\(\int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx\) [343]

Optimal result

Integrand size = 21, antiderivative size = 121 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^9}{9 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6 \log (F)}{18 d}+\frac {b^2 F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log ^2(F)}{18 d}-\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log ^3(F)}{18 d} \]

[Out]

1/9*F^(a+b/(d*x+c)^3)*(d*x+c)^9/d+1/18*b*F^(a+b/(d*x+c)^3)*(d*x+c)^6*ln(F)/d+1/18*b^2*F^(a+b/(d*x+c)^3)*(d*x+c
)^3*ln(F)^2/d-1/18*b^3*F^a*Ei(b*ln(F)/(d*x+c)^3)*ln(F)^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=-\frac {b^3 F^a \log ^3(F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right )}{18 d}+\frac {b^2 \log ^2(F) (c+d x)^3 F^{a+\frac {b}{(c+d x)^3}}}{18 d}+\frac {(c+d x)^9 F^{a+\frac {b}{(c+d x)^3}}}{9 d}+\frac {b \log (F) (c+d x)^6 F^{a+\frac {b}{(c+d x)^3}}}{18 d} \]

[In]

Int[F^(a + b/(c + d*x)^3)*(c + d*x)^8,x]

[Out]

(F^(a + b/(c + d*x)^3)*(c + d*x)^9)/(9*d) + (b*F^(a + b/(c + d*x)^3)*(c + d*x)^6*Log[F])/(18*d) + (b^2*F^(a +
b/(c + d*x)^3)*(c + d*x)^3*Log[F]^2)/(18*d) - (b^3*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^3]*Log[F]^3)/(18*d)

Rule 2241

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] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2245

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] - Dist[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]))

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^9}{9 d}+\frac {1}{3} (b \log (F)) \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^9}{9 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6 \log (F)}{18 d}+\frac {1}{6} \left (b^2 \log ^2(F)\right ) \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^9}{9 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6 \log (F)}{18 d}+\frac {b^2 F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log ^2(F)}{18 d}+\frac {1}{6} \left (b^3 \log ^3(F)\right ) \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{c+d x} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^9}{9 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6 \log (F)}{18 d}+\frac {b^2 F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log ^2(F)}{18 d}-\frac {b^3 F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log ^3(F)}{18 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\frac {F^a \left (2 F^{\frac {b}{(c+d x)^3}} (c+d x)^9+b \log (F) \left (F^{\frac {b}{(c+d x)^3}} (c+d x)^6+b \log (F) \left (F^{\frac {b}{(c+d x)^3}} (c+d x)^3-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log (F)\right )\right )\right )}{18 d} \]

[In]

Integrate[F^(a + b/(c + d*x)^3)*(c + d*x)^8,x]

[Out]

(F^a*(2*F^(b/(c + d*x)^3)*(c + d*x)^9 + b*Log[F]*(F^(b/(c + d*x)^3)*(c + d*x)^6 + b*Log[F]*(F^(b/(c + d*x)^3)*
(c + d*x)^3 - b*ExpIntegralEi[(b*Log[F])/(c + d*x)^3]*Log[F]))))/(18*d)

Maple [F]

\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{8}d x\]

[In]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^8,x)

[Out]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^8,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (113) = 226\).

Time = 0.30 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.73 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=-\frac {F^{a} b^{3} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \left (F\right )^{3} - {\left (2 \, d^{9} x^{9} + 18 \, c d^{8} x^{8} + 72 \, c^{2} d^{7} x^{7} + 168 \, c^{3} d^{6} x^{6} + 252 \, c^{4} d^{5} x^{5} + 252 \, c^{5} d^{4} x^{4} + 168 \, c^{6} d^{3} x^{3} + 72 \, c^{7} d^{2} x^{2} + 18 \, c^{8} d x + 2 \, c^{9} + {\left (b^{2} d^{3} x^{3} + 3 \, b^{2} c d^{2} x^{2} + 3 \, b^{2} c^{2} d x + b^{2} c^{3}\right )} \log \left (F\right )^{2} + {\left (b d^{6} x^{6} + 6 \, b c d^{5} x^{5} + 15 \, b c^{2} d^{4} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 15 \, b c^{4} d^{2} x^{2} + 6 \, b c^{5} d x + b c^{6}\right )} \log \left (F\right )\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{18 \, d} \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^8,x, algorithm="fricas")

