\(\int \frac {F^{a+b x}}{x^{3/2}} \, dx\) [35]

Optimal result

Integrand size = 13, antiderivative size = 54 \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=-\frac {2 F^{a+b x}}{\sqrt {x}}+2 \sqrt {b} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \sqrt {\log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2208, 2211, 2235} \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=2 \sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )-\frac {2 F^{a+b x}}{\sqrt {x}} \]

[In]

Int[F^(a + b*x)/x^(3/2),x]

[Out]

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

Rule 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=-\frac {2 F^a \left (F^{b x}-\Gamma \left (\frac {1}{2},-b x \log (F)\right ) \sqrt {-b x \log (F)}\right )}{\sqrt {x}} \]

[In]

Integrate[F^(a + b*x)/x^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19

method	result	size
meijerg	\(-\frac {F^{a} \left (-b \right )^{\frac {3}{2}} \sqrt {\ln \left (F \right )}\, \left (-\frac {2 \,{\mathrm e}^{x b \ln \left (F \right )}}{\sqrt {x}\, \sqrt {-b}\, \sqrt {\ln \left (F \right )}}+\frac {2 \sqrt {b}\, \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{\sqrt {-b}}\right )}{b}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} x \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) + F^{b x + a} \sqrt {x}\right )}}{x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=\int \frac {F^{a + b x}}{x^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral(F**(a + b*x)/x**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.44 \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=-\frac {\sqrt {-b x \log \left (F\right )} F^{a} \Gamma \left (-\frac {1}{2}, -b x \log \left (F\right )\right )}{\sqrt {x}} \]

[In]

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

[Out]

-sqrt(-b*x*log(F))*F^a*gamma(-1/2, -b*x*log(F))/sqrt(x)

Giac [F]

\[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=\int { \frac {F^{b x + a}}{x^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(F^(b*x + a)/x^(3/2), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {F^{a+b x}}{x^{3/2}} \, dx=\frac {2\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )\,\sqrt {-b\,x\,\ln \left (F\right )}}{\sqrt {x}}-\frac {2\,F^a\,F^{b\,x}}{\sqrt {x}} \]

[In]

int(F^(a + b*x)/x^(3/2),x)

[Out]

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