\(\int F^{a+b (c+d x)^n} (c+d x)^m \, dx\) [359]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (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} \[ \int F^{a+b (c+d x)^n} (c+d x)^m \, dx=-\frac {F^a (c+d x)^{m+1} \left (-b \log (F) (c+d x)^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b (c+d x)^n \log (F)\right )}{d n} \]

[In]

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

[Out]

-((F^a*(c + d*x)^(1 + m)*Gamma[(1 + m)/n, -(b*(c + d*x)^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^((1 + m)/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)^{1+m} \Gamma \left (\frac {1+m}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-\frac {1+m}{n}}}{d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^m \, dx=-\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1+m}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-\frac {1+m}{n}}}{d n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.52 \[ \int F^{a+b (c+d x)^n} (c+d x)^m \, dx=\frac {F^{a+\frac {b\,{\left (c+d\,x\right )}^n}{2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{-\frac {\frac {m}{2}-\frac {n}{2}+\frac {1}{2}}{n},\frac {\frac {m}{2}+\frac {1}{2}}{n}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}{d\,\left (m+1\right )\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}^{\frac {m}{2\,n}+\frac {1}{2\,n}+\frac {1}{2}}} \]

[In]

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

[Out]

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