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

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

[Out]

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

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Rubi [A]  time = 0.05, 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} \[ \frac {F^a (c+d x)^{m+1} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {m+1}{2}} \text {Gamma}\left (\frac {1}{2} (-m-1),-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

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

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

Antiderivative was successfully verified.

[In]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d x + c\right )}^{m} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)^m*F^((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(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 \[ \int {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int F^{a +\frac {b}{\left (d x +c \right )^{2}}} \left (d x +c \right )^{m}\, dx \]

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 \[ \int {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.76, size = 73, normalized size = 1.20 \[ \frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \relax (F)}{2\,{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {m}{4}+\frac {3}{4},-\frac {m}{4}-\frac {1}{4}}\left (\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}\right )\,{\left (\frac {b\,\ln \relax (F)}{{\left (c+d\,x\right )}^2}\right )}^{\frac {m}{4}-\frac {1}{4}}}{d\,\left (m+1\right )} \]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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