Integrand size = 10, antiderivative size = 77 \[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=-\frac {a^2 \operatorname {ExpIntegralEi}(2 \log (a+b x))}{b^3}+\frac {a \operatorname {ExpIntegralEi}(3 \log (a+b x))}{b^3}-\frac {\operatorname {ExpIntegralEi}(4 \log (a+b x))}{3 b^3}+\frac {a^3 \operatorname {LogIntegral}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \operatorname {LogIntegral}(a+b x) \] Output:
-a^2*Ei(2*ln(b*x+a))/b^3+a*Ei(3*ln(b*x+a))/b^3-1/3*Ei(4*ln(b*x+a))/b^3+1/3 *a^3*Li(b*x+a)/b^3+1/3*x^3*Li(b*x+a)
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=-\frac {-a^3 \operatorname {ExpIntegralEi}(\log (a+b x))+3 a^2 \operatorname {ExpIntegralEi}(2 \log (a+b x))-3 a \operatorname {ExpIntegralEi}(3 \log (a+b x))+\operatorname {ExpIntegralEi}(4 \log (a+b x))}{3 b^3}+\frac {1}{3} x^3 \operatorname {LogIntegral}(a+b x) \] Input:
Integrate[x^2*LogIntegral[a + b*x],x]
Output:
-1/3*(-(a^3*ExpIntegralEi[Log[a + b*x]]) + 3*a^2*ExpIntegralEi[2*Log[a + b *x]] - 3*a*ExpIntegralEi[3*Log[a + b*x]] + ExpIntegralEi[4*Log[a + b*x]])/ b^3 + (x^3*LogIntegral[a + b*x])/3
Time = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7052, 2846, 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.
\(\displaystyle \int x^2 \operatorname {LogIntegral}(a+b x) \, dx\) |
\(\Big \downarrow \) 7052 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {LogIntegral}(a+b x)-\frac {1}{3} b \int \frac {x^3}{\log (a+b x)}dx\) |
\(\Big \downarrow \) 2846 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {LogIntegral}(a+b x)-\frac {1}{3} b \int \left (-\frac {a^3}{b^3 \log (a+b x)}+\frac {3 (a+b x) a^2}{b^3 \log (a+b x)}-\frac {3 (a+b x)^2 a}{b^3 \log (a+b x)}+\frac {(a+b x)^3}{b^3 \log (a+b x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {LogIntegral}(a+b x)-\frac {1}{3} b \left (-\frac {a^3 \operatorname {LogIntegral}(a+b x)}{b^4}+\frac {3 a^2 \operatorname {ExpIntegralEi}(2 \log (a+b x))}{b^4}-\frac {3 a \operatorname {ExpIntegralEi}(3 \log (a+b x))}{b^4}+\frac {\operatorname {ExpIntegralEi}(4 \log (a+b x))}{b^4}\right )\) |
Input:
Int[x^2*LogIntegral[a + b*x],x]
Output:
(x^3*LogIntegral[a + b*x])/3 - (b*((3*a^2*ExpIntegralEi[2*Log[a + b*x]])/b ^4 - (3*a*ExpIntegralEi[3*Log[a + b*x]])/b^4 + ExpIntegralEi[4*Log[a + b*x ]]/b^4 - (a^3*LogIntegral[a + b*x])/b^4))/3
Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) ]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q/(a + b*Log[c*(d + e* x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] & & IGtQ[q, 0]
Int[LogIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] : > Simp[(c + d*x)^(m + 1)*(LogIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/(d *(m + 1)) Int[(c + d*x)^(m + 1)/Log[a + b*x], x], x] /; FreeQ[{a, b, c, d , m}, x] && NeQ[m, -1]
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95
method | result | size |
parts | \(\frac {x^{3} \operatorname {Li}\left (b x +a \right )}{3}-\frac {a^{3} \operatorname {expIntegral}_{1}\left (-\ln \left (b x +a \right )\right )-3 a^{2} \operatorname {expIntegral}_{1}\left (-2 \ln \left (b x +a \right )\right )+3 a \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (b x +a \right )\right )-\operatorname {expIntegral}_{1}\left (-4 \ln \left (b x +a \right )\right )}{3 b^{3}}\) | \(73\) |
