\(\int F^{a+b (c+d x)^3} (c+d x)^m \, dx\) [280]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int F^{a+b (c+d x)^3} (c+d x)^m \, dx=-\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1+m}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{\frac {1}{3} (-1-m)}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, 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 = {2250} \[ \int F^{a+b (c+d x)^3} (c+d x)^m \, dx=-\frac {F^a (c+d x)^{m+1} \left (-b \log (F) (c+d x)^3\right )^{\frac {1}{3} (-m-1)} \Gamma \left (\frac {m+1}{3},-b (c+d x)^3 \log (F)\right )}{3 d} \]

[In]

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

[Out]

-1/3*(F^a*(c + d*x)^(1 + m)*Gamma[(1 + m)/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^((-1 - m)/3))/
d

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1+m}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{\frac {1}{3} (-1-m)}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^3} (c+d x)^m \, dx=-\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1+m}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{\frac {1}{3} (-1-m)}}{3 d} \]

[In]

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

[Out]

-1/3*(F^a*(c + d*x)^(1 + m)*Gamma[(1 + m)/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^((-1 - m)/3))/
d

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int F^{a+b (c+d x)^3} (c+d x)^m \, dx=\frac {e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-b \log \left (F\right )\right ) + a \log \left (F\right )\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )}{3 \, b d \log \left (F\right )} \]

[In]

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

[Out]

1/3*e^(-1/3*(m - 2)*log(-b*log(F)) + a*log(F))*gamma(1/3*m + 1/3, -(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x +
b*c^3)*log(F))/(b*d*log(F))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int F^{a+b (c+d x)^3} (c+d x)^m \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}{2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {1}{3}-\frac {m}{6},\frac {m}{6}+\frac {1}{6}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}{d\,\left (m+1\right )\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}^{\frac {m}{6}+\frac {2}{3}}} \]

[In]

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

[Out]

(F^a*exp((b*log(F)*(c + d*x)^3)/2)*(c + d*x)^(m + 1)*whittakerM(1/3 - m/6, m/6 + 1/6, b*log(F)*(c + d*x)^3))/(
d*(m + 1)*(b*log(F)*(c + d*x)^3)^(m/6 + 2/3))