log[(d*(a^(1/3) + b^(1/3)*x))/( -(b^(1/3)*c) + a^(1/3)*d)]*Log[c + d*x] - 2*(-1)^(2/3)*b^(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] + 2*(-1)^(1/3)*b^(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*a^(2/3)*d*(c + d*x*Log [x] - d*x*Log[c + d*x]))/(c^2*x) - 2*b^(2/3)*PolyLog[2, (b^(1/3)*(c + d*x) )/(b^(1/3)*c - a^(1/3)*d)] + 2*(-1)^(1/3)*b^(2/3)*PolyLog[2, ((-1)^(2/3)*b ^(1/3)*(c + d*x))/((-1)^(2/3)*b^(1/3)*c - a^(1/3)*d)] - 2*(-1)^(2/3)*b^(2/ 3)*PolyLog[2, ((-1)^(1/3)*b^(1/3)*(c + d*x))/((-1)^(1/3)*b^(1/3)*c + a^(1/ 3)*d)])/(6*a^(5/3))
Time = 1.21 (sec) , antiderivative size = 423, 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.
| \(\displaystyle \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {\log (c+d x)}{a x^3}-\frac {b \log (c+d x)}{a \left (a+b x^3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}+\frac {\sqrt [3]{-1} b^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{b} (c+d x)}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}-\frac {(-1)^{2/3} b^{2/3} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (c+d x)}{\sqrt [3]{-1} \sqrt [3]{b} c+\sqrt [3]{a} d}\right )}{3 a^{5/3}}-\frac {b^{2/3} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}-\frac {(-1)^{2/3} b^{2/3} \log (c+d x) \log \left (\frac {d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} c}\right )}{3 a^{5/3}}+\frac {\sqrt [3]{-1} b^{2/3} \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{(-1)^{2/3} \sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 a^{5/3}}-\frac {d^2 \log (x)}{2 a c^2}+\frac {d^2 \log (c+d x)}{2 a c^2}-\frac {\log (c+d x)}{2 a x^2}-\frac {d}{2 a c x}\) |
Input:
Int[Log[c + d*x]/(x^3*(a + b*x^3)),x]
Output:
-1/2*d/(a*c*x) - (d^2*Log[x])/(2*a*c^2) + (d^2*Log[c + d*x])/(2*a*c^2) - L og[c + d*x]/(2*a*x^2) - (b^(2/3)*Log[-((d*(a^(1/3) + b^(1/3)*x))/(b^(1/3)* c - a^(1/3)*d))]*Log[c + d*x])/(3*a^(5/3)) - ((-1)^(2/3)*b^(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*a^(5/3)) + ((-1)^(1/3)*b^(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*a^(5/3)) - (b^(2/3)*PolyLog[2, (b^(1/3)*(c + d*x))/(b^(1/3)*c - a^(1/3)*d)])/(3*a^( 5/3)) + ((-1)^(1/3)*b^(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*a^(5/3)) - ((-1)^(2/3)*b^(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 *a^(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 = 1.63 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.35
| method | result | size |
| derivativedivides | \(d^{2} \left (\frac {-\frac {\ln \left (-d x \right )}{2 c^{2}}-\frac {1}{2 c d x}-\frac {\ln \left (d x +c \right ) \left (d x +c \right ) \left (-d x +c \right )}{2 c^{2} d^{2} x^{2}}}{a}-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}}{3 a}\right )\) | \(150\) |
| default | \(d^{2} \left (\frac {-\frac {\ln \left (-d x \right )}{2 c^{2}}-\frac {1}{2 c d x}-\frac {\ln \left (d x +c \right ) \left (d x +c \right ) \left (-d x +c \right )}{2 c^{2} d^{2} x^{2}}}{a}-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}}{3 a}\right )\) | \(150\) |
| risch | \(-\frac {d^{2} \ln \left (-d x \right )}{2 a \,c^{2}}-\frac {d}{2 a c x}+\frac {d^{2} \ln \left (d x +c \right )}{2 a \,c^{2}}-\frac {\ln \left (d x +c \right )}{2 a \,x^{2}}-\frac {d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} c +c^{2}}\right )}{3 a}\) | \(154\) |
Input:
int(ln(d*x+c)/x^3/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
d^2*((-1/2/c^2*ln(-d*x)-1/2/c/d/x-1/2*ln(d*x+c)*(d*x+c)*(-d*x+c)/c^2/d^2/x ^2)/a-1/3*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 )
\[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\int { \frac {\log \left (d x + c\right )}{{\left (b x^{3} + a\right )} x^{3}} \,d x } \] Input:
integrate(log(d*x+c)/x^3/(b*x^3+a),x, algorithm="fricas")
Output:
integral(log(d*x + c)/(b*x^6 + a*x^3), x)
Timed out. \[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(ln(d*x+c)/x**3/(b*x**3+a),x)
Output:
Timed out
\[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\int { \frac {\log \left (d x + c\right )}{{\left (b x^{3} + a\right )} x^{3}} \,d x } \] Input:
integrate(log(d*x+c)/x^3/(b*x^3+a),x, algorithm="maxima")
Output:
integrate(log(d*x + c)/((b*x^3 + a)*x^3), x)
\[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\int { \frac {\log \left (d x + c\right )}{{\left (b x^{3} + a\right )} x^{3}} \,d x } \] Input:
integrate(log(d*x+c)/x^3/(b*x^3+a),x, algorithm="giac")
Output:
integrate(log(d*x + c)/((b*x^3 + a)*x^3), x)
Timed out. \[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\int \frac {\ln \left (c+d\,x\right )}{x^3\,\left (b\,x^3+a\right )} \,d x \] Input:
int(log(c + d*x)/(x^3*(a + b*x^3)),x)
Output:
int(log(c + d*x)/(x^3*(a + b*x^3)), x)
\[ \int \frac {\log (c+d x)}{x^3 \left (a+b x^3\right )} \, dx=\int \frac {\mathrm {log}\left (d x +c \right )}{b \,x^{6}+a \,x^{3}}d x \] Input:
int(log(d*x+c)/x^3/(b*x^3+a),x)
Output:
int(log(c + d*x)/(a*x**3 + b*x**6),x)