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

Optimal. Leaf size=54 \[ -\frac {F^a (c+d x)^3 \Gamma \left (\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{-3/n}}{d n} \]

[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))

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Rubi [A]
time = 0.02, antiderivative size = 54, 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 = {2250} \begin {gather*} -\frac {F^a (c+d x)^3 \left (-b \log (F) (c+d x)^n\right )^{-3/n} \text {Gamma}\left (\frac {3}{n},-b \log (F) (c+d x)^n\right )}{d n} \end {gather*}

Antiderivative was successfully verified.

[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*} \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}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 54, normalized size = 1.00 \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[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)))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int F^{a +b \left (d x +c \right )^{n}} \left (d x +c \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} F^{a + \frac {b}{c^{3}}} c^{2} x & \text {for}\: d = 0 \wedge n = -3 \\F^{a + b c^{n}} c^{2} x & \text {for}\: d = 0 \\\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{2}\, dx & \text {for}\: n = -3 \\\frac {3 F^{a} F^{b \left (c + d x\right )^{n}} b c^{3} n \left (c + d x\right )^{n} \log {\left (F \right )}}{3 d n + 9 d} - \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} b c^{2} d n x \left (c + d x\right )^{n} \log {\left (F \right )}}{3 d n + 9 d} - \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} b c d^{2} n x^{2} \left (c + d x\right )^{n} \log {\left (F \right )}}{3 d n + 9 d} - \frac {F^{a} F^{b \left (c + d x\right )^{n}} b d^{3} n x^{3} \left (c + d x\right )^{n} \log {\left (F \right )}}{3 d n + 9 d} - \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} c^{3} n}{3 d n + 9 d} + \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} c^{3}}{3 d n + 9 d} + \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} c^{2} d n x}{3 d n + 9 d} + \frac {9 F^{a} F^{b \left (c + d x\right )^{n}} c^{2} d x}{3 d n + 9 d} + \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} c d^{2} n x^{2}}{3 d n + 9 d} + \frac {9 F^{a} F^{b \left (c + d x\right )^{n}} c d^{2} x^{2}}{3 d n + 9 d} + \frac {F^{a} F^{b \left (c + d x\right )^{n}} d^{3} n x^{3}}{3 d n + 9 d} + \frac {3 F^{a} F^{b \left (c + d x\right )^{n}} d^{3} x^{3}}{3 d n + 9 d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((F**(a + b/c**3)*c**2*x, Eq(d, 0) & Eq(n, -3)), (F**(a + b*c**n)*c**2*x, Eq(d, 0)), (Integral(F**(a
+ b/(c + d*x)**3)*(c + d*x)**2, x), Eq(n, -3)), (3*F**a*F**(b*(c + d*x)**n)*b*c**3*n*(c + d*x)**n*log(F)/(3*d*
n + 9*d) - 3*F**a*F**(b*(c + d*x)**n)*b*c**2*d*n*x*(c + d*x)**n*log(F)/(3*d*n + 9*d) - 3*F**a*F**(b*(c + d*x)*
*n)*b*c*d**2*n*x**2*(c + d*x)**n*log(F)/(3*d*n + 9*d) - F**a*F**(b*(c + d*x)**n)*b*d**3*n*x**3*(c + d*x)**n*lo
g(F)/(3*d*n + 9*d) - 3*F**a*F**(b*(c + d*x)**n)*c**3*n/(3*d*n + 9*d) + 3*F**a*F**(b*(c + d*x)**n)*c**3/(3*d*n
+ 9*d) + 3*F**a*F**(b*(c + d*x)**n)*c**2*d*n*x/(3*d*n + 9*d) + 9*F**a*F**(b*(c + d*x)**n)*c**2*d*x/(3*d*n + 9*
d) + 3*F**a*F**(b*(c + d*x)**n)*c*d**2*n*x**2/(3*d*n + 9*d) + 9*F**a*F**(b*(c + d*x)**n)*c*d**2*x**2/(3*d*n +
9*d) + F**a*F**(b*(c + d*x)**n)*d**3*n*x**3/(3*d*n + 9*d) + 3*F**a*F**(b*(c + d*x)**n)*d**3*x**3/(3*d*n + 9*d)
, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 3.90, size = 73, normalized size = 1.35 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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