\(\int \frac {F^{a+b x}}{x^{7/2}} \, dx\) [37]

Optimal result

Integrand size = 13, antiderivative size = 100 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {8}{15} b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2208, 2211, 2235} \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\frac {8}{15} \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )-\frac {8 b^2 \log ^2(F) F^{a+b x}}{15 \sqrt {x}}-\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b \log (F) F^{a+b x}}{15 x^{3/2}} \]

[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 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}}{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 ) \text {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] (verified)

Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.61 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^a \left (-4 \Gamma \left (\frac {1}{2},-b x \log (F)\right ) (-b x \log (F))^{5/2}+F^{b x} \left (3+2 b x \log (F)+4 b^2 x^2 \log ^2(F)\right )\right )}{15 x^{5/2}} \]

[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))

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84

method	result	size
meijerg	\(-\frac {F^{a} \left (-b \right )^{\frac {7}{2}} \ln \left (F \right )^{\frac {5}{2}} \left (-\frac {2 \left (\frac {4 b^{2} x^{2} \ln \left (F \right )^{2}}{3}+\frac {2 x b \ln \left (F \right )}{3}+1\right ) {\mathrm e}^{x b \ln \left (F \right )}}{5 x^{\frac {5}{2}} \left (-b \right )^{\frac {5}{2}} \ln \left (F \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{b}\)	\(84\)

[In]

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

[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*x*b*ln(F)+1)*exp(x
*b*ln(F))+8/15/(-b)^(5/2)*b^(5/2)*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} b^{2} x^{3} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) \log \left (F\right )^{2} + {\left (4 \, b^{2} x^{2} \log \left (F\right )^{2} + 2 \, b x \log \left (F\right ) + 3\right )} F^{b x + a} \sqrt {x}\right )}}{15 \, x^{3}} \]

[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

Sympy [F]

\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int \frac {F^{a + b x}}{x^{\frac {7}{2}}}\, dx \]

[In]

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

[Out]

Integral(F**(a + b*x)/x**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.24 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\left (-b x \log \left (F\right )\right )^{\frac {5}{2}} F^{a} \Gamma \left (-\frac {5}{2}, -b x \log \left (F\right )\right )}{x^{\frac {5}{2}}} \]

[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)

Giac [F]

\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int { \frac {F^{b x + a}}{x^{\frac {7}{2}}} \,d x } \]

[In]

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

[Out]

integrate(F^(b*x + a)/x^(7/2), x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\frac {2\,F^{a+b\,x}}{5}+\frac {4\,F^{a+b\,x}\,b\,x\,\ln \left (F\right )}{15}+\frac {8\,F^{a+b\,x}\,b^2\,x^2\,{\ln \left (F\right )}^2}{15}-\frac {8\,F^a\,b^2\,x^2\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )\,{\ln \left (F\right )}^2\,\sqrt {-\pi \,b\,x\,\ln \left (F\right )}}{15}}{x^{5/2}} \]

[In]

int(F^(a + b*x)/x^(7/2),x)

[Out]

-((2*F^(a + b*x))/5 + (4*F^(a + b*x)*b*x*log(F))/15 + (8*F^(a + b*x)*b^2*x^2*log(F)^2)/15 - (8*F^a*b^2*x^2*erf
c((-b*x*log(F))^(1/2))*log(F)^2*(-b*x*pi*log(F))^(1/2))/15)/x^(5/2)