Integrand size = 16, antiderivative size = 76 \[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\frac {4 e^{-\frac {4 a}{b n}} x^4 \left (c x^n\right )^{-4/n} \operatorname {ExpIntegralEi}\left (\frac {4 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b^2 n^2}-\frac {x^4}{b n \left (a+b \log \left (c x^n\right )\right )} \] Output:
4*x^4*Ei(4*(a+b*ln(c*x^n))/b/n)/b^2/exp(4*a/b/n)/n^2/((c*x^n)^(4/n))-x^4/b /n/(a+b*ln(c*x^n))
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\frac {x^4 \left (4 e^{-\frac {4 a}{b n}} \left (c x^n\right )^{-4/n} \operatorname {ExpIntegralEi}\left (\frac {4 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )-\frac {b n}{a+b \log \left (c x^n\right )}\right )}{b^2 n^2} \] Input:
Integrate[x^3/(a + b*Log[c*x^n])^2,x]
Output:
(x^4*((4*ExpIntegralEi[(4*(a + b*Log[c*x^n]))/(b*n)])/(E^((4*a)/(b*n))*(c* x^n)^(4/n)) - (b*n)/(a + b*Log[c*x^n])))/(b^2*n^2)
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2743, 2747, 2609}
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.
\(\displaystyle \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx\) |
\(\Big \downarrow \) 2743 |
\(\displaystyle \frac {4 \int \frac {x^3}{a+b \log \left (c x^n\right )}dx}{b n}-\frac {x^4}{b n \left (a+b \log \left (c x^n\right )\right )}\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {4 x^4 \left (c x^n\right )^{-4/n} \int \frac {\left (c x^n\right )^{4/n}}{a+b \log \left (c x^n\right )}d\log \left (c x^n\right )}{b n^2}-\frac {x^4}{b n \left (a+b \log \left (c x^n\right )\right )}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {4 x^4 e^{-\frac {4 a}{b n}} \left (c x^n\right )^{-4/n} \operatorname {ExpIntegralEi}\left (\frac {4 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b^2 n^2}-\frac {x^4}{b n \left (a+b \log \left (c x^n\right )\right )}\) |
Input:
Int[x^3/(a + b*Log[c*x^n])^2,x]
Output:
(4*x^4*ExpIntegralEi[(4*(a + b*Log[c*x^n]))/(b*n)])/(b^2*E^((4*a)/(b*n))*n ^2*(c*x^n)^(4/n)) - x^4/(b*n*(a + b*Log[c*x^n]))
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol ] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Simp[(m + 1)/(b*n*(p + 1)) Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol ] :> Simp[(d*x)^(m + 1)/(d*n*(c*x^n)^((m + 1)/n)) Subst[Int[E^(((m + 1)/n )*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.65 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.66
method | result | size |
risch | \(-\frac {2 x^{4}}{\left (i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (x^{n}\right ) b +2 b \ln \left (c \right )+2 a \right ) b n}-\frac {4 x^{4} \left (x^{n}\right )^{-\frac {4}{n}} c^{-\frac {4}{n}} {\mathrm e}^{-\frac {2 \left (i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 a \right )}{n b}} \operatorname {expIntegral}_{1}\left (-4 \ln \left (x \right )-\frac {2 \left (i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 b \ln \left (c \right )+2 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a \right )}{n b}\right )}{n^{2} b^{2}}\) | \(354\) |
Input:
int(x^3/(a+b*ln(c*x^n))^2,x,method=_RETURNVERBOSE)
Output:
-2*x^4/(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n )*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*ln(x ^n)*b+2*b*ln(c)+2*a)/b/n-4/n^2/b^2*x^4*(x^n)^(-4/n)*c^(-4/n)*exp(-2*(I*Pi* b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I *Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*a)/n/b)*Ei(1,-4*l n(x)-2*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n )*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln (c)+2*b*(ln(x^n)-n*ln(x))+2*a)/n/b)
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.33 \[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=-\frac {{\left (b n x^{4} e^{\left (\frac {4 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} - 4 \, {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )} \operatorname {log\_integral}\left (x^{4} e^{\left (\frac {4 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {4 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b^{3} n^{3} \log \left (x\right ) + b^{3} n^{2} \log \left (c\right ) + a b^{2} n^{2}} \] Input:
integrate(x^3/(a+b*log(c*x^n))^2,x, algorithm="fricas")
Output:
-(b*n*x^4*e^(4*(b*log(c) + a)/(b*n)) - 4*(b*n*log(x) + b*log(c) + a)*log_i ntegral(x^4*e^(4*(b*log(c) + a)/(b*n))))*e^(-4*(b*log(c) + a)/(b*n))/(b^3* n^3*log(x) + b^3*n^2*log(c) + a*b^2*n^2)
\[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int \frac {x^{3}}{\left (a + b \log {\left (c x^{n} \right )}\right )^{2}}\, dx \] Input:
integrate(x**3/(a+b*ln(c*x**n))**2,x)
Output:
Integral(x**3/(a + b*log(c*x**n))**2, x)
\[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int { \frac {x^{3}}{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(x^3/(a+b*log(c*x^n))^2,x, algorithm="maxima")
Output:
-x^4/(b^2*n*log(c) + b^2*n*log(x^n) + a*b*n) + 4*integrate(x^3/(b^2*n*log( c) + b^2*n*log(x^n) + a*b*n), x)
Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (77) = 154\).
