3.37 \(\int \frac{F^{a+b x}}{x^{7/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac{8}{15} \sqrt{\pi } b^{5/2} F^a \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )-\frac{8 b^2 \log ^2(F) F^{a+b x}}{15 \sqrt{x}}-\frac{2 F^{a+b x}}{5 x^{5/2}}-\frac{4 b \log (F) F^{a+b x}}{15 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0894239, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2177, 2180, 2204} \[ \frac{8}{15} \sqrt{\pi } b^{5/2} F^a \log ^{\frac{5}{2}}(F) \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )-\frac{8 b^2 \log ^2(F) F^{a+b x}}{15 \sqrt{x}}-\frac{2 F^{a+b x}}{5 x^{5/2}}-\frac{4 b \log (F) F^{a+b x}}{15 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*x)/x^(7/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0500716, size = 61, normalized size = 0.61 \[ -\frac{2 F^a \left (F^{b x} \left (4 b^2 x^2 \log ^2(F)+2 b x \log (F)+3\right )-4 (-b x \log (F))^{5/2} \text{Gamma}\left (\frac{1}{2},-b x \log (F)\right )\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*x)/x^(7/2),x]

[Out]

(-2*F^a*(-4*Gamma[1/2, -(b*x*Log[F])]*(-(b*x*Log[F]))^(5/2) + F^(b*x)*(3 + 2*b*x*Log[F] + 4*b^2*x^2*Log[F]^2))
)/(15*x^(5/2))

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Maple [A]  time = 0.012, size = 84, normalized size = 0.8 \begin{align*} -{\frac{{F}^{a}}{b} \left ( -b \right ) ^{{\frac{7}{2}}} \left ( \ln \left ( F \right ) \right ) ^{{\frac{5}{2}}} \left ( -{\frac{2\,{{\rm e}^{b\ln \left ( F \right ) x}}}{5} \left ({\frac{4\,{b}^{2}{x}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}{3}}+{\frac{2\,b\ln \left ( F \right ) x}{3}}+1 \right ){x}^{-{\frac{5}{2}}} \left ( -b \right ) ^{-{\frac{5}{2}}} \left ( \ln \left ( F \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,\sqrt{\pi }}{15}{b}^{{\frac{5}{2}}}{\it erfi} \left ( \sqrt{b}\sqrt{x}\sqrt{\ln \left ( F \right ) } \right ) \left ( -b \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*x+a)/x^(7/2),x)

[Out]

-F^a*(-b)^(7/2)*ln(F)^(5/2)/b*(-2/5/x^(5/2)/(-b)^(5/2)/ln(F)^(5/2)*(4/3*b^2*x^2*ln(F)^2+2/3*b*ln(F)*x+1)*exp(b
*ln(F)*x)+8/15/(-b)^(5/2)*b^(5/2)*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))

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Maxima [A]  time = 1.25492, size = 32, normalized size = 0.32 \begin{align*} -\frac{\left (-b x \log \left (F\right )\right )^{\frac{5}{2}} F^{a} \Gamma \left (-\frac{5}{2}, -b x \log \left (F\right )\right )}{x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(7/2),x, algorithm="maxima")

[Out]

-(-b*x*log(F))^(5/2)*F^a*gamma(-5/2, -b*x*log(F))/x^(5/2)

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Fricas [A]  time = 1.52355, size = 205, normalized size = 2.05 \begin{align*} -\frac{2 \,{\left (4 \, \sqrt{\pi } \sqrt{-b \log \left (F\right )} F^{a} b^{2} x^{3} \operatorname{erf}\left (\sqrt{-b \log \left (F\right )} \sqrt{x}\right ) \log \left (F\right )^{2} +{\left (4 \, b^{2} x^{2} \log \left (F\right )^{2} + 2 \, b x \log \left (F\right ) + 3\right )} F^{b x + a} \sqrt{x}\right )}}{15 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(7/2),x, algorithm="fricas")

[Out]

-2/15*(4*sqrt(pi)*sqrt(-b*log(F))*F^a*b^2*x^3*erf(sqrt(-b*log(F))*sqrt(x))*log(F)^2 + (4*b^2*x^2*log(F)^2 + 2*
b*x*log(F) + 3)*F^(b*x + a)*sqrt(x))/x^3

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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**(b*x+a)/x**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{b x + a}}{x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)/x^(7/2),x, algorithm="giac")

[Out]

integrate(F^(b*x + a)/x^(7/2), x)