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

Optimal. Leaf size=53 \[ \frac{(c+d x)^2 F^{a+\frac{b}{(c+d x)^2}}}{2 d}-\frac{b F^a \log (F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[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)

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Rubi [A]  time = 0.0713583, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2214, 2210} \[ \frac{(c+d x)^2 F^{a+\frac{b}{(c+d x)^2}}}{2 d}-\frac{b F^a \log (F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)*(c + d*x),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 2214

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]))

Rule 2210

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]

Rubi steps

\begin{align*} \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}+(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]  time = 0.0228826, size = 47, normalized size = 0.89 \[ \frac{F^a \left ((c+d x)^2 F^{\frac{b}{(c+d x)^2}}-b \log (F) \text{Ei}\left (\frac{b \log (F)}{(c+d x)^2}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.03, size = 86, normalized size = 1.6 \begin{align*}{\frac{d{F}^{a}{x}^{2}}{2}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}cx+{\frac{{F}^{a}{c}^{2}}{2\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}b\ln \left ( F \right ) }{2\,d}{\it Ei} \left ( 1,-{\frac{b\ln \left ( F \right ) }{ \left ( dx+c \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (F^{a} d x^{2} + 2 \, F^{a} c x\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac{{\left (F^{a} b d^{2} x^{2} \log \left (F\right ) + 2 \, F^{a} b c d x \log \left (F\right )\right )} F^{\frac{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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.59978, size = 211, normalized size = 3.98 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)