Time = 0.13 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.43 \[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=-\frac {b n x^{4}}{b^{3} n^{3} \log \left (x\right ) + b^{3} n^{2} \log \left (c\right ) + a b^{2} n^{2}} + \frac {4 \, b n {\rm Ei}\left (\frac {4 \, \log \left (c\right )}{n} + \frac {4 \, a}{b n} + 4 \, \log \left (x\right )\right ) e^{\left (-\frac {4 \, a}{b n}\right )} \log \left (x\right )}{{\left (b^{3} n^{3} \log \left (x\right ) + b^{3} n^{2} \log \left (c\right ) + a b^{2} n^{2}\right )} c^{\frac {4}{n}}} + \frac {4 \, b {\rm Ei}\left (\frac {4 \, \log \left (c\right )}{n} + \frac {4 \, a}{b n} + 4 \, \log \left (x\right )\right ) e^{\left (-\frac {4 \, a}{b n}\right )} \log \left (c\right )}{{\left (b^{3} n^{3} \log \left (x\right ) + b^{3} n^{2} \log \left (c\right ) + a b^{2} n^{2}\right )} c^{\frac {4}{n}}} + \frac {4 \, a {\rm Ei}\left (\frac {4 \, \log \left (c\right )}{n} + \frac {4 \, a}{b n} + 4 \, \log \left (x\right )\right ) e^{\left (-\frac {4 \, a}{b n}\right )}}{{\left (b^{3} n^{3} \log \left (x\right ) + b^{3} n^{2} \log \left (c\right ) + a b^{2} n^{2}\right )} c^{\frac {4}{n}}} \] Input:
integrate(x^3/(a+b*log(c*x^n))^2,x, algorithm="giac")
Output:
-b*n*x^4/(b^3*n^3*log(x) + b^3*n^2*log(c) + a*b^2*n^2) + 4*b*n*Ei(4*log(c) /n + 4*a/(b*n) + 4*log(x))*e^(-4*a/(b*n))*log(x)/((b^3*n^3*log(x) + b^3*n^ 2*log(c) + a*b^2*n^2)*c^(4/n)) + 4*b*Ei(4*log(c)/n + 4*a/(b*n) + 4*log(x)) *e^(-4*a/(b*n))*log(c)/((b^3*n^3*log(x) + b^3*n^2*log(c) + a*b^2*n^2)*c^(4 /n)) + 4*a*Ei(4*log(c)/n + 4*a/(b*n) + 4*log(x))*e^(-4*a/(b*n))/((b^3*n^3* log(x) + b^3*n^2*log(c) + a*b^2*n^2)*c^(4/n))
Timed out. \[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int \frac {x^3}{{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \] Input:
int(x^3/(a + b*log(c*x^n))^2,x)
Output:
int(x^3/(a + b*log(c*x^n))^2, x)
\[ \int \frac {x^3}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int \frac {x^{3}}{\mathrm {log}\left (x^{n} c \right )^{2} b^{2}+2 \,\mathrm {log}\left (x^{n} c \right ) a b +a^{2}}d x \] Input:
int(x^3/(a+b*log(c*x^n))^2,x)
Output:
int(x**3/(log(x**n*c)**2*b**2 + 2*log(x**n*c)*a*b + a**2),x)