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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.116043, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2214, 2206, 2211, 2204} \[ -\frac{2 \sqrt{\pi } b^{3/2} F^a \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )}{3 d}+\frac{(c+d x)^3 F^{a+\frac{b}{(c+d x)^2}}}{3 d}+\frac{2 b \log (F) (c+d x) F^{a+\frac{b}{(c+d x)^2}}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0759613, size = 79, normalized size = 0.77 \[ \frac{F^a \left ((c+d x) F^{\frac{b}{(c+d x)^2}} \left (2 b \log (F)+(c+d x)^2\right )-2 \sqrt{\pi } b^{3/2} \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (F)}}{c+d x}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 169, normalized size = 1.7 \begin{align*}{\frac{{d}^{2}{F}^{a}{x}^{3}}{3}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+d{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}c{x}^{2}+{F}^{a}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}{c}^{2}x+{\frac{{F}^{a}{c}^{3}}{3\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,{F}^{a}b\ln \left ( F \right ) x}{3}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{2\,{F}^{a}b\ln \left ( F \right ) c}{3\,d}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}-{\frac{2\,{F}^{a}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}\sqrt{\pi }}{3\,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)^2,x)

[Out]

1/3*d^2*F^a*F^(b/(d*x+c)^2)*x^3+d*F^a*F^(b/(d*x+c)^2)*c*x^2+F^a*F^(b/(d*x+c)^2)*c^2*x+1/3/d*F^a*F^(b/(d*x+c)^2
)*c^3+2/3*F^a*b*ln(F)*F^(b/(d*x+c)^2)*x+2/3/d*F^a*b*ln(F)*F^(b/(d*x+c)^2)*c-2/3/d*F^a*b^2*ln(F)^2*Pi^(1/2)/(-b
*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \,{\left (F^{a} d^{2} x^{3} + 3 \, F^{a} c d x^{2} +{\left (3 \, F^{a} c^{2} + 2 \, F^{a} b \log \left (F\right )\right )} x\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} + \int \frac{2 \,{\left (2 \, F^{a} b^{2} d x \log \left (F\right )^{2} - F^{a} b c^{3} \log \left (F\right )\right )} F^{\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{3 \,{\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)^2,x, algorithm="maxima")

[Out]

1/3*(F^a*d^2*x^3 + 3*F^a*c*d*x^2 + (3*F^a*c^2 + 2*F^a*b*log(F))*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate
(2/3*(2*F^a*b^2*d*x*log(F)^2 - F^a*b*c^3*log(F))*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)

________________________________________________________________________________________

Fricas [A]  time = 1.55369, size = 306, normalized size = 3. \begin{align*} \frac{2 \, \sqrt{\pi } F^{a} b 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 ) +{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3} + 2 \,{\left (b d x + b c\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}}}}{3 \, 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)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} 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)^2,x, algorithm="giac")

[Out]

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