\(\int \frac {F^{a+b x}}{x^{9/2}} \, dx\) [38]

Optimal result

Integrand size = 13, antiderivative size = 123 \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=-\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{105 x^{3/2}}-\frac {16 b^3 F^{a+b x} \log ^3(F)}{105 \sqrt {x}}+\frac {16}{105} b^{7/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {7}{2}}(F) \]

[Out]

-2/7*F^(b*x+a)/x^(7/2)-4/35*b*F^(b*x+a)*ln(F)/x^(5/2)-8/105*b^2*F^(b*x+a)*ln(F)^2/x^(3/2)+16/105*b^(7/2)*F^a*e
rfi(b^(1/2)*x^(1/2)*ln(F)^(1/2))*ln(F)^(7/2)*Pi^(1/2)-16/105*b^3*F^(b*x+a)*ln(F)^3/x^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2208, 2211, 2235} \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=\frac {16}{105} \sqrt {\pi } b^{7/2} F^a \log ^{\frac {7}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )-\frac {16 b^3 \log ^3(F) F^{a+b x}}{105 \sqrt {x}}-\frac {8 b^2 \log ^2(F) F^{a+b x}}{105 x^{3/2}}-\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b \log (F) F^{a+b x}}{35 x^{5/2}} \]

[In]

Int[F^(a + b*x)/x^(9/2),x]

[Out]

(-2*F^(a + b*x))/(7*x^(7/2)) - (4*b*F^(a + b*x)*Log[F])/(35*x^(5/2)) - (8*b^2*F^(a + b*x)*Log[F]^2)/(105*x^(3/
2)) - (16*b^3*F^(a + b*x)*Log[F]^3)/(105*Sqrt[x]) + (16*b^(7/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqrt[x]*Sqrt[Log[F]]
]*Log[F]^(7/2))/105

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}}{7 x^{7/2}}+\frac {1}{7} (2 b \log (F)) \int \frac {F^{a+b x}}{x^{7/2}} \, dx \\ & = -\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}+\frac {1}{35} \left (4 b^2 \log ^2(F)\right ) \int \frac {F^{a+b x}}{x^{5/2}} \, dx \\ & = -\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{105 x^{3/2}}+\frac {1}{105} \left (8 b^3 \log ^3(F)\right ) \int \frac {F^{a+b x}}{x^{3/2}} \, dx \\ & = -\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{105 x^{3/2}}-\frac {16 b^3 F^{a+b x} \log ^3(F)}{105 \sqrt {x}}+\frac {1}{105} \left (16 b^4 \log ^4(F)\right ) \int \frac {F^{a+b x}}{\sqrt {x}} \, dx \\ & = -\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{105 x^{3/2}}-\frac {16 b^3 F^{a+b x} \log ^3(F)}{105 \sqrt {x}}+\frac {1}{105} \left (32 b^4 \log ^4(F)\right ) \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 F^{a+b x}}{7 x^{7/2}}-\frac {4 b F^{a+b x} \log (F)}{35 x^{5/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{105 x^{3/2}}-\frac {16 b^3 F^{a+b x} \log ^3(F)}{105 \sqrt {x}}+\frac {16}{105} b^{7/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {7}{2}}(F) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59 \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=-\frac {2 F^a \left (8 \Gamma \left (\frac {1}{2},-b x \log (F)\right ) (-b x \log (F))^{7/2}+F^{b x} \left (15+6 b x \log (F)+4 b^2 x^2 \log ^2(F)+8 b^3 x^3 \log ^3(F)\right )\right )}{105 x^{7/2}} \]

[In]

Integrate[F^(a + b*x)/x^(9/2),x]

[Out]

(-2*F^a*(8*Gamma[1/2, -(b*x*Log[F])]*(-(b*x*Log[F]))^(7/2) + F^(b*x)*(15 + 6*b*x*Log[F] + 4*b^2*x^2*Log[F]^2 +
8*b^3*x^3*Log[F]^3)))/(105*x^(7/2))

