3.22 \(\int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^m \, dx\)

Optimal. Leaf size=71 \[ \frac{\left ((d+e x)^2\right )^m F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{-2 m} \text{Gamma}\left (2 m+1,-\frac{b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.097998, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071 \[ \frac{\left ((d+e x)^2\right )^m F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{-2 m} \text{Gamma}\left (2 m+1,-\frac{b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 18.4941, size = 65, normalized size = 0.92 \[ \frac{F^{\frac{c \left (a e - b d\right )}{e}} \left (\frac{b c \left (- d - e x\right ) \log{\left (F \right )}}{e}\right )^{- 2 m} \left (\left (d + e x\right )^{2}\right )^{m} \Gamma{\left (2 m + 1,\frac{b c \left (- d - e x\right ) \log{\left (F \right )}}{e} \right )}}{b c \log{\left (F \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(F**(c*(b*x+a))*(e**2*x**2+2*d*e*x+d**2)**m,x)

[Out]

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

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Mathematica [A]  time = 0.04669, size = 73, normalized size = 1.03 \[ -\frac{(d+e x) \left ((d+e x)^2\right )^m F^{a c-\frac{b c d}{e}} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{-2 m-1} \text{Gamma}\left (2 m+1,-\frac{b c \log (F) (d+e x)}{e}\right )}{e} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.138, size = 0, normalized size = 0. \[ \int{F}^{c \left ( bx+a \right ) } \left ({e}^{2}{x}^{2}+2\,dex+{d}^{2} \right ) ^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}^{m} F^{b c x + a c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(F**(c*(b*x+a))*(e**2*x**2+2*d*e*x+d**2)**m,x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}^{m} F^{{\left (b x + a\right )} c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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