3.362 \(\int F^{a+b (c+d x)^n} (c+d x) \, dx\)

Optimal. Leaf size=54 \[ -\frac {F^a (c+d x)^2 \left (-b \log (F) (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-b (c+d x)^n \log (F)\right )}{d n} \]

[Out]

-F^a*(d*x+c)^2*GAMMA(2/n,-b*(d*x+c)^n*ln(F))/d/n/((-b*(d*x+c)^n*ln(F))^(2/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 = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2218} \[ -\frac {F^a (c+d x)^2 \left (-b \log (F) (c+d x)^n\right )^{-2/n} \text {Gamma}\left (\frac {2}{n},-b \log (F) (c+d x)^n\right )}{d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 54, normalized size = 1.00 \[ -\frac {F^a (c+d x)^2 \left (-b \log (F) (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-b (c+d x)^n \log (F)\right )}{d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d x + c\right )} F^{{\left (d x + c\right )}^{n} b + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right ) F^{b \left (d x +c \right )^{n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int F^{a+b\,{\left (c+d\,x\right )}^n}\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} F^{a + \frac {b}{c^{2}}} c x & \text {for}\: d = 0 \wedge n = -2 \\F^{a + b c^{n}} c x & \text {for}\: d = 0 \\\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )\, dx & \text {for}\: n = -2 \\\frac {2 F^{a} F^{b \left (c + d x\right )^{n}} b c^{2} n \left (c + d x\right )^{n} \log {\relax (F )}}{2 d n + 4 d} + \frac {6 F^{a} F^{b \left (c + d x\right )^{n}} b c^{2} \left (c + d x\right )^{n} \log {\relax (F )}}{2 d n + 4 d} - \frac {2 F^{a} F^{b \left (c + d x\right )^{n}} b c d n x \left (c + d x\right )^{n} \log {\relax (F )}}{2 d n + 4 d} - \frac {F^{a} F^{b \left (c + d x\right )^{n}} b d^{2} n x^{2} \left (c + d x\right )^{n} \log {\relax (F )}}{2 d n + 4 d} - \frac {2 F^{a} F^{b \left (c + d x\right )^{n}} c^{2} n}{2 d n + 4 d} - \frac {4 F^{a} F^{b \left (c + d x\right )^{n}} c^{2}}{2 d n + 4 d} + \frac {2 F^{a} F^{b \left (c + d x\right )^{n}} c d n x}{2 d n + 4 d} + \frac {4 F^{a} F^{b \left (c + d x\right )^{n}} c d x}{2 d n + 4 d} + \frac {F^{a} F^{b \left (c + d x\right )^{n}} d^{2} n x^{2}}{2 d n + 4 d} + \frac {2 F^{a} F^{b \left (c + d x\right )^{n}} d^{2} x^{2}}{2 d n + 4 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((F**(a + b/c**2)*c*x, Eq(d, 0) & Eq(n, -2)), (F**(a + b*c**n)*c*x, Eq(d, 0)), (Integral(F**(a + b/(c
+ d*x)**2)*(c + d*x), x), Eq(n, -2)), (2*F**a*F**(b*(c + d*x)**n)*b*c**2*n*(c + d*x)**n*log(F)/(2*d*n + 4*d)
+ 6*F**a*F**(b*(c + d*x)**n)*b*c**2*(c + d*x)**n*log(F)/(2*d*n + 4*d) - 2*F**a*F**(b*(c + d*x)**n)*b*c*d*n*x*(
c + d*x)**n*log(F)/(2*d*n + 4*d) - F**a*F**(b*(c + d*x)**n)*b*d**2*n*x**2*(c + d*x)**n*log(F)/(2*d*n + 4*d) -
2*F**a*F**(b*(c + d*x)**n)*c**2*n/(2*d*n + 4*d) - 4*F**a*F**(b*(c + d*x)**n)*c**2/(2*d*n + 4*d) + 2*F**a*F**(b
*(c + d*x)**n)*c*d*n*x/(2*d*n + 4*d) + 4*F**a*F**(b*(c + d*x)**n)*c*d*x/(2*d*n + 4*d) + F**a*F**(b*(c + d*x)**
n)*d**2*n*x**2/(2*d*n + 4*d) + 2*F**a*F**(b*(c + d*x)**n)*d**2*x**2/(2*d*n + 4*d), True))

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