3.3.83 \(\int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx\) [283]

3.3.83.1 Optimal result
3.3.83.2 Mathematica [A] (verified)
3.3.83.3 Rubi [A] (verified)
3.3.83.4 Maple [C] (verified)
3.3.83.5 Fricas [F]
3.3.83.6 Sympy [F(-1)]
3.3.83.7 Maxima [F]
3.3.83.8 Giac [F]
3.3.83.9 Mupad [F(-1)]

3.3.83.1 Optimal result

Integrand size = 19, antiderivative size = 371 \[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=-\frac {c^2 x}{3 b d^2}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {x^3 \log (c+d x)}{3 b}-\frac {a \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2} \]

output
-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*l 
n(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
 
3.3.83.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.01 \[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=-\frac {c^2 x}{3 b d^2}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}+\frac {c^3 \log (c+d x)}{3 b d^3}+\frac {x^3 \log (c+d x)}{3 b}-\frac {a \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (-\frac {(-1)^{2/3} d \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right ) \log (c+d x)}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2} \]

input
Integrate[(x^5*Log[c + d*x])/(a + b*x^3),x]
 
output
-1/3*(c^2*x)/(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[-(((-1)^(2/3)*d*(a^ 
(1/3) - (-1)^(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)
 
3.3.83.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 371, 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.

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

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (\frac {x^2 \log (c+d x)}{b}-\frac {a x^2 \log (c+d x)}{b \left (a+b x^3\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c+\sqrt [3]{-1} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (c+d x)}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-\sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (-\frac {d \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} c-(-1)^{2/3} \sqrt [3]{a} d}\right )}{3 b^2}-\frac {a \log (c+d x) \log \left (\frac {\sqrt [3]{-1} d \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{a} d+\sqrt [3]{b} c}\right )}{3 b^2}+\frac {c^3 \log (c+d x)}{3 b d^3}-\frac {c^2 x}{3 b d^2}+\frac {x^3 \log (c+d x)}{3 b}+\frac {c x^2}{6 b d}-\frac {x^3}{9 b}\)

input
Int[(x^5*Log[c + d*x])/(a + b*x^3),x]
 
output
-1/3*(c^2*x)/(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)
 

3.3.83.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2863
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]
 
3.3.83.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.65 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.41

method result size
risch \(-\frac {c^{2} x}{3 b \,d^{2}}-\frac {11 c^{3}}{18 d^{3} b}+\frac {c \,x^{2}}{6 b d}+\frac {x^{3} \ln \left (d x +c \right )}{3 b}+\frac {c^{3} \ln \left (d x +c \right )}{3 b \,d^{3}}-\frac {x^{3}}{9 b}-\frac {a \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\left (\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )\right )\right )}{3 b^{2}}\) \(153\)
derivativedivides \(\frac {\frac {d^{3} \left (c^{2} \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )-2 c \left (\frac {\left (d x +c \right )^{2} \ln \left (d x +c \right )}{2}-\frac {\left (d x +c \right )^{2}}{4}\right )+\frac {\left (d x +c \right )^{3} \ln \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{3}}{9}\right )}{b}-\frac {a \,d^{6} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\left (\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )\right )\right )}{3 b^{2}}}{d^{6}}\) \(170\)
default \(\frac {\frac {d^{3} \left (c^{2} \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )-2 c \left (\frac {\left (d x +c \right )^{2} \ln \left (d x +c \right )}{2}-\frac {\left (d x +c \right )^{2}}{4}\right )+\frac {\left (d x +c \right )^{3} \ln \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{3}}{9}\right )}{b}-\frac {a \,d^{6} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\left (\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )\right )\right )}{3 b^{2}}}{d^{6}}\) \(170\)
parts \(\frac {x^{3} \ln \left (d x +c \right )}{3 b}-\frac {\ln \left (d x +c \right ) a \ln \left (b \,x^{3}+a \right )}{3 b^{2}}-\frac {d \left (\frac {x^{3}}{3 b d}-\frac {x^{2} c}{2 b \,d^{2}}+\frac {x \,c^{2}}{b \,d^{3}}-\frac {c^{3} \ln \left (d x +c \right )}{b \,d^{4}}-\frac {a \ln \left (d x +c \right ) \ln \left (b \,x^{3}+a \right )}{b^{2} d}+\frac {a \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 c b \,\textit {\_Z}^{2}+3 b \,c^{2} \textit {\_Z} +a \,d^{3}-b \,c^{3}\right )}{\sum }\left (\ln \left (d x +c \right ) \ln \left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c}{\textit {\_R1}}\right )\right )\right )}{b^{2} d}\right )}{3}\) \(193\)

input
int(x^5*ln(d*x+c)/(b*x^3+a),x,method=_RETURNVERBOSE)
 
output
-1/3*c^2*x/b/d^2-11/18/d^3/b*c^3+1/6*c*x^2/b/d+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-1/3*a/b^2*sum(ln(d*x+c)*ln((-d*x+_R1-c)/_R1)+di 
log((-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) 
)
 
3.3.83.5 Fricas [F]

\[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \]

input
integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="fricas")
 
output
integral(x^5*log(d*x + c)/(b*x^3 + a), x)
 
3.3.83.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=\text {Timed out} \]

input
integrate(x**5*ln(d*x+c)/(b*x**3+a),x)
 
output
Timed out
 
3.3.83.7 Maxima [F]

\[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \]

input
integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="maxima")
 
output
integrate(x^5*log(d*x + c)/(b*x^3 + a), x)
 
3.3.83.8 Giac [F]

\[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=\int { \frac {x^{5} \log \left (d x + c\right )}{b x^{3} + a} \,d x } \]

input
integrate(x^5*log(d*x+c)/(b*x^3+a),x, algorithm="giac")
 
output
integrate(x^5*log(d*x + c)/(b*x^3 + a), x)
 
3.3.83.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \log (c+d x)}{a+b x^3} \, dx=\int \frac {x^5\,\ln \left (c+d\,x\right )}{b\,x^3+a} \,d x \]

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