\(\int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx\) [287]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx=\frac {F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2241} \[ \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx=\frac {F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d} \]

[In]

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

[Out]

(F^a*ExpIntegralEi[b*(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]

Rubi steps \begin{align*} \text {integral}& = \frac {F^a \text {Ei}\left (b (c+d x)^3 \log (F)\right )}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx=\frac {F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx=\frac {F^{a} {\rm Ei}\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )}{3 \, d} \]

[In]

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

[Out]

1/3*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))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx=\frac {F^a\,\mathrm {ei}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}{3\,d} \]

[In]

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

[Out]

(F^a*ei(b*log(F)*(c + d*x)^3))/(3*d)