3.36 \(\int \frac{F^{a+b x}}{x^{5/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{4}{3} \sqrt{\pi } b^{3/2} F^a \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )-\frac{2 F^{a+b x}}{3 x^{3/2}}-\frac{4 b \log (F) F^{a+b x}}{3 \sqrt{x}} \]

[Out]

(-2*F^(a + b*x))/(3*x^(3/2)) - (4*b*F^(a + b*x)*Log[F])/(3*Sqrt[x]) + (4*b^(3/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqr
t[x]*Sqrt[Log[F]]]*Log[F]^(3/2))/3

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Rubi [A]  time = 0.064935, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2177, 2180, 2204} \[ \frac{4}{3} \sqrt{\pi } b^{3/2} F^a \log ^{\frac{3}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )-\frac{2 F^{a+b x}}{3 x^{3/2}}-\frac{4 b \log (F) F^{a+b x}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*x)/x^(5/2),x]

[Out]

(-2*F^(a + b*x))/(3*x^(3/2)) - (4*b*F^(a + b*x)*Log[F])/(3*Sqrt[x]) + (4*b^(3/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqr
t[x]*Sqrt[Log[F]]]*Log[F]^(3/2))/3

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 x}}{x^{5/2}} \, dx &=-\frac{2 F^{a+b x}}{3 x^{3/2}}+\frac{1}{3} (2 b \log (F)) \int \frac{F^{a+b x}}{x^{3/2}} \, dx\\ &=-\frac{2 F^{a+b x}}{3 x^{3/2}}-\frac{4 b F^{a+b x} \log (F)}{3 \sqrt{x}}+\frac{1}{3} \left (4 b^2 \log ^2(F)\right ) \int \frac{F^{a+b x}}{\sqrt{x}} \, dx\\ &=-\frac{2 F^{a+b x}}{3 x^{3/2}}-\frac{4 b F^{a+b x} \log (F)}{3 \sqrt{x}}+\frac{1}{3} \left (8 b^2 \log ^2(F)\right ) \operatorname{Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 F^{a+b x}}{3 x^{3/2}}-\frac{4 b F^{a+b x} \log (F)}{3 \sqrt{x}}+\frac{4}{3} b^{3/2} F^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right ) \log ^{\frac{3}{2}}(F)\\ \end{align*}

Mathematica [A]  time = 0.0368925, size = 49, normalized size = 0.64 \[ -\frac{2 F^a \left (2 (-b x \log (F))^{3/2} \text{Gamma}\left (\frac{1}{2},-b x \log (F)\right )+F^{b x} (2 b x \log (F)+1)\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*x)/x^(5/2),x]

[Out]

(-2*F^a*(2*Gamma[1/2, -(b*x*Log[F])]*(-(b*x*Log[F]))^(3/2) + F^(b*x)*(1 + 2*b*x*Log[F])))/(3*x^(3/2))

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Maple [A]  time = 0.011, size = 72, normalized size = 0.9 \begin{align*} -{\frac{{F}^{a}}{b} \left ( -b \right ) ^{{\frac{5}{2}}} \left ( \ln \left ( F \right ) \right ) ^{{\frac{3}{2}}} \left ( -{\frac{ \left ( 4\,b\ln \left ( F \right ) x+2 \right ){{\rm e}^{b\ln \left ( F \right ) x}}}{3}{x}^{-{\frac{3}{2}}} \left ( -b \right ) ^{-{\frac{3}{2}}} \left ( \ln \left ( F \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,\sqrt{\pi }}{3}{b}^{{\frac{3}{2}}}{\it erfi} \left ( \sqrt{b}\sqrt{x}\sqrt{\ln \left ( F \right ) } \right ) \left ( -b \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*x+a)/x^(5/2),x)

[Out]

-F^a*(-b)^(5/2)*ln(F)^(3/2)/b*(-2/3/x^(3/2)/(-b)^(3/2)/ln(F)^(3/2)*(2*b*ln(F)*x+1)*exp(b*ln(F)*x)+4/3/(-b)^(3/
2)*b^(3/2)*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))

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Maxima [A]  time = 1.25236, size = 32, normalized size = 0.42 \begin{align*} -\frac{\left (-b x \log \left (F\right )\right )^{\frac{3}{2}} F^{a} \Gamma \left (-\frac{3}{2}, -b x \log \left (F\right )\right )}{x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(5/2),x, algorithm="maxima")

[Out]

-(-b*x*log(F))^(3/2)*F^a*gamma(-3/2, -b*x*log(F))/x^(3/2)

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Fricas [A]  time = 1.54017, size = 170, normalized size = 2.21 \begin{align*} -\frac{2 \,{\left (2 \, \sqrt{\pi } \sqrt{-b \log \left (F\right )} F^{a} b x^{2} \operatorname{erf}\left (\sqrt{-b \log \left (F\right )} \sqrt{x}\right ) \log \left (F\right ) +{\left (2 \, b x \log \left (F\right ) + 1\right )} F^{b x + a} \sqrt{x}\right )}}{3 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 175.103, size = 39, normalized size = 0.51 \begin{align*} - \frac{4 F^{a} F^{b x} b \log{\left (F \right )}}{3 \sqrt{x}} - \frac{2 F^{a} F^{b x}}{3 x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(b*x+a)/x**(5/2),x)

[Out]

-4*F**a*F**(b*x)*b*log(F)/(3*sqrt(x)) - 2*F**a*F**(b*x)/(3*x**(3/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{b x + a}}{x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(5/2),x, algorithm="giac")

[Out]

integrate(F^(b*x + a)/x^(5/2), x)