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

Optimal. Leaf size=102 \[ \frac{2 \sqrt{\pi } b^{3/2} F^a \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{3 d}-\frac{F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac{2 b \log (F) F^{a+b (c+d x)^2}}{3 d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.145172, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2214, 2204} \[ \frac{2 \sqrt{\pi } b^{3/2} F^a \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{3 d}-\frac{F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac{2 b \log (F) F^{a+b (c+d x)^2}}{3 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-F^(a + b*(c + d*x)^2)/(3*d*(c + d*x)^3) - (2*b*F^(a + b*(c + d*x)^2)*Log[F])/(3*d*(c + d*x)) + (2*b^(3/2)*F^a
*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(3/2))/(3*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 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+b (c+d x)^2}}{(c+d x)^4} \, dx &=-\frac{F^{a+b (c+d x)^2}}{3 d (c+d x)^3}+\frac{1}{3} (2 b \log (F)) \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac{2 b F^{a+b (c+d x)^2} \log (F)}{3 d (c+d x)}+\frac{1}{3} \left (4 b^2 \log ^2(F)\right ) \int F^{a+b (c+d x)^2} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac{2 b F^{a+b (c+d x)^2} \log (F)}{3 d (c+d x)}+\frac{2 b^{3/2} F^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} (c+d x) \sqrt{\log (F)}\right ) \log ^{\frac{3}{2}}(F)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0972484, size = 81, normalized size = 0.79 \[ \frac{F^a \left (2 \sqrt{\pi } b^{3/2} \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )-\frac{F^{b (c+d x)^2} \left (2 b \log (F) (c+d x)^2+1\right )}{(c+d x)^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*(2*b^(3/2)*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(3/2) - (F^(b*(c + d*x)^2)*(1 + 2*b*(c +
d*x)^2*Log[F]))/(c + d*x)^3))/(3*d)

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Maple [A]  time = 0.049, size = 96, normalized size = 0.9 \begin{align*} -{\frac{{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{3\,d \left ( dx+c \right ) ^{3}}}-{\frac{2\,b\ln \left ( F \right ){F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{3\, \left ( dx+c \right ) d}}+{\frac{2\,{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}\sqrt{\pi }{F}^{a}}{3\,d}{\it Erf} \left ( \sqrt{-b\ln \left ( F \right ) } \left ( dx+c \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]

-1/3/d/(d*x+c)^3*F^(b*(d*x+c)^2)*F^a-2/3/d*b*ln(F)/(d*x+c)*F^(b*(d*x+c)^2)*F^a+2/3/d*b^2*ln(F)^2*Pi^(1/2)*F^a/
(-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^{{\left (d x + c\right )}^{2} b + a}}{{\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^((d*x + c)^2*b + a)/(d*x + c)^4, x)

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Fricas [A]  time = 1.56351, size = 373, normalized size = 3.66 \begin{align*} -\frac{2 \, \sqrt{\pi }{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \sqrt{-b d^{2} \log \left (F\right )} F^{a} \operatorname{erf}\left (\frac{\sqrt{-b d^{2} \log \left (F\right )}{\left (d x + c\right )}}{d}\right ) \log \left (F\right ) +{\left (2 \,{\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{3 \,{\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} \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]

-1/3*(2*sqrt(pi)*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log
(F))*(d*x + c)/d)*log(F) + (2*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2
+ a))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^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^{{\left (d x + c\right )}^{2} b + a}}{{\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^((d*x + c)^2*b + a)/(d*x + c)^4, x)