3.254 \(\int F^{a+b (c+d x)^2} (c+d x)^m \, dx\)

Optimal. Leaf size=61 \[ -\frac {F^a (c+d x)^{m+1} \left (-b \log (F) (c+d x)^2\right )^{\frac {1}{2} (-m-1)} \Gamma \left (\frac {m+1}{2},-b (c+d x)^2 \log (F)\right )}{2 d} \]

[Out]

-1/2*F^a*(d*x+c)^(1+m)*GAMMA(1/2+1/2*m,-b*(d*x+c)^2*ln(F))*(-b*(d*x+c)^2*ln(F))^(-1/2-1/2*m)/d

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2218} \[ -\frac {F^a (c+d x)^{m+1} \left (-b \log (F) (c+d x)^2\right )^{\frac {1}{2} (-m-1)} \text {Gamma}\left (\frac {m+1}{2},-b \log (F) (c+d x)^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(F^a*(c + d*x)^(1 + m)*Gamma[(1 + m)/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^((-1 - m)/2))/(2*d
)

Rule 2218

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

Rubi steps

\begin {align*} \int F^{a+b (c+d x)^2} (c+d x)^m \, dx &=-\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1+m}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{\frac {1}{2} (-1-m)}}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 1.00 \[ -\frac {F^a (c+d x)^{m+1} \left (-b \log (F) (c+d x)^2\right )^{\frac {1}{2} (-m-1)} \Gamma \left (\frac {m+1}{2},-b (c+d x)^2 \log (F)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(F^a*(c + d*x)^(1 + m)*Gamma[(1 + m)/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^((-1 - m)/2))/
d

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fricas [A]  time = 0.42, size = 59, normalized size = 0.97 \[ \frac {e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-b \log \relax (F)\right ) + a \log \relax (F)\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F)\right )}{2 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{m} F^{{\left (d x + c\right )}^{2} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int F^{a +\left (d x +c \right )^{2} b} \left (d x +c \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{m} F^{{\left (d x + c\right )}^{2} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.81, size = 75, normalized size = 1.23 \[ \frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2}{2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {1}{4}-\frac {m}{4},\frac {m}{4}+\frac {1}{4}}\left (b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2\right )}{d\,\left (m+1\right )\,{\left (b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2\right )}^{\frac {m}{4}+\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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