\(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx\) [319]

Optimal result

Integrand size = 19, antiderivative size = 53 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2}{2 d}-\frac {b F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2245, 2241} \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx=\frac {(c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{2 d}-\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[In]

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

[Out]

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

Rule 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2}{2 d}+(b \log (F)) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2}{2 d}-\frac {b F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^2-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)\right )}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.62

method	result	size
risch	\(\frac {d \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{2}}{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2}}{2 d}+\frac {F^{a} b \ln \left (F \right ) \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{\left (d x +c \right )^{2}}\right )}{2 d}\)	\(86\)

[In]

int(F^(a+b/(d*x+c)^2)*(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.81 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx=-\frac {F^{a} b {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \left (F\right ) - {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(F^a*b*Ei(b*log(F)/(d^2*x^2 + 2*c*d*x + c^2))*log(F) - (d^2*x^2 + 2*c*d*x + c^2)*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)))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

1/2*(F^a*d*x^2 + 2*F^a*c*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate((F^a*b*d^2*x^2*log(F) + 2*F^a*b*c*d*x*
log(F))*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^2}{2\,d}+\frac {F^a\,b\,\ln \left (F\right )\,\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}{2\,d} \]

[In]

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

[Out]

(F^a*F^(b/(c + d*x)^2)*(c + d*x)^2)/(2*d) + (F^a*b*log(F)*expint(-(b*log(F))/(c + d*x)^2))/(2*d)