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

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

[Out]

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

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Rubi [A]  time = 0.168491, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2214, 2206, 2211, 2204} \[ -\frac{4 \sqrt{\pi } b^{5/2} F^a \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{15 d}+\frac{4 b^2 \log ^2(F) (c+d x) F^{a+\frac{b}{(c+d x)^2}}}{15 d}+\frac{(c+d x)^5 F^{a+\frac{b}{(c+d x)^2}}}{5 d}+\frac{2 b \log (F) (c+d x)^3 F^{a+\frac{b}{(c+d x)^2}}}{15 d} \]

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)*(c + d*x)^5)/(5*d) + (2*b*F^(a + b/(c + d*x)^2)*(c + d*x)^3*Log[F])/(15*d) + (4*b^2*F^(
a + b/(c + d*x)^2)*(c + d*x)*Log[F]^2)/(15*d) - (4*b^(5/2)*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]
*Log[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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[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 F^{a+\frac{b}{(c+d x)^2}} (c+d x)^4 \, dx &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac{1}{5} (2 b \log (F)) \int F^{a+\frac{b}{(c+d x)^2}} (c+d x)^2 \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac{2 b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac{1}{15} \left (4 b^2 \log ^2(F)\right ) \int F^{a+\frac{b}{(c+d x)^2}} \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac{2 b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac{4 b^2 F^{a+\frac{b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}+\frac{1}{15} \left (8 b^3 \log ^3(F)\right ) \int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^2} \, dx\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac{2 b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac{4 b^2 F^{a+\frac{b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac{\left (8 b^3 \log ^3(F)\right ) \operatorname{Subst}\left (\int F^{a+b x^2} \, dx,x,\frac{1}{c+d x}\right )}{15 d}\\ &=\frac{F^{a+\frac{b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac{2 b F^{a+\frac{b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac{4 b^2 F^{a+\frac{b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac{4 b^{5/2} F^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right ) \log ^{\frac{5}{2}}(F)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.0963702, size = 97, normalized size = 0.71 \[ \frac{F^a \left ((c+d x) F^{\frac{b}{(c+d x)^2}} \left (4 b^2 \log ^2(F)+2 b \log (F) (c+d x)^2+3 (c+d x)^4\right )-4 \sqrt{\pi } b^{5/2} \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.046, size = 324, normalized size = 2.4 \begin{align*}{\frac{{d}^{4}{F}^{a}{x}^{5}}{5}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{d}^{3}{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}c{x}^{4}+2\,{d}^{2}{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}{c}^{2}{x}^{3}+2\,d{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}{c}^{3}{x}^{2}+{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}{c}^{4}x+{\frac{{F}^{a}{c}^{5}}{5\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,{d}^{2}{F}^{a}b\ln \left ( F \right ){x}^{3}}{15}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,d{F}^{a}b\ln \left ( F \right ) c{x}^{2}}{5}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,{F}^{a}b\ln \left ( F \right ){c}^{2}x}{5}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,{F}^{a}b\ln \left ( F \right ){c}^{3}}{15\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{4\,{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}x}{15}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{4\,{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}c}{15\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}-{\frac{4\,{F}^{a}{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}\sqrt{\pi }}{15\,d}{\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]

1/5*d^4*F^a*F^(b/(d*x+c)^2)*x^5+d^3*F^a*F^(b/(d*x+c)^2)*c*x^4+2*d^2*F^a*F^(b/(d*x+c)^2)*c^2*x^3+2*d*F^a*F^(b/(
d*x+c)^2)*c^3*x^2+F^a*F^(b/(d*x+c)^2)*c^4*x+1/5/d*F^a*F^(b/(d*x+c)^2)*c^5+2/15*d^2*F^a*b*ln(F)*F^(b/(d*x+c)^2)
*x^3+2/5*d*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c*x^2+2/5*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c^2*x+2/15/d*F^a*b*ln(F)*F^(b/(d*
x+c)^2)*c^3+4/15*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*x+4/15/d*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c-4/15/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*} \frac{1}{15} \,{\left (3 \, F^{a} d^{4} x^{5} + 15 \, F^{a} c d^{3} x^{4} + 2 \,{\left (15 \, F^{a} c^{2} d^{2} + F^{a} b d^{2} \log \left (F\right )\right )} x^{3} + 6 \,{\left (5 \, F^{a} c^{3} d + F^{a} b c d \log \left (F\right )\right )} x^{2} +{\left (15 \, F^{a} c^{4} + 6 \, F^{a} b c^{2} \log \left (F\right ) + 4 \, F^{a} b^{2} \log \left (F\right )^{2}\right )} x\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac{2 \,{\left (4 \, F^{a} b^{3} d x \log \left (F\right )^{3} - 3 \, F^{a} b c^{5} \log \left (F\right ) - 2 \, F^{a} b^{2} c^{3} \log \left (F\right )^{2}\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{15 \,{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{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]

1/15*(3*F^a*d^4*x^5 + 15*F^a*c*d^3*x^4 + 2*(15*F^a*c^2*d^2 + F^a*b*d^2*log(F))*x^3 + 6*(5*F^a*c^3*d + F^a*b*c*
d*log(F))*x^2 + (15*F^a*c^4 + 6*F^a*b*c^2*log(F) + 4*F^a*b^2*log(F)^2)*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + in
tegrate(2/15*(4*F^a*b^3*d*x*log(F)^3 - 3*F^a*b*c^5*log(F) - 2*F^a*b^2*c^3*log(F)^2)*F^(b/(d^2*x^2 + 2*c*d*x +
c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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Fricas [A]  time = 1.6879, size = 458, normalized size = 3.37 \begin{align*} \frac{4 \, \sqrt{\pi } F^{a} b^{2} d \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 ) \log \left (F\right )^{2} +{\left (3 \, d^{5} x^{5} + 15 \, c d^{4} x^{4} + 30 \, c^{2} d^{3} x^{3} + 30 \, c^{3} d^{2} x^{2} + 15 \, c^{4} d x + 3 \, c^{5} + 4 \,{\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} + 2 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\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}}}}{15 \, d} \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]

1/15*(4*sqrt(pi)*F^a*b^2*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c))*log(F)^2 + (3*d^5*x^5 + 15
*c*d^4*x^4 + 30*c^2*d^3*x^3 + 30*c^3*d^2*x^2 + 15*c^4*d*x + 3*c^5 + 4*(b^2*d*x + b^2*c)*log(F)^2 + 2*(b*d^3*x^
3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*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)))/d

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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{\left (d x + c\right )}^{4} F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}\,{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((d*x + c)^4*F^(a + b/(d*x + c)^2), x)