\(\int F^{a+b (c+d x)^n} (c+d x)^2 \, dx\) [361]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int F^{a+b (c+d x)^n} (c+d x)^2 \, dx=-\frac {F^a (c+d x)^3 \Gamma \left (\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-3/n}}{d n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, 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} \[ \int F^{a+b (c+d x)^n} (c+d x)^2 \, dx=-\frac {F^a (c+d x)^3 \left (-b \log (F) (c+d x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-b (c+d x)^n \log (F)\right )}{d n} \]

[In]

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

[Out]

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

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)^3 \Gamma \left (\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-3/n}}{d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^2 \, dx=-\frac {F^a (c+d x)^3 \Gamma \left (\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-3/n}}{d n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((d^2*x^2 + 2*c*d*x + c^2)*F^((d*x + c)^n*b + a), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int F^{a+b (c+d x)^n} (c+d x)^2 \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n}{2}}\,{\left (c+d\,x\right )}^3\,{\mathrm {M}}_{\frac {1}{2}-\frac {3}{2\,n},\frac {3}{2\,n}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}{3\,d\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}^{\frac {3}{2\,n}+\frac {1}{2}}} \]

[In]

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

[Out]

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