3.336 \(\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^8} \, dx\)

Optimal. Leaf size=149 \[ \frac{15 \sqrt{\pi } F^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}+\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^3}-\frac{15 F^{a+\frac{b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5} \]

[Out]

(15*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(16*b^(7/2)*d*Log[F]^(7/2)) - (15*F^(a + b/(c + d*x)^
2))/(8*b^3*d*(c + d*x)*Log[F]^3) + (5*F^(a + b/(c + d*x)^2))/(4*b^2*d*(c + d*x)^3*Log[F]^2) - F^(a + b/(c + d*
x)^2)/(2*b*d*(c + d*x)^5*Log[F])

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Rubi [A]  time = 0.207721, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2212, 2211, 2204} \[ \frac{15 \sqrt{\pi } F^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}+\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^3}-\frac{15 F^{a+\frac{b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^8,x]

[Out]

(15*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(16*b^(7/2)*d*Log[F]^(7/2)) - (15*F^(a + b/(c + d*x)^
2))/(8*b^3*d*(c + d*x)*Log[F]^3) + (5*F^(a + b/(c + d*x)^2))/(4*b^2*d*(c + d*x)^3*Log[F]^2) - F^(a + b/(c + d*
x)^2)/(2*b*d*(c + d*x)^5*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)^8} \, dx &=-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac{5 \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^6} \, dx}{2 b \log (F)}\\ &=\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac{15 \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^4} \, dx}{4 b^2 \log ^2(F)}\\ &=-\frac{15 F^{a+\frac{b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac{15 \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{8 b^3 \log ^3(F)}\\ &=-\frac{15 F^{a+\frac{b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac{15 \operatorname{Subst}\left (\int F^{a+b x^2} \, dx,x,\frac{1}{c+d x}\right )}{8 b^3 d \log ^3(F)}\\ &=\frac{15 F^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}-\frac{15 F^{a+\frac{b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac{5 F^{a+\frac{b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac{F^{a+\frac{b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.128488, size = 111, normalized size = 0.74 \[ \frac{F^a \left (15 \sqrt{\pi } \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )-\frac{2 \sqrt{b} \sqrt{\log (F)} F^{\frac{b}{(c+d x)^2}} \left (4 b^2 \log ^2(F)-10 b \log (F) (c+d x)^2+15 (c+d x)^4\right )}{(c+d x)^5}\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^8,x]

[Out]

(F^a*(15*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)] - (2*Sqrt[b]*F^(b/(c + d*x)^2)*Sqrt[Log[F]]*(15*(c +
d*x)^4 - 10*b*(c + d*x)^2*Log[F] + 4*b^2*Log[F]^2))/(c + d*x)^5))/(16*b^(7/2)*d*Log[F]^(7/2))

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Maple [A]  time = 0.113, size = 142, normalized size = 1. \begin{align*} -{\frac{{F}^{a}}{2\,d \left ( dx+c \right ) ^{5}b\ln \left ( F \right ) }{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{5\,{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d \left ( dx+c \right ) ^{3}}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}-{\frac{15\,{F}^{a}}{8\,d{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3} \left ( dx+c \right ) }{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{15\,{F}^{a}\sqrt{\pi }}{16\,d{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}}{\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)^8,x)

[Out]

-1/2/d*F^a*F^(b/(d*x+c)^2)/(d*x+c)^5/b/ln(F)+5/4/d*F^a/b^2/ln(F)^2*F^(b/(d*x+c)^2)/(d*x+c)^3-15/8/d*F^a/b^3/ln
(F)^3*F^(b/(d*x+c)^2)/(d*x+c)+15/16/d*F^a/b^3/ln(F)^3*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 )}^{8}}\,{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)^8,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^8, x)

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Fricas [B]  time = 1.68478, size = 670, normalized size = 4.5 \begin{align*} -\frac{15 \, \sqrt{\pi }{\left (d^{6} x^{5} + 5 \, c d^{5} x^{4} + 10 \, c^{2} d^{4} x^{3} + 10 \, c^{3} d^{3} x^{2} + 5 \, c^{4} d^{2} x + c^{5} 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 \,{\left (4 \, b^{3} \log \left (F\right )^{3} - 10 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (F\right )^{2} + 15 \,{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \log \left (F\right )\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}}}}{16 \,{\left (b^{4} d^{6} x^{5} + 5 \, b^{4} c d^{5} x^{4} + 10 \, b^{4} c^{2} d^{4} x^{3} + 10 \, b^{4} c^{3} d^{3} x^{2} + 5 \, b^{4} c^{4} d^{2} x + b^{4} c^{5} d\right )} \log \left (F\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^8,x, algorithm="fricas")

[Out]

-1/16*(15*sqrt(pi)*(d^6*x^5 + 5*c*d^5*x^4 + 10*c^2*d^4*x^3 + 10*c^3*d^3*x^2 + 5*c^4*d^2*x + c^5*d)*F^a*sqrt(-b
*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) + 2*(4*b^3*log(F)^3 - 10*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^
2)*log(F)^2 + 15*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*log(F))*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^4*d^6*x^5 + 5*b^4*c*d^5*x^4 + 10*b^4*c^2*d^4*x^3 + 10*b^
4*c^3*d^3*x^2 + 5*b^4*c^4*d^2*x + b^4*c^5*d)*log(F)^4)

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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)**8,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 )}^{8}}\,{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)^8,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^8, x)