3.276 \(\int \frac{F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx\)

Optimal. Leaf size=136 \[ \frac{4 \sqrt{\pi } b^{5/2} F^a \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{15 d}-\frac{4 b^2 \log ^2(F) F^{a+b (c+d x)^2}}{15 d (c+d x)}-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac{2 b \log (F) F^{a+b (c+d x)^2}}{15 d (c+d x)^3} \]

[Out]

-F^(a + b*(c + d*x)^2)/(5*d*(c + d*x)^5) - (2*b*F^(a + b*(c + d*x)^2)*Log[F])/(15*d*(c + d*x)^3) - (4*b^2*F^(a
+ b*(c + d*x)^2)*Log[F]^2)/(15*d*(c + d*x)) + (4*b^(5/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Lo
g[F]^(5/2))/(15*d)

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Rubi [A]  time = 0.218594, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2214, 2204} \[ \frac{4 \sqrt{\pi } b^{5/2} F^a \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{15 d}-\frac{4 b^2 \log ^2(F) F^{a+b (c+d x)^2}}{15 d (c+d x)}-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac{2 b \log (F) F^{a+b (c+d x)^2}}{15 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^6,x]

[Out]

-F^(a + b*(c + d*x)^2)/(5*d*(c + d*x)^5) - (2*b*F^(a + b*(c + d*x)^2)*Log[F])/(15*d*(c + d*x)^3) - (4*b^2*F^(a
+ b*(c + d*x)^2)*Log[F]^2)/(15*d*(c + d*x)) + (4*b^(5/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Lo
g[F]^(5/2))/(15*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)^6} \, dx &=-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}+\frac{1}{5} (2 b \log (F)) \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac{2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}+\frac{1}{15} \left (4 b^2 \log ^2(F)\right ) \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac{2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac{4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac{1}{15} \left (8 b^3 \log ^3(F)\right ) \int F^{a+b (c+d x)^2} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac{2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac{4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac{4 b^{5/2} F^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} (c+d x) \sqrt{\log (F)}\right ) \log ^{\frac{5}{2}}(F)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.119602, size = 97, normalized size = 0.71 \[ \frac{F^a \left (4 \sqrt{\pi } b^{5/2} \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )-\frac{F^{b (c+d x)^2} \left (4 b^2 \log ^2(F) (c+d x)^4+2 b \log (F) (c+d x)^2+3\right )}{(c+d x)^5}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^6,x]

[Out]

(F^a*(4*b^(5/2)*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(5/2) - (F^(b*(c + d*x)^2)*(3 + 2*b*(c +
d*x)^2*Log[F] + 4*b^2*(c + d*x)^4*Log[F]^2))/(c + d*x)^5))/(15*d)

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Maple [A]  time = 0.064, size = 129, normalized size = 1. \begin{align*} -{\frac{{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{5\,d \left ( dx+c \right ) ^{5}}}-{\frac{2\,b\ln \left ( F \right ){F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{15\,d \left ( dx+c \right ) ^{3}}}-{\frac{4\,{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{15\, \left ( dx+c \right ) d}}+{\frac{4\,{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}\sqrt{\pi }{F}^{a}}{15\,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)^6,x)

[Out]

-1/5/d/(d*x+c)^5*F^(b*(d*x+c)^2)*F^a-2/15/d*b*ln(F)/(d*x+c)^3*F^(b*(d*x+c)^2)*F^a-4/15/d*b^2*ln(F)^2/(d*x+c)*F
^(b*(d*x+c)^2)*F^a+4/15/d*b^3*ln(F)^3*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 )}^{6}}\,{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)^6,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^6, x)

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Fricas [B]  time = 1.5253, size = 621, normalized size = 4.57 \begin{align*} -\frac{4 \, \sqrt{\pi }{\left (b^{2} d^{5} x^{5} + 5 \, b^{2} c d^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} x^{3} + 10 \, b^{2} c^{3} d^{2} x^{2} + 5 \, b^{2} c^{4} d x + b^{2} c^{5}\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 )^{2} +{\left (4 \,{\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 2 \,{\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 3 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{15 \,{\left (d^{7} x^{5} + 5 \, c d^{6} x^{4} + 10 \, c^{2} d^{5} x^{3} + 10 \, c^{3} d^{4} x^{2} + 5 \, c^{4} d^{3} x + c^{5} 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)^6,x, algorithm="fricas")

[Out]

-1/15*(4*sqrt(pi)*(b^2*d^5*x^5 + 5*b^2*c*d^4*x^4 + 10*b^2*c^2*d^3*x^3 + 10*b^2*c^3*d^2*x^2 + 5*b^2*c^4*d*x + b
^2*c^5)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d)*log(F)^2 + (4*(b^2*d^5*x^4 + 4*b^2*c*d^4*
x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c^3*d^2*x + b^2*c^4*d)*log(F)^2 + 2*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F)
+ 3*d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^7*x^5 + 5*c*d^6*x^4 + 10*c^2*d^5*x^3 + 10*c^3*d^4*x^2 + 5*c^
4*d^3*x + c^5*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)**6,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 )}^{6}}\,{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)^6,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^6, x)