\(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx\) [314]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {1+m}{2}}}{2 d} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2250} \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{m+1} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[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 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {1+m}{2}}}{2 d} \]

[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)

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )}{2\,{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {m}{4}+\frac {3}{4},-\frac {m}{4}-\frac {1}{4}}\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )\,{\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{\frac {m}{4}-\frac {1}{4}}}{d\,\left (m+1\right )} \]

[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))