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

Optimal. Leaf size=111 \[ \frac{3 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{8 b^{5/2} d \log ^{\frac{5}{2}}(F)}-\frac{3 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac{(c+d x)^3 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.153824, antiderivative size = 111, 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 = {2212, 2204} \[ \frac{3 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{8 b^{5/2} d \log ^{\frac{5}{2}}(F)}-\frac{3 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac{(c+d x)^3 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0939819, size = 90, normalized size = 0.81 \[ \frac{F^a \left (3 \sqrt{\pi } \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )+2 \sqrt{b} \sqrt{\log (F)} (c+d x) F^{b (c+d x)^2} \left (2 b \log (F) (c+d x)^2-3\right )\right )}{8 b^{5/2} d \log ^{\frac{5}{2}}(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.056, size = 300, normalized size = 2.7 \begin{align*}{\frac{{d}^{2}{x}^{3}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+{\frac{3\,cd{x}^{2}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+{\frac{3\,{c}^{2}x{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+{\frac{{c}^{3}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,d\ln \left ( F \right ) b}}-{\frac{3\,c{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d}}-{\frac{3\,x{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}}-{\frac{3\,\sqrt{\pi }{F}^{a}}{8\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d}{\it Erf} \left ( -d\sqrt{-b\ln \left ( F \right ) }x+{bc\ln \left ( F \right ){\frac{1}{\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^2/ln(F)/b*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+3/2*d*c/ln(F)/b*x^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)*
F^(c^2*b)*F^a+3/2*c^2/ln(F)/b*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+1/2/d*c^3/ln(F)/b*F^(b*d^2*x^2)*F^(2
*b*c*d*x)*F^(c^2*b)*F^a-3/4/d*c/ln(F)^2/b^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a-3/4/ln(F)^2/b^2*x*F^(b*d
^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a-3/8/d/ln(F)^2/b^2*Pi^(1/2)*F^a/(-b*ln(F))^(1/2)*erf(-d*(-b*ln(F))^(1/2)*x+
b*c*ln(F)/(-b*ln(F))^(1/2))

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Maxima [B]  time = 1.78637, size = 1463, normalized size = 13.18 \begin{align*} \text{result too large to display} \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]

-2*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*d*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^2/((b*d^2*log
(F))^(3/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*d^2*log(F)/(b*d^2*lo
g(F))^(3/2))*F^a*c^3*d/sqrt(b*d^2*log(F)) + 3*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*d^2*(erf(sqrt(-(b*d^2*x + b*
c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^3/((b*d^2*log(F))^(5/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*F^(
(b*d^2*x + b*c*d)^2/(b*d^2))*b^2*c*d^3*log(F)^2/(b*d^2*log(F))^(5/2) - (b*d^2*x + b*c*d)^3*gamma(3/2, -(b*d^2*
x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*d^2*log(F))^(5/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F^a*
c^2*d^2/sqrt(b*d^2*log(F)) - 2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*d^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(
b*d^2))) - 1)*log(F)^4/((b*d^2*log(F))^(7/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 3*F^((b*d^2*x + b*c*
d)^2/(b*d^2))*b^3*c^2*d^4*log(F)^3/(b*d^2*log(F))^(7/2) - 3*(b*d^2*x + b*c*d)^3*b*c*d*gamma(3/2, -(b*d^2*x + b
*c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*d^2*log(F))^(7/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + b^2*d^4*
gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^2/(b*d^2*log(F))^(7/2))*F^a*c*d^3/sqrt(b*d^2*log(F)) + 1/
2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^4*c^4*d^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^5/((b*d^
2*log(F))^(9/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 4*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^4*c^3*d^5*log
(F)^4/(b*d^2*log(F))^(9/2) - 6*(b*d^2*x + b*c*d)^3*b^2*c^2*d^2*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))
*log(F)^5/((b*d^2*log(F))^(9/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 4*b^3*c*d^5*gamma(2, -(b*d^2*x
+ b*c*d)^2*log(F)/(b*d^2))*log(F)^3/(b*d^2*log(F))^(9/2) - (b*d^2*x + b*c*d)^5*gamma(5/2, -(b*d^2*x + b*c*d)^2
*log(F)/(b*d^2))*log(F)^5/((b*d^2*log(F))^(9/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*d^4/sqrt(b*d
^2*log(F)) + 1/2*sqrt(pi)*F^(b*c^2 + a)*c^4*erf(sqrt(-b*log(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log
(F))*F^(b*c^2)*d)

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Fricas [A]  time = 1.53707, size = 331, normalized size = 2.98 \begin{align*} -\frac{3 \, \sqrt{\pi } \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 ) - 2 \,{\left (2 \,{\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} - 3 \,{\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{8 \, b^{3} d^{2} \log \left (F\right )^{3}} \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]

-1/8*(3*sqrt(pi)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d) - 2*(2*(b^2*d^4*x^3 + 3*b^2*c*d^
3*x^2 + 3*b^2*c^2*d^2*x + b^2*c^3*d)*log(F)^2 - 3*(b*d^2*x + b*c*d)*log(F))*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 +
a))/(b^3*d^2*log(F)^3)

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

Integral(F**(a + b*(c + d*x)**2)*(c + d*x)**4, x)

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Giac [A]  time = 1.33989, size = 150, normalized size = 1.35 \begin{align*} \frac{{\left (2 \, b d^{2}{\left (x + \frac{c}{d}\right )}^{3} \log \left (F\right ) - 3 \, x - \frac{3 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{4 \, b^{2} \log \left (F\right )^{2}} - \frac{3 \, \sqrt{\pi } F^{a} \operatorname{erf}\left (-\sqrt{-b \log \left (F\right )} d{\left (x + \frac{c}{d}\right )}\right )}{8 \, \sqrt{-b \log \left (F\right )} b^{2} d \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="giac")

[Out]

1/4*(2*b*d^2*(x + c/d)^3*log(F) - 3*x - 3*c/d)*e^(b*d^2*x^2*log(F) + 2*b*c*d*x*log(F) + b*c^2*log(F) + a*log(F
))/(b^2*log(F)^2) - 3/8*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*d*(x + c/d))/(sqrt(-b*log(F))*b^2*d*log(F)^2)