Integrand size = 21, antiderivative size = 121 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=-\frac {F^{a+b (c+d x)^3}}{9 d (c+d x)^9}-\frac {b F^{a+b (c+d x)^3} \log (F)}{18 d (c+d x)^6}-\frac {b^2 F^{a+b (c+d x)^3} \log ^2(F)}{18 d (c+d x)^3}+\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log ^3(F)}{18 d} \]
-1/9*F^(a+b*(d*x+c)^3)/d/(d*x+c)^9-1/18*b*F^(a+b*(d*x+c)^3)*ln(F)/d/(d*x+c )^6-1/18*b^2*F^(a+b*(d*x+c)^3)*ln(F)^2/d/(d*x+c)^3+1/18*b^3*F^a*Ei(b*(d*x+ c)^3*ln(F))*ln(F)^3/d
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=\frac {F^a \left (b^3 \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log ^3(F)+\frac {F^{b (c+d x)^3} \left (-2-b (c+d x)^3 \log (F)-b^2 (c+d x)^6 \log ^2(F)\right )}{(c+d x)^9}\right )}{18 d} \]
(F^a*(b^3*ExpIntegralEi[b*(c + d*x)^3*Log[F]]*Log[F]^3 + (F^(b*(c + d*x)^3 )*(-2 - b*(c + d*x)^3*Log[F] - b^2*(c + d*x)^6*Log[F]^2))/(c + d*x)^9))/(1 8*d)
Time = 0.49 (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.
\(\displaystyle \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \int \frac {F^{b (c+d x)^3+a}}{(c+d x)^7}dx-\frac {F^{a+b (c+d x)^3}}{9 d (c+d x)^9}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \int \frac {F^{b (c+d x)^3+a}}{(c+d x)^4}dx-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\right )-\frac {F^{a+b (c+d x)^3}}{9 d (c+d x)^9}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{b (c+d x)^3+a}}{c+d x}dx-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\right )-\frac {F^{a+b (c+d x)^3}}{9 d (c+d x)^9}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d}-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\right )-\frac {F^{a+b (c+d x)^3}}{9 d (c+d x)^9}\) |
-1/9*F^(a + b*(c + d*x)^3)/(d*(c + d*x)^9) + (b*Log[F]*(-1/6*F^(a + b*(c + d*x)^3)/(d*(c + d*x)^6) + (b*Log[F]*(-1/3*F^(a + b*(c + d*x)^3)/(d*(c + d *x)^3) + (b*F^a*ExpIntegralEi[b*(c + d*x)^3*Log[F]]*Log[F])/(3*d)))/2))/3
3.3.90.3.1 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]))
\[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{10}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (113) = 226\).
Time = 0.28 (sec) , antiderivative size = 431, normalized size of antiderivative = 3.56 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=\frac {{\left (b^{3} d^{9} x^{9} + 9 \, b^{3} c d^{8} x^{8} + 36 \, b^{3} c^{2} d^{7} x^{7} + 84 \, b^{3} c^{3} d^{6} x^{6} + 126 \, b^{3} c^{4} d^{5} x^{5} + 126 \, b^{3} c^{5} d^{4} x^{4} + 84 \, b^{3} c^{6} d^{3} x^{3} + 36 \, b^{3} c^{7} d^{2} x^{2} + 9 \, b^{3} c^{8} d x + b^{3} c^{9}\right )} F^{a} {\rm Ei}\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right )^{3} - {\left ({\left (b^{2} d^{6} x^{6} + 6 \, b^{2} c d^{5} x^{5} + 15 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{3} d^{3} x^{3} + 15 \, b^{2} c^{4} d^{2} x^{2} + 6 \, b^{2} c^{5} d x + b^{2} c^{6}\right )} \log \left (F\right )^{2} + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right ) + 2\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{18 \, {\left (d^{10} x^{9} + 9 \, c d^{9} x^{8} + 36 \, c^{2} d^{8} x^{7} + 84 \, c^{3} d^{7} x^{6} + 126 \, c^{4} d^{6} x^{5} + 126 \, c^{5} d^{5} x^{4} + 84 \, c^{6} d^{4} x^{3} + 36 \, c^{7} d^{3} x^{2} + 9 \, c^{8} d^{2} x + c^{9} d\right )}} \]
1/18*((b^3*d^9*x^9 + 9*b^3*c*d^8*x^8 + 36*b^3*c^2*d^7*x^7 + 84*b^3*c^3*d^6 *x^6 + 126*b^3*c^4*d^5*x^5 + 126*b^3*c^5*d^4*x^4 + 84*b^3*c^6*d^3*x^3 + 36 *b^3*c^7*d^2*x^2 + 9*b^3*c^8*d*x + b^3*c^9)*F^a*Ei((b*d^3*x^3 + 3*b*c*d^2* x^2 + 3*b*c^2*d*x + b*c^3)*log(F))*log(F)^3 - ((b^2*d^6*x^6 + 6*b^2*c*d^5* x^5 + 15*b^2*c^2*d^4*x^4 + 20*b^2*c^3*d^3*x^3 + 15*b^2*c^4*d^2*x^2 + 6*b^2 *c^5*d*x + b^2*c^6)*log(F)^2 + (b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F) + 2)*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a) )/(d^10*x^9 + 9*c*d^9*x^8 + 36*c^2*d^8*x^7 + 84*c^3*d^7*x^6 + 126*c^4*d^6* x^5 + 126*c^5*d^5*x^4 + 84*c^6*d^4*x^3 + 36*c^7*d^3*x^2 + 9*c^8*d^2*x + c^ 9*d)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{3}}}{\left (c + d x\right )^{10}}\, dx \]
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
Time = 0.57 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^{10}} \, dx=-\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}{18\,d}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^3}\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {1}{6\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}+\frac {1}{6\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^6}+\frac {1}{3\,b^3\,{\ln \left (F\right )}^3\,{\left (c+d\,x\right )}^9}\right )}{3\,d} \]