Integrand size = 21, antiderivative size = 121 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6}{6 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \log (F)}{12 d}+\frac {b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log ^2(F)}{12 d}-\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log ^3(F)}{12 d} \]
1/6*F^(a+b/(d*x+c)^2)*(d*x+c)^6/d+1/12*b*F^(a+b/(d*x+c)^2)*(d*x+c)^4*ln(F) /d+1/12*b^2*F^(a+b/(d*x+c)^2)*(d*x+c)^2*ln(F)^2/d-1/12*b^3*F^a*Ei(b*ln(F)/ (d*x+c)^2)*ln(F)^3/d
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\frac {F^a \left (2 F^{\frac {b}{(c+d x)^2}} (c+d x)^6+b \log (F) \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^4+b \log (F) \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^2-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)\right )\right )\right )}{12 d} \]
(F^a*(2*F^(b/(c + d*x)^2)*(c + d*x)^6 + b*Log[F]*(F^(b/(c + d*x)^2)*(c + d *x)^4 + b*Log[F]*(F^(b/(c + d*x)^2)*(c + d*x)^2 - b*ExpIntegralEi[(b*Log[F ])/(c + d*x)^2]*Log[F]))))/(12*d)
Time = 0.42 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2643, 2643, 2643, 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.
\(\displaystyle \int (c+d x)^5 F^{a+\frac {b}{(c+d x)^2}} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3dx+\frac {(c+d x)^6 F^{a+\frac {b}{(c+d x)^2}}}{6 d}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)dx+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\right )+\frac {(c+d x)^6 F^{a+\frac {b}{(c+d x)^2}}}{6 d}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x}dx+\frac {(c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{2 d}\right )+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\right )+\frac {(c+d x)^6 F^{a+\frac {b}{(c+d x)^2}}}{6 d}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (\frac {(c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{2 d}-\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d}\right )+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\right )+\frac {(c+d x)^6 F^{a+\frac {b}{(c+d x)^2}}}{6 d}\) |
(F^(a + b/(c + d*x)^2)*(c + d*x)^6)/(6*d) + (b*Log[F]*((F^(a + b/(c + d*x) ^2)*(c + d*x)^4)/(4*d) + (b*Log[F]*((F^(a + b/(c + d*x)^2)*(c + d*x)^2)/(2 *d) - (b*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F])/(2*d)))/2))/3
3.4.17.3.1 Defintions of rubi rules used
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. \(394\) vs. \(2(113)=226\).
Time = 1.03 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.26
method | result | size |
risch | \(\frac {F^{a} d^{5} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{6}}{6}+F^{a} d^{4} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{5}+\frac {5 F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{4}}{2}+\frac {10 F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x^{3}}{3}+\frac {5 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4} x^{2}}{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{5} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{6}}{6 d}+\frac {F^{a} d^{3} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{4}}{12}+\frac {F^{a} d^{2} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{3}}{3}+\frac {F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{2}}{2}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x}{3}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4}}{12 d}+\frac {F^{a} d \,b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{2}}{12}+\frac {F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c x}{6}+\frac {F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2}}{12 d}+\frac {F^{a} b^{3} \ln \left (F \right )^{3} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{\left (d x +c \right )^{2}}\right )}{12 d}\) | \(395\) |
1/6*F^a*d^5*F^(b/(d*x+c)^2)*x^6+F^a*d^4*F^(b/(d*x+c)^2)*c*x^5+5/2*F^a*d^3* F^(b/(d*x+c)^2)*c^2*x^4+10/3*F^a*d^2*F^(b/(d*x+c)^2)*c^3*x^3+5/2*F^a*d*F^( b/(d*x+c)^2)*c^4*x^2+F^a*F^(b/(d*x+c)^2)*c^5*x+1/6*F^a/d*F^(b/(d*x+c)^2)*c ^6+1/12*F^a*d^3*b*ln(F)*F^(b/(d*x+c)^2)*x^4+1/3*F^a*d^2*b*ln(F)*F^(b/(d*x+ c)^2)*c*x^3+1/2*F^a*d*b*ln(F)*F^(b/(d*x+c)^2)*c^2*x^2+1/3*F^a*b*ln(F)*F^(b /(d*x+c)^2)*c^3*x+1/12*F^a/d*b*ln(F)*F^(b/(d*x+c)^2)*c^4+1/12*F^a*d*b^2*ln (F)^2*F^(b/(d*x+c)^2)*x^2+1/6*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c*x+1/12*F^a /d*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c^2+1/12*F^a/d*b^3*ln(F)^3*Ei(1,-b*ln(F)/(d *x+c)^2)
Time = 0.27 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.86 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=-\frac {F^{a} b^{3} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \left (F\right )^{3} - {\left (2 \, d^{6} x^{6} + 12 \, c d^{5} x^{5} + 30 \, c^{2} d^{4} x^{4} + 40 \, c^{3} d^{3} x^{3} + 30 \, c^{4} d^{2} x^{2} + 12 \, c^{5} d x + 2 \, c^{6} + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (F\right )^{2} + {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{12 \, d} \]
-1/12*(F^a*b^3*Ei(b*log(F)/(d^2*x^2 + 2*c*d*x + c^2))*log(F)^3 - (2*d^6*x^ 6 + 12*c*d^5*x^5 + 30*c^2*d^4*x^4 + 40*c^3*d^3*x^3 + 30*c^4*d^2*x^2 + 12*c ^5*d*x + 2*c^6 + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(F)^2 + (b*d^4*x ^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*log(F))*F^((a* d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/d
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{5}\, dx \]
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\int { {\left (d x + c\right )}^{5} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
1/12*(2*F^a*d^5*x^6 + 12*F^a*c*d^4*x^5 + (30*F^a*c^2*d^3 + F^a*b*d^3*log(F ))*x^4 + 4*(10*F^a*c^3*d^2 + F^a*b*c*d^2*log(F))*x^3 + (30*F^a*c^4*d + 6*F ^a*b*c^2*d*log(F) + F^a*b^2*d*log(F)^2)*x^2 + 2*(6*F^a*c^5 + 2*F^a*b*c^3*l og(F) + F^a*b^2*c*log(F)^2)*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate (1/6*(F^a*b^3*d^2*x^2*log(F)^3 + 2*F^a*b^3*c*d*x*log(F)^3 - 2*F^a*b*c^6*lo g(F) - F^a*b^2*c^4*log(F)^2)*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3* c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\int { {\left (d x + c\right )}^{5} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
Time = 0.45 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \, dx=\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}{6}+F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (\frac {{\left (c+d\,x\right )}^2}{6\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^4}{6\,b^2\,{\ln \left (F\right )}^2}+\frac {{\left (c+d\,x\right )}^6}{3\,b^3\,{\ln \left (F\right )}^3}\right )\right )}{2\,d} \]