[Out]

-1/18*(F^a*b^3*Ei(b*log(F)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*log(F)^3 - (2*d^9*x^9 + 18*c*d^8*x^8 + 7
2*c^2*d^7*x^7 + 168*c^3*d^6*x^6 + 252*c^4*d^5*x^5 + 252*c^5*d^4*x^4 + 168*c^6*d^3*x^3 + 72*c^7*d^2*x^2 + 18*c^
8*d*x + 2*c^9 + (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*log(F)^2 + (b*d^6*x^6 + 6*b*c*d^5*x^
5 + 15*b*c^2*d^4*x^4 + 20*b*c^3*d^3*x^3 + 15*b*c^4*d^2*x^2 + 6*b*c^5*d*x + b*c^6)*log(F))*F^((a*d^3*x^3 + 3*a*
c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)))/d

Sympy [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{8}\, dx \]

[In]

integrate(F**(a+b/(d*x+c)**3)*(d*x+c)**8,x)

[Out]

Integral(F**(a + b/(c + d*x)**3)*(c + d*x)**8, x)

Maxima [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\int { {\left (d x + c\right )}^{8} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^8,x, algorithm="maxima")

[Out]

1/18*(2*F^a*d^8*x^9 + 18*F^a*c*d^7*x^8 + 72*F^a*c^2*d^6*x^7 + (168*F^a*c^3*d^5 + F^a*b*d^5*log(F))*x^6 + 6*(42
*F^a*c^4*d^4 + F^a*b*c*d^4*log(F))*x^5 + 3*(84*F^a*c^5*d^3 + 5*F^a*b*c^2*d^3*log(F))*x^4 + (168*F^a*c^6*d^2 +
20*F^a*b*c^3*d^2*log(F) + F^a*b^2*d^2*log(F)^2)*x^3 + 3*(24*F^a*c^7*d + 5*F^a*b*c^4*d*log(F) + F^a*b^2*c*d*log
(F)^2)*x^2 + 3*(6*F^a*c^8 + 2*F^a*b*c^5*log(F) + F^a*b^2*c^2*log(F)^2)*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*
d*x + c^3)) + integrate(1/6*(F^a*b^3*d^3*x^3*log(F)^3 - 2*F^a*b*c^9*log(F) + 3*F^a*b^3*c*d^2*x^2*log(F)^3 - F^
a*b^2*c^6*log(F)^2 + 3*F^a*b^3*c^2*d*x*log(F)^3)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(d^4*x^4 + 4*
c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)

Giac [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\int { {\left (d x + c\right )}^{8} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^8,x, algorithm="giac")

[Out]

integrate((d*x + c)^8*F^(a + b/(d*x + c)^3), x)

Mupad [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^8 \, dx=\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}{6}+F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,\left (\frac {{\left (c+d\,x\right )}^3}{6\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^6}{6\,b^2\,{\ln \left (F\right )}^2}+\frac {{\left (c+d\,x\right )}^9}{3\,b^3\,{\ln \left (F\right )}^3}\right )\right )}{3\,d} \]

[In]

int(F^(a + b/(c + d*x)^3)*(c + d*x)^8,x)

[Out]

(F^a*b^3*log(F)^3*(expint(-(b*log(F))/(c + d*x)^3)/6 + F^(b/(c + d*x)^3)*((c + d*x)^3/(6*b*log(F)) + (c + d*x)
^6/(6*b^2*log(F)^2) + (c + d*x)^9/(3*b^3*log(F)^3))))/(3*d)