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.78

method	result	size
meijerg	\(-\frac {F^{a} \left (-b \right )^{\frac {9}{2}} \ln \left (F \right )^{\frac {7}{2}} \left (-\frac {2 \left (\frac {8 b^{3} x^{3} \ln \left (F \right )^{3}}{15}+\frac {4 b^{2} x^{2} \ln \left (F \right )^{2}}{15}+\frac {2 x b \ln \left (F \right )}{5}+1\right ) {\mathrm e}^{x b \ln \left (F \right )}}{7 x^{\frac {7}{2}} \left (-b \right )^{\frac {7}{2}} \ln \left (F \right )^{\frac {7}{2}}}+\frac {16 b^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{105 \left (-b \right )^{\frac {7}{2}}}\right )}{b}\)	\(96\)

[In]

int(F^(b*x+a)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

-F^a*(-b)^(9/2)*ln(F)^(7/2)/b*(-2/7/x^(7/2)/(-b)^(7/2)/ln(F)^(7/2)*(8/15*b^3*x^3*ln(F)^3+4/15*b^2*x^2*ln(F)^2+
2/5*x*b*ln(F)+1)*exp(x*b*ln(F))+16/105/(-b)^(7/2)*b^(7/2)*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.70 \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=-\frac {2 \, {\left (8 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} b^{3} x^{4} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) \log \left (F\right )^{3} + {\left (8 \, b^{3} x^{3} \log \left (F\right )^{3} + 4 \, b^{2} x^{2} \log \left (F\right )^{2} + 6 \, b x \log \left (F\right ) + 15\right )} F^{b x + a} \sqrt {x}\right )}}{105 \, x^{4}} \]

[In]

integrate(F^(b*x+a)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(8*sqrt(pi)*sqrt(-b*log(F))*F^a*b^3*x^4*erf(sqrt(-b*log(F))*sqrt(x))*log(F)^3 + (8*b^3*x^3*log(F)^3 + 4
*b^2*x^2*log(F)^2 + 6*b*x*log(F) + 15)*F^(b*x + a)*sqrt(x))/x^4

Sympy [F]

\[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=\int \frac {F^{a + b x}}{x^{\frac {9}{2}}}\, dx \]

[In]

integrate(F**(b*x+a)/x**(9/2),x)

[Out]

Integral(F**(a + b*x)/x**(9/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.20 \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=-\frac {\left (-b x \log \left (F\right )\right )^{\frac {7}{2}} F^{a} \Gamma \left (-\frac {7}{2}, -b x \log \left (F\right )\right )}{x^{\frac {7}{2}}} \]

[In]

integrate(F^(b*x+a)/x^(9/2),x, algorithm="maxima")

[Out]

-(-b*x*log(F))^(7/2)*F^a*gamma(-7/2, -b*x*log(F))/x^(7/2)

Giac [F]

\[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=\int { \frac {F^{b x + a}}{x^{\frac {9}{2}}} \,d x } \]

[In]

integrate(F^(b*x+a)/x^(9/2),x, algorithm="giac")

[Out]

integrate(F^(b*x + a)/x^(9/2), x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b x}}{x^{9/2}} \, dx=-\frac {\frac {2\,F^{a+b\,x}}{7}+\frac {4\,F^{a+b\,x}\,b\,x\,\ln \left (F\right )}{35}+\frac {8\,F^{a+b\,x}\,b^2\,x^2\,{\ln \left (F\right )}^2}{105}+\frac {16\,F^{a+b\,x}\,b^3\,x^3\,{\ln \left (F\right )}^3}{105}-\frac {16\,F^a\,b^3\,x^3\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )\,{\ln \left (F\right )}^3\,\sqrt {-\pi \,b\,x\,\ln \left (F\right )}}{105}}{x^{7/2}} \]

[In]

int(F^(a + b*x)/x^(9/2),x)

[Out]

-((2*F^(a + b*x))/7 + (4*F^(a + b*x)*b*x*log(F))/35 + (8*F^(a + b*x)*b^2*x^2*log(F)^2)/105 + (16*F^(a + b*x)*b
^3*x^3*log(F)^3)/105 - (16*F^a*b^3*x^3*erfc((-b*x*log(F))^(1/2))*log(F)^3*(-b*x*pi*log(F))^(1/2))/105)/x^(7/2)