Integrand size = 13, antiderivative size = 108 \[ \int F^{a+b x} x^{5/2} \, dx=-\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{8 b^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 F^{a+b x} \sqrt {x}}{4 b^3 \log ^3(F)}-\frac {5 F^{a+b x} x^{3/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{5/2}}{b \log (F)} \] Output:
-15/8*F^a*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2))/b^(7/2)/ln(F)^(7/2)+1 5/4*F^(b*x+a)*x^(1/2)/b^3/ln(F)^3-5/2*F^(b*x+a)*x^(3/2)/b^2/ln(F)^2+F^(b*x +a)*x^(5/2)/b/ln(F)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.33 \[ \int F^{a+b x} x^{5/2} \, dx=\frac {F^a \sqrt {x} \Gamma \left (\frac {7}{2},-b x \log (F)\right )}{b^3 \log ^3(F) \sqrt {-b x \log (F)}} \] Input:
Integrate[F^(a + b*x)*x^(5/2),x]
Output:
(F^a*Sqrt[x]*Gamma[7/2, -(b*x*Log[F])])/(b^3*Log[F]^3*Sqrt[-(b*x*Log[F])])
Time = 0.63 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2607, 2607, 2607, 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.
\(\displaystyle \int x^{5/2} F^{a+b x} \, dx\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {x^{5/2} F^{a+b x}}{b \log (F)}-\frac {5 \int F^{a+b x} x^{3/2}dx}{2 b \log (F)}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {x^{5/2} F^{a+b x}}{b \log (F)}-\frac {5 \left (\frac {x^{3/2} F^{a+b x}}{b \log (F)}-\frac {3 \int F^{a+b x} \sqrt {x}dx}{2 b \log (F)}\right )}{2 b \log (F)}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {x^{5/2} F^{a+b x}}{b \log (F)}-\frac {5 \left (\frac {x^{3/2} F^{a+b x}}{b \log (F)}-\frac {3 \left (\frac {\sqrt {x} F^{a+b x}}{b \log (F)}-\frac {\int \frac {F^{a+b x}}{\sqrt {x}}dx}{2 b \log (F)}\right )}{2 b \log (F)}\right )}{2 b \log (F)}\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle \frac {x^{5/2} F^{a+b x}}{b \log (F)}-\frac {5 \left (\frac {x^{3/2} F^{a+b x}}{b \log (F)}-\frac {3 \left (\frac {\sqrt {x} F^{a+b x}}{b \log (F)}-\frac {\int F^{a+b x}d\sqrt {x}}{b \log (F)}\right )}{2 b \log (F)}\right )}{2 b \log (F)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {x^{5/2} F^{a+b x}}{b \log (F)}-\frac {5 \left (\frac {x^{3/2} F^{a+b x}}{b \log (F)}-\frac {3 \left (\frac {\sqrt {x} F^{a+b x}}{b \log (F)}-\frac {\sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{2 b^{3/2} \log ^{\frac {3}{2}}(F)}\right )}{2 b \log (F)}\right )}{2 b \log (F)}\) |
Input:
Int[F^(a + b*x)*x^(5/2),x]
Output:
(F^(a + b*x)*x^(5/2))/(b*Log[F]) - (5*((F^(a + b*x)*x^(3/2))/(b*Log[F]) - (3*(-1/2*(F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqrt[x]*Sqrt[Log[F]]])/(b^(3/2)*Log[F] ^(3/2)) + (F^(a + b*x)*Sqrt[x])/(b*Log[F])))/(2*b*Log[F])))/(2*b*Log[F])
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] - Simp[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] && !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 = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81
method | result | size |
meijerg | \(-\frac {F^{a} \left (\frac {\sqrt {x}\, \left (-b \right )^{\frac {7}{2}} \sqrt {\ln \left (F \right )}\, \left (28 b^{2} x^{2} \ln \left (F \right )^{2}-70 \ln \left (F \right ) b x +105\right ) {\mathrm e}^{\ln \left (F \right ) b x}}{28 b^{3}}-\frac {15 \left (-b \right )^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{8 b^{\frac {7}{2}}}\right )}{\left (-b \right )^{\frac {5}{2}} \ln \left (F \right )^{\frac {7}{2}} b}\) | \(87\) |
Input:
int(F^(b*x+a)*x^(5/2),x,method=_RETURNVERBOSE)
Output:
-F^a/(-b)^(5/2)/ln(F)^(7/2)/b*(1/28*x^(1/2)*(-b)^(7/2)*ln(F)^(1/2)*(28*b^2 *x^2*ln(F)^2-70*ln(F)*b*x+105)/b^3*exp(ln(F)*b*x)-15/8*(-b)^(7/2)/b^(7/2)* Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int F^{a+b x} x^{5/2} \, dx=\frac {15 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) + 2 \, {\left (4 \, b^{3} x^{2} \log \left (F\right )^{3} - 10 \, b^{2} x \log \left (F\right )^{2} + 15 \, b \log \left (F\right )\right )} F^{b x + a} \sqrt {x}}{8 \, b^{4} \log \left (F\right )^{4}} \] Input:
integrate(F^(b*x+a)*x^(5/2),x, algorithm="fricas")
Output:
1/8*(15*sqrt(pi)*sqrt(-b*log(F))*F^a*erf(sqrt(-b*log(F))*sqrt(x)) + 2*(4*b ^3*x^2*log(F)^3 - 10*b^2*x*log(F)^2 + 15*b*log(F))*F^(b*x + a)*sqrt(x))/(b ^4*log(F)^4)
\[ \int F^{a+b x} x^{5/2} \, dx=\int F^{a + b x} x^{\frac {5}{2}}\, dx \] Input:
integrate(F**(b*x+a)*x**(5/2),x)
Output:
Integral(F**(a + b*x)*x**(5/2), x)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.22 \[ \int F^{a+b x} x^{5/2} \, dx=-\frac {F^{a} x^{\frac {7}{2}} \Gamma \left (\frac {7}{2}, -b x \log \left (F\right )\right )}{\left (-b x \log \left (F\right )\right )^{\frac {7}{2}}} \] Input:
integrate(F^(b*x+a)*x^(5/2),x, algorithm="maxima")
Output:
-F^a*x^(7/2)*gamma(7/2, -b*x*log(F))/(-b*x*log(F))^(7/2)
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.76 \[ \int F^{a+b x} x^{5/2} \, dx=\frac {15 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} \sqrt {x}\right )}{8 \, \sqrt {-b \log \left (F\right )} b^{3} \log \left (F\right )^{3}} + \frac {{\left (4 \, b^{2} x^{\frac {5}{2}} \log \left (F\right )^{2} - 10 \, b x^{\frac {3}{2}} \log \left (F\right ) + 15 \, \sqrt {x}\right )} e^{\left (b x \log \left (F\right ) + a \log \left (F\right )\right )}}{4 \, b^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^(b*x+a)*x^(5/2),x, algorithm="giac")
Output:
15/8*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*sqrt(x))/(sqrt(-b*log(F))*b^3*log(F )^3) + 1/4*(4*b^2*x^(5/2)*log(F)^2 - 10*b*x^(3/2)*log(F) + 15*sqrt(x))*e^( b*x*log(F) + a*log(F))/(b^3*log(F)^3)
Time = 17.45 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.67 \[ \int F^{a+b x} x^{5/2} \, dx=\frac {F^a\,x^{5/2}\,\left (F^{b\,x}\,\left (\frac {15\,\sqrt {-b\,x\,\ln \left (F\right )}}{4}+\frac {5\,{\left (-b\,x\,\ln \left (F\right )\right )}^{3/2}}{2}+{\left (-b\,x\,\ln \left (F\right )\right )}^{5/2}\right )+\frac {15\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )}{8}\right )}{b\,\ln \left (F\right )\,{\left (-b\,x\,\ln \left (F\right )\right )}^{5/2}} \] Input:
int(F^(a + b*x)*x^(5/2),x)
Output:
(F^a*x^(5/2)*(F^(b*x)*((15*(-b*x*log(F))^(1/2))/4 + (5*(-b*x*log(F))^(3/2) )/2 + (-b*x*log(F))^(5/2)) + (15*pi^(1/2)*erfc((-b*x*log(F))^(1/2)))/8))/( b*log(F)*(-b*x*log(F))^(5/2))
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.86 \[ \int F^{a+b x} x^{5/2} \, dx=\frac {f^{a} \left (15 \sqrt {\pi }\, \mathrm {erf}\left (\sqrt {x}\, \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, i \right ) i +8 \sqrt {x}\, f^{b x} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} x^{2}-20 \sqrt {x}\, f^{b x} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b x +30 \sqrt {x}\, f^{b x} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\right )}{8 \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{3} b^{3}} \] Input:
int(F^(b*x+a)*x^(5/2),x)
Output:
(f**a*(15*sqrt(pi)*erf(sqrt(x)*sqrt(b)*sqrt(log(f))*i)*i + 8*sqrt(x)*f**(b *x)*sqrt(b)*sqrt(log(f))*log(f)**2*b**2*x**2 - 20*sqrt(x)*f**(b*x)*sqrt(b) *sqrt(log(f))*log(f)*b*x + 30*sqrt(x)*f**(b*x)*sqrt(b)*sqrt(log(f))))/(8*s qrt(b)*sqrt(log(f))*log(f)**3*b**3)