derivativedivides | \(\frac {-\frac {\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a^{3}}{3}+\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a^{2} \left (b x +a \right )-\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a \left (b x +a \right )^{2}+\frac {\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) \left (b x +a \right )^{3}}{3}-\frac {a^{3} \operatorname {expIntegral}_{1}\left (-\ln \left (b x +a \right )\right )}{3}+a^{2} \operatorname {expIntegral}_{1}\left (-2 \ln \left (b x +a \right )\right )-a \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (b x +a \right )\right )+\frac {\operatorname {expIntegral}_{1}\left (-4 \ln \left (b x +a \right )\right )}{3}}{b^{3}}\) | \(121\) |
default | \(\frac {-\frac {\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a^{3}}{3}+\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a^{2} \left (b x +a \right )-\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) a \left (b x +a \right )^{2}+\frac {\operatorname {expIntegral}\left (\ln \left (b x +a \right )\right ) \left (b x +a \right )^{3}}{3}-\frac {a^{3} \operatorname {expIntegral}_{1}\left (-\ln \left (b x +a \right )\right )}{3}+a^{2} \operatorname {expIntegral}_{1}\left (-2 \ln \left (b x +a \right )\right )-a \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (b x +a \right )\right )+\frac {\operatorname {expIntegral}_{1}\left (-4 \ln \left (b x +a \right )\right )}{3}}{b^{3}}\) | \(121\) |
Input:
int(x^2*Li(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/3*x^3*Li(b*x+a)-1/3/b^3*(a^3*Ei(1,-ln(b*x+a))-3*a^2*Ei(1,-2*ln(b*x+a))+3 *a*Ei(1,-3*ln(b*x+a))-Ei(1,-4*ln(b*x+a)))
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.52 \[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=-\frac {3 \, a^{2} \operatorname {log\_integral}\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) - 3 \, a \operatorname {log\_integral}\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right ) - {\left (b^{3} x^{3} + a^{3}\right )} \operatorname {log\_integral}\left (b x + a\right ) + \operatorname {log\_integral}\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )}{3 \, b^{3}} \] Input:
integrate(x^2*log_integral(b*x+a),x, algorithm="fricas")
Output:
-1/3*(3*a^2*log_integral(b^2*x^2 + 2*a*b*x + a^2) - 3*a*log_integral(b^3*x ^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3) - (b^3*x^3 + a^3)*log_integral(b*x + a ) + log_integral(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)) /b^3
\[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=\int x^{2} \operatorname {Li}{\left (a + b x \right )}\, dx \] Input:
integrate(x**2*Li(b*x+a),x)
Output:
Integral(x**2*Li(a + b*x), x)
\[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=\int { x^{2} \operatorname {log\_integral}\left (b x + a\right ) \,d x } \] Input:
integrate(x^2*log_integral(b*x+a),x, algorithm="maxima")
Output:
integrate(x^2*log_integral(b*x + a), x)
\[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=\int { x^{2} \operatorname {log\_integral}\left (b x + a\right ) \,d x } \] Input:
integrate(x^2*log_integral(b*x+a),x, algorithm="giac")
Output:
integrate(x^2*log_integral(b*x + a), x)
Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.16 \[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=\int x^2\,\mathrm {logint}\left (a+b\,x\right ) \,d x \] Input:
int(x^2*logint(a + b*x),x)
Output:
int(x^2*logint(a + b*x), x)
\[ \int x^2 \operatorname {LogIntegral}(a+b x) \, dx=\int \mathit {ei} \left (\mathrm {log}\left (b x +a \right )\right ) x^{2}d x \] Input:
int(x^2*Li(b*x+a),x)
Output:
int(ei(log(a + b*x))*x**2,x)