Integrand size = 13, antiderivative size = 100 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {8}{15} b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F) \] Output:
-2/5*F^(b*x+a)/x^(5/2)-4/15*b*F^(b*x+a)*ln(F)/x^(3/2)-8/15*b^2*F^(b*x+a)*l n(F)^2/x^(1/2)+8/15*b^(5/2)*F^a*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)) *ln(F)^(5/2)
Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.61 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^a \left (-4 \Gamma \left (\frac {1}{2},-b x \log (F)\right ) (-b x \log (F))^{5/2}+F^{b x} \left (3+2 b x \log (F)+4 b^2 x^2 \log ^2(F)\right )\right )}{15 x^{5/2}} \] Input:
Integrate[F^(a + b*x)/x^(7/2),x]
Output:
(-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))
Time = 0.55 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2608, 2608, 2608, 2611, 2633}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {F^{a+b x}}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} b \log (F) \int \frac {F^{a+b x}}{x^{5/2}}dx-\frac {2 F^{a+b x}}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \int \frac {F^{a+b x}}{x^{3/2}}dx-\frac {2 F^{a+b x}}{3 x^{3/2}}\right )-\frac {2 F^{a+b x}}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (2 b \log (F) \int \frac {F^{a+b x}}{\sqrt {x}}dx-\frac {2 F^{a+b x}}{\sqrt {x}}\right )-\frac {2 F^{a+b x}}{3 x^{3/2}}\right )-\frac {2 F^{a+b x}}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (4 b \log (F) \int F^{a+b x}d\sqrt {x}-\frac {2 F^{a+b x}}{\sqrt {x}}\right )-\frac {2 F^{a+b x}}{3 x^{3/2}}\right )-\frac {2 F^{a+b x}}{5 x^{5/2}}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (2 \sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )-\frac {2 F^{a+b x}}{\sqrt {x}}\right )-\frac {2 F^{a+b x}}{3 x^{3/2}}\right )-\frac {2 F^{a+b x}}{5 x^{5/2}}\) |
Input:
Int[F^(a + b*x)/x^(7/2),x]
Output:
(-2*F^(a + b*x))/(5*x^(5/2)) + (2*b*Log[F]*((-2*F^(a + b*x))/(3*x^(3/2)) + (2*b*((-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]])*Log[F])/3))/5
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] - Simp[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] && In tegerQ[2*m] && !TrueQ[$UseGamma]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : > Simp[2/d Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d *x]], x] /; FreeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
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]
Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84
method | result | size |
meijerg | \(-\frac {F^{a} \left (-b \right )^{\frac {7}{2}} \ln \left (F \right )^{\frac {5}{2}} \left (-\frac {2 \left (\frac {4 b^{2} x^{2} \ln \left (F \right )^{2}}{3}+\frac {2 \ln \left (F \right ) b x}{3}+1\right ) {\mathrm e}^{\ln \left (F \right ) b x}}{5 x^{\frac {5}{2}} \left (-b \right )^{\frac {5}{2}} \ln \left (F \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{b}\) | \(84\) |
Input:
int(F^(b*x+a)/x^(7/2),x,method=_RETURNVERBOSE)
Output:
-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*ln(F)*b*x+1)*exp(ln(F)*b*x)+8/15/(-b)^(5/2)*b^(5/2)*Pi^( 1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} b^{2} x^{3} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) \log \left (F\right )^{2} + {\left (4 \, b^{2} x^{2} \log \left (F\right )^{2} + 2 \, b x \log \left (F\right ) + 3\right )} F^{b x + a} \sqrt {x}\right )}}{15 \, x^{3}} \] Input:
integrate(F^(b*x+a)/x^(7/2),x, algorithm="fricas")
Output:
-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
\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int \frac {F^{a + b x}}{x^{\frac {7}{2}}}\, dx \] Input:
integrate(F**(b*x+a)/x**(7/2),x)
Output:
Integral(F**(a + b*x)/x**(7/2), x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.24 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\left (-b x \log \left (F\right )\right )^{\frac {5}{2}} F^{a} \Gamma \left (-\frac {5}{2}, -b x \log \left (F\right )\right )}{x^{\frac {5}{2}}} \] Input:
integrate(F^(b*x+a)/x^(7/2),x, algorithm="maxima")
Output:
-(-b*x*log(F))^(5/2)*F^a*gamma(-5/2, -b*x*log(F))/x^(5/2)
\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int { \frac {F^{b x + a}}{x^{\frac {7}{2}}} \,d x } \] Input:
integrate(F^(b*x+a)/x^(7/2),x, algorithm="giac")
Output:
integrate(F^(b*x + a)/x^(7/2), x)
Time = 22.81 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\frac {2\,F^{a+b\,x}}{5}+\frac {4\,F^{a+b\,x}\,b\,x\,\ln \left (F\right )}{15}+\frac {8\,F^{a+b\,x}\,b^2\,x^2\,{\ln \left (F\right )}^2}{15}-\frac {8\,F^a\,b^2\,x^2\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )\,{\ln \left (F\right )}^2\,\sqrt {-\pi \,b\,x\,\ln \left (F\right )}}{15}}{x^{5/2}} \] Input:
int(F^(a + b*x)/x^(7/2),x)
Output:
-((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*erfc((-b*x*log(F))^(1/2))*log(F)^2*(-b*x* pi*log(F))^(1/2))/15)/x^(5/2)
\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=f^{a} \left (\int \frac {f^{b x}}{\sqrt {x}\, x^{3}}d x \right ) \] Input:
int(F^(b*x+a)/x^(7/2),x)
Output:
f**a*int(f**(b*x)/(sqrt(x)*x**3),x)