3.32 \(\int F^{a+b x} x^{3/2} \, dx\)

Optimal. Leaf size=85 \[ \frac{3 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )}{4 b^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 \sqrt{x} F^{a+b x}}{2 b^2 \log ^2(F)}+\frac{x^{3/2} F^{a+b x}}{b \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.06581, antiderivative size = 85, 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 = {2176, 2180, 2204} \[ \frac{3 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )}{4 b^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 \sqrt{x} F^{a+b x}}{2 b^2 \log ^2(F)}+\frac{x^{3/2} F^{a+b x}}{b \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*x)*x^(3/2),x]

[Out]

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && 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 F^{a+b x} x^{3/2} \, dx &=\frac{F^{a+b x} x^{3/2}}{b \log (F)}-\frac{3 \int F^{a+b x} \sqrt{x} \, dx}{2 b \log (F)}\\ &=-\frac{3 F^{a+b x} \sqrt{x}}{2 b^2 \log ^2(F)}+\frac{F^{a+b x} x^{3/2}}{b \log (F)}+\frac{3 \int \frac{F^{a+b x}}{\sqrt{x}} \, dx}{4 b^2 \log ^2(F)}\\ &=-\frac{3 F^{a+b x} \sqrt{x}}{2 b^2 \log ^2(F)}+\frac{F^{a+b x} x^{3/2}}{b \log (F)}+\frac{3 \operatorname{Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt{x}\right )}{2 b^2 \log ^2(F)}\\ &=\frac{3 F^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} \sqrt{x} \sqrt{\log (F)}\right )}{4 b^{5/2} \log ^{\frac{5}{2}}(F)}-\frac{3 F^{a+b x} \sqrt{x}}{2 b^2 \log ^2(F)}+\frac{F^{a+b x} x^{3/2}}{b \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0059449, size = 36, normalized size = 0.42 \[ \frac{F^a \sqrt{-b x \log (F)} \text{Gamma}\left (\frac{5}{2},-b x \log (F)\right )}{b^3 \sqrt{x} \log ^3(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*x)*x^(3/2),x]

[Out]

(F^a*Gamma[5/2, -(b*x*Log[F])]*Sqrt[-(b*x*Log[F])])/(b^3*Sqrt[x]*Log[F]^3)

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Maple [A]  time = 0.012, size = 75, normalized size = 0.9 \begin{align*} -{\frac{{F}^{a}}{b} \left ( -{\frac{ \left ( -10\,b\ln \left ( F \right ) x+15 \right ){{\rm e}^{b\ln \left ( F \right ) x}}}{10\,{b}^{2}}\sqrt{x} \left ( -b \right ) ^{{\frac{5}{2}}}\sqrt{\ln \left ( F \right ) }}+{\frac{3\,\sqrt{\pi }}{4} \left ( -b \right ) ^{{\frac{5}{2}}}{\it erfi} \left ( \sqrt{b}\sqrt{x}\sqrt{\ln \left ( F \right ) } \right ){b}^{-{\frac{5}{2}}}} \right ) \left ( -b \right ) ^{-{\frac{3}{2}}} \left ( \ln \left ( F \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*x+a)*x^(3/2),x)

[Out]

-F^a/(-b)^(3/2)/ln(F)^(5/2)/b*(-1/10*x^(1/2)*(-b)^(5/2)*ln(F)^(1/2)*(-10*b*ln(F)*x+15)/b^2*exp(b*ln(F)*x)+3/4*
(-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 = 1.21725, size = 32, normalized size = 0.38 \begin{align*} -\frac{F^{a} x^{\frac{5}{2}} \Gamma \left (\frac{5}{2}, -b x \log \left (F\right )\right )}{\left (-b x \log \left (F\right )\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.49949, size = 188, normalized size = 2.21 \begin{align*} -\frac{3 \, \sqrt{\pi } \sqrt{-b \log \left (F\right )} F^{a} \operatorname{erf}\left (\sqrt{-b \log \left (F\right )} \sqrt{x}\right ) - 2 \,{\left (2 \, b^{2} x \log \left (F\right )^{2} - 3 \, b \log \left (F\right )\right )} F^{b x + a} \sqrt{x}}{4 \, b^{3} \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.24248, size = 95, normalized size = 1.12 \begin{align*} -\frac{3 \, \sqrt{\pi } F^{a} \operatorname{erf}\left (-\sqrt{-b \log \left (F\right )} \sqrt{x}\right )}{4 \, \sqrt{-b \log \left (F\right )} b^{2} \log \left (F\right )^{2}} + \frac{{\left (2 \, b x^{\frac{3}{2}} \log \left (F\right ) - 3 \, \sqrt{x}\right )} e^{\left (b x \log \left (F\right ) + a \log \left (F\right )\right )}}{2 \, b^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(3/2),x, algorithm="giac")

[Out]

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