3.365 \(\int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx\)

Optimal. Leaf size=52 \[ -\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right )}{d n (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 52, 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} \[ -\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},-b \log (F) (c+d x)^n\right )}{d n (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 52, normalized size = 1.00 \[ -\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right )}{d n (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {F^{{\left (d x + c\right )}^{n} b + a}}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \]

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

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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mupad [B]  time = 3.70, size = 71, normalized size = 1.37 \[ -\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \relax (F)\,{\left (c+d\,x\right )}^n}{2}}\,{\left (b\,\ln \relax (F)\,{\left (c+d\,x\right )}^n\right )}^{\frac {1}{2\,n}-\frac {1}{2}}\,{\mathrm {M}}_{\frac {1}{2\,n}+\frac {1}{2},-\frac {1}{2\,n}}\left (b\,\ln \relax (F)\,{\left (c+d\,x\right )}^n\right )}{d\,\left (c+d\,x\right )} \]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {\tilde {\infty } F^{a}}{x} & \text {for}\: c = 0 \wedge d = 0 \wedge n = 1 \\\tilde {\infty } F^{0^{n} b + a} x & \text {for}\: c = - d x \\\frac {F^{a + b c^{n}} x}{c^{2}} & \text {for}\: d = 0 \\\int \frac {F^{a + b \left (c + d x\right )}}{\left (c + d x\right )^{2}}\, dx & \text {for}\: n = 1 \\\frac {F^{a} F^{b \left (c + d x\right )^{n}} b n \left (c + d x\right )^{n} \log {\relax (F )}}{c d n - c d + d^{2} n x - d^{2} x} - \frac {F^{a} F^{b \left (c + d x\right )^{n}} n}{c d n - c d + d^{2} n x - d^{2} x} + \frac {F^{a} F^{b \left (c + d x\right )^{n}}}{c d n - c d + d^{2} n x - d^{2} x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*F**a/x, Eq(c, 0) & Eq(d, 0) & Eq(n, 1)), (zoo*F**(0**n*b + a)*x, Eq(c, -d*x)), (F**(a + b*c**n)
*x/c**2, Eq(d, 0)), (Integral(F**(a + b*(c + d*x))/(c + d*x)**2, x), Eq(n, 1)), (F**a*F**(b*(c + d*x)**n)*b*n*
(c + d*x)**n*log(F)/(c*d*n - c*d + d**2*n*x - d**2*x) - F**a*F**(b*(c + d*x)**n)*n/(c*d*n - c*d + d**2*n*x - d
**2*x) + F**a*F**(b*(c + d*x)**n)/(c*d*n - c*d + d**2*n*x - d**2*x), True))

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