\(\int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx\) [345]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3}{3 d}-\frac {b F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log (F)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^3}}}{3 d}-\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right )}{3 d} \]

[In]

Int[F^(a + b/(c + d*x)^3)*(c + d*x)^2,x]

[Out]

(F^(a + b/(c + d*x)^3)*(c + d*x)^3)/(3*d) - (b*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^3]*Log[F])/(3*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)^3}{3 d}+(b \log (F)) \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)^3}{3 d}-\frac {b F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log (F)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^3}} (c+d x)^3-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log (F)\right )}{3 d} \]

[In]

Integrate[F^(a + b/(c + d*x)^3)*(c + d*x)^2,x]

[Out]

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

Maple [F]

\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{2}d x\]

[In]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^2,x)

[Out]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^2,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (49) = 98\).

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.66 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=-\frac {F^{a} b {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \left (F\right ) - {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{3 \, d} \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/3*(F^a*b*Ei(b*log(F)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*log(F) - (d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x
+ c^3)*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]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{2}\, dx \]

[In]

integrate(F**(a+b/(d*x+c)**3)*(d*x+c)**2,x)

[Out]

Integral(F**(a + b/(c + d*x)**3)*(c + d*x)**2, x)

Maxima [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\int { {\left (d x + c\right )}^{2} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^2,x, algorithm="maxima")

[Out]

1/3*(F^a*d^2*x^3 + 3*F^a*c*d*x^2 + 3*F^a*c^2*x)*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + integrate((F
^a*b*d^3*x^3*log(F) + 3*F^a*b*c*d^2*x^2*log(F) + 3*F^a*b*c^2*d*x*log(F))*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]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\int { {\left (d x + c\right )}^{2} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*F^(a + b/(d*x + c)^3), x)

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^3}{3\,d}+\frac {F^a\,b\,\ln \left (F\right )\,\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}{3\,d} \]

[In]

int(F^(a + b/(c + d*x)^3)*(c + d*x)^2,x)

[Out]

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