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]

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

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Rubi [A]  time = 0.07, antiderivative size = 85, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.42 \[ \frac {F^a \sqrt {-b x \log (F)} \Gamma \left (\frac {5}{2},-b x \log (F)\right )}{b^3 \sqrt {x} \log ^3(F)} \]

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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fricas [A]  time = 0.44, size = 65, normalized size = 0.76 \[ -\frac {3 \, \sqrt {\pi } \sqrt {-b \log \relax (F)} F^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (F)} \sqrt {x}\right ) - 2 \, {\left (2 \, b^{2} x \log \relax (F)^{2} - 3 \, b \log \relax (F)\right )} F^{b x + a} \sqrt {x}}{4 \, b^{3} \log \relax (F)^{3}} \]

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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giac [A]  time = 0.42, size = 70, normalized size = 0.82 \[ -\frac {3 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (F)} \sqrt {x}\right )}{4 \, \sqrt {-b \log \relax (F)} b^{2} \log \relax (F)^{2}} + \frac {{\left (2 \, b x^{\frac {3}{2}} \log \relax (F) - 3 \, \sqrt {x}\right )} e^{\left (b x \log \relax (F) + a \log \relax (F)\right )}}{2 \, b^{2} \log \relax (F)^{2}} \]

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)

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maple [A]  time = 0.02, size = 75, normalized size = 0.88 \[ -\frac {\left (-\frac {\left (-b \right )^{\frac {5}{2}} \left (-10 b x \ln \relax (F )+15\right ) \sqrt {x}\, {\mathrm e}^{b x \ln \relax (F )} \sqrt {\ln \relax (F )}}{10 b^{2}}+\frac {3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, \erfi \left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \relax (F )}\right )}{4 b^{\frac {5}{2}}}\right ) F^{a}}{\left (-b \right )^{\frac {3}{2}} b \ln \relax (F )^{\frac {5}{2}}} \]

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*x*ln(F)+15)/b^2*exp(b*x*ln(F))+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 = 0.65, size = 24, normalized size = 0.28 \[ -\frac {F^{a} x^{\frac {5}{2}} \Gamma \left (\frac {5}{2}, -b x \log \relax (F)\right )}{\left (-b x \log \relax (F)\right )^{\frac {5}{2}}} \]

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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mupad [B]  time = 3.42, size = 75, normalized size = 0.88 \[ \frac {F^a\,F^{b\,x}\,x^{3/2}}{b\,\ln \relax (F)}-\frac {3\,F^a\,F^{b\,x}\,\sqrt {x}}{2\,b^2\,{\ln \relax (F)}^2}+\frac {3\,F^a\,x^{3/2}\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \relax (F)}\right )}{4\,b\,\ln \relax (F)\,{\left (-b\,x\,\ln \relax (F)\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*x)*x^(3/2),x)

[Out]

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

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sympy [A]  time = 130.97, size = 37, normalized size = 0.44 \[ - \frac {4 F^{a} F^{b x} b x^{\frac {7}{2}} \log {\relax (F )}}{35} + \frac {2 F^{a} F^{b x} x^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*F**a*F**(b*x)*b*x**(7/2)*log(F)/35 + 2*F**a*F**(b*x)*x**(5/2)/5

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