3.334 \(\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^4} \, dx\)

Optimal. Leaf size=81 \[ \frac{\sqrt{\pi } F^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac{3}{2}}(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)} \]

[Out]

(F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(4*b^(3/2)*d*Log[F]^(3/2)) - F^(a + b/(c + d*x)^2)/(2*b*
d*(c + d*x)*Log[F])

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Rubi [A]  time = 0.102135, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2212, 2211, 2204} \[ \frac{\sqrt{\pi } F^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac{3}{2}}(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^4,x]

[Out]

(F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(4*b^(3/2)*d*Log[F]^(3/2)) - F^(a + b/(c + d*x)^2)/(2*b*
d*(c + d*x)*Log[F])

Rule 2212

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

Rule 2211

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^4} \, dx &=-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x) \log (F)}-\frac{\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{2 b \log (F)}\\ &=-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x) \log (F)}+\frac{\operatorname{Subst}\left (\int F^{a+b x^2} \, dx,x,\frac{1}{c+d x}\right )}{2 b d \log (F)}\\ &=\frac{F^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac{3}{2}}(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x) \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0461247, size = 81, normalized size = 1. \[ \frac{\sqrt{\pi } F^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac{3}{2}}(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^4,x]

[Out]

(F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(4*b^(3/2)*d*Log[F]^(3/2)) - F^(a + b/(c + d*x)^2)/(2*b*
d*(c + d*x)*Log[F])

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Maple [A]  time = 0.054, size = 76, normalized size = 0.9 \begin{align*} -{\frac{{F}^{a}}{2\, \left ( dx+c \right ) db\ln \left ( F \right ) }{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{{F}^{a}\sqrt{\pi }}{4\,\ln \left ( F \right ) bd}{\it Erf} \left ({\frac{1}{dx+c}\sqrt{-b\ln \left ( F \right ) }} \right ){\frac{1}{\sqrt{-b\ln \left ( F \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)^4,x)

[Out]

-1/2/d*F^a*F^(b/(d*x+c)^2)/(d*x+c)/b/ln(F)+1/4/d*F^a/b/ln(F)*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d
*x+c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^4, x)

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Fricas [A]  time = 1.54413, size = 275, normalized size = 3.4 \begin{align*} -\frac{\sqrt{\pi }{\left (d^{2} x + c d\right )} F^{a} \sqrt{-\frac{b \log \left (F\right )}{d^{2}}} \operatorname{erf}\left (\frac{d \sqrt{-\frac{b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) + 2 \, 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}}} b \log \left (F\right )}{4 \,{\left (b^{2} d^{2} x + b^{2} c d\right )} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/4*(sqrt(pi)*(d^2*x + c*d)*F^a*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(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))*b*log(F))/((b^2*d^2*x + b^2*c*d)*log(F)^2)

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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)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^4, x)