Integrand size = 21, antiderivative size = 119 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\frac {F^{a+\frac {b}{c+d x}} (c+d x)^3}{3 d}+\frac {b F^{a+\frac {b}{c+d x}} (c+d x)^2 \log (F)}{6 d}+\frac {b^2 F^{a+\frac {b}{c+d x}} (c+d x) \log ^2(F)}{6 d}-\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right ) \log ^3(F)}{6 d} \]
1/3*F^(a+b/(d*x+c))*(d*x+c)^3/d+1/6*b*F^(a+b/(d*x+c))*(d*x+c)^2*ln(F)/d+1/ 6*b^2*F^(a+b/(d*x+c))*(d*x+c)*ln(F)^2/d-1/6*b^3*F^a*Ei(b*ln(F)/(d*x+c))*ln (F)^3/d
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.64 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\frac {F^a \left (-b^3 \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right ) \log ^3(F)+F^{\frac {b}{c+d x}} (c+d x) \left (2 (c+d x)^2+b (c+d x) \log (F)+b^2 \log ^2(F)\right )\right )}{6 d} \]
(F^a*(-(b^3*ExpIntegralEi[(b*Log[F])/(c + d*x)]*Log[F]^3) + F^(b/(c + d*x) )*(c + d*x)*(2*(c + d*x)^2 + b*(c + d*x)*Log[F] + b^2*Log[F]^2)))/(6*d)
Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2643, 2643, 2635, 2639}
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.
\(\displaystyle \int (c+d x)^2 F^{a+\frac {b}{c+d x}} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \int F^{a+\frac {b}{c+d x}} (c+d x)dx+\frac {(c+d x)^3 F^{a+\frac {b}{c+d x}}}{3 d}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \int F^{a+\frac {b}{c+d x}}dx+\frac {(c+d x)^2 F^{a+\frac {b}{c+d x}}}{2 d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{c+d x}}}{3 d}\) |
\(\Big \downarrow \) 2635 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x}dx+\frac {(c+d x) F^{a+\frac {b}{c+d x}}}{d}\right )+\frac {(c+d x)^2 F^{a+\frac {b}{c+d x}}}{2 d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{c+d x}}}{3 d}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (\frac {(c+d x) F^{a+\frac {b}{c+d x}}}{d}-\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{c+d x}\right )}{d}\right )+\frac {(c+d x)^2 F^{a+\frac {b}{c+d x}}}{2 d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{c+d x}}}{3 d}\) |
(F^(a + b/(c + d*x))*(c + d*x)^3)/(3*d) + (b*Log[F]*((F^(a + b/(c + d*x))* (c + d*x)^2)/(2*d) + (b*Log[F]*((F^(a + b/(c + d*x))*(c + d*x))/d - (b*F^a *ExpIntegralEi[(b*Log[F])/(c + d*x)]*Log[F])/d))/2))/3
3.4.4.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x] - Simp[b*n*Log[F] Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] && ILtQ[n, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_ Symbol] :> Simp[F^a*(ExpIntegralEi[b*(c + d*x)^n*Log[F]]/(f*n)), x] /; Free Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) Int[(c + d*x)^(m + n)*F^(a + b*(c + d*x) ^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[ -4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))
Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(111)=222\).
Time = 0.33 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.97
method | result | size |
risch | \(\frac {d^{2} F^{a} F^{\frac {b}{d x +c}} x^{3}}{3}+d \,F^{a} F^{\frac {b}{d x +c}} c \,x^{2}+F^{a} F^{\frac {b}{d x +c}} c^{2} x +\frac {F^{a} F^{\frac {b}{d x +c}} c^{3}}{3 d}+\frac {d b \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}} x^{2}}{6}+\frac {b \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}} c x}{3}+\frac {b \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}} c^{2}}{6 d}+\frac {b^{2} \ln \left (F \right )^{2} F^{a} F^{\frac {b}{d x +c}} x}{6}+\frac {b^{2} \ln \left (F \right )^{2} F^{a} F^{\frac {b}{d x +c}} c}{6 d}+\frac {b^{3} \ln \left (F \right )^{3} F^{a} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}\right )}{6 d}\) | \(234\) |
1/3*d^2*F^a*F^(b/(d*x+c))*x^3+d*F^a*F^(b/(d*x+c))*c*x^2+F^a*F^(b/(d*x+c))* c^2*x+1/3/d*F^a*F^(b/(d*x+c))*c^3+1/6*d*b*ln(F)*F^a*F^(b/(d*x+c))*x^2+1/3* b*ln(F)*F^a*F^(b/(d*x+c))*c*x+1/6/d*b*ln(F)*F^a*F^(b/(d*x+c))*c^2+1/6*b^2* ln(F)^2*F^a*F^(b/(d*x+c))*x+1/6/d*b^2*ln(F)^2*F^a*F^(b/(d*x+c))*c+1/6/d*b^ 3*ln(F)^3*F^a*Ei(1,-b*ln(F)/(d*x+c))
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=-\frac {F^{a} b^{3} {\rm Ei}\left (\frac {b \log \left (F\right )}{d x + c}\right ) \log \left (F\right )^{3} - {\left (2 \, d^{3} x^{3} + 6 \, c d^{2} x^{2} + 6 \, c^{2} d x + 2 \, c^{3} + {\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{6 \, d} \]
-1/6*(F^a*b^3*Ei(b*log(F)/(d*x + c))*log(F)^3 - (2*d^3*x^3 + 6*c*d^2*x^2 + 6*c^2*d*x + 2*c^3 + (b^2*d*x + b^2*c)*log(F)^2 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*F^((a*d*x + a*c + b)/(d*x + c)))/d
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\int F^{a + \frac {b}{c + d x}} \left (c + d x\right )^{2}\, dx \]
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\int { {\left (d x + c\right )}^{2} F^{a + \frac {b}{d x + c}} \,d x } \]
1/6*(2*F^a*d^2*x^3 + (F^a*b*d*log(F) + 6*F^a*c*d)*x^2 + (F^a*b^2*log(F)^2 + 2*F^a*b*c*log(F) + 6*F^a*c^2)*x)*F^(b/(d*x + c)) + integrate(1/6*(F^a*b^ 3*d*x*log(F)^3 - F^a*b^2*c^2*log(F)^2 - 2*F^a*b*c^3*log(F))*F^(b/(d*x + c) )/(d^2*x^2 + 2*c*d*x + c^2), x)
\[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\int { {\left (d x + c\right )}^{2} F^{a + \frac {b}{d x + c}} \,d x } \]
Time = 0.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int F^{a+\frac {b}{c+d x}} (c+d x)^2 \, dx=\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{c+d\,x}\right )}{6}+F^{\frac {b}{c+d\,x}}\,\left (\frac {c+d\,x}{6\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^2}{6\,b^2\,{\ln \left (F\right )}^2}+\frac {{\left (c+d\,x\right )}^3}{3\,b^3\,{\ln \left (F\right )}^3}\right )\right )}{d} \]