\(\int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx\) [288]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}+\frac {b F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log (F)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (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} \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d}-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3} \]

[In]

Int[F^(a + b*(c + d*x)^3)/(c + d*x)^4,x]

[Out]

-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)

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+b (c+d x)^3}}{3 d (c+d x)^3}+(b \log (F)) \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx \\ & = -\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}+\frac {b F^a \text {Ei}\left (b (c+d x)^3 \log (F)\right ) \log (F)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\frac {F^a \left (-\frac {F^{b (c+d x)^3}}{(c+d x)^3}+b \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log (F)\right )}{3 d} \]

[In]

Integrate[F^(a + b*(c + d*x)^3)/(c + d*x)^4,x]

[Out]

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

Maple [F]

\[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{4}}d x\]

[In]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^4,x)

[Out]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^4,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.77 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\frac {{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} F^{a} {\rm Ei}\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right ) - F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} \]

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^4,x, algorithm="fricas")

[Out]

1/3*((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^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))*log(F) - F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2
*x + c^3*d)

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{3}}}{\left (c + d x\right )^{4}}\, dx \]

[In]

integrate(F**(a+b*(d*x+c)**3)/(d*x+c)**4,x)

[Out]

Integral(F**(a + b*(c + d*x)**3)/(c + d*x)**4, x)

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{4}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^4, x)

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{4}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^4,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^4, x)

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx=-\frac {F^a\,\left (F^{b\,{\left (c+d\,x\right )}^3}+b\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )\,{\left (c+d\,x\right )}^3\right )}{3\,d\,{\left (c+d\,x\right )}^3} \]

[In]

int(F^(a + b*(c + d*x)^3)/(c + d*x)^4,x)

[Out]

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