3.31 \(\int F^{a+b x} x^{5/2} \, dx\)

Optimal. Leaf size=108 \[ -\frac {15 \sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{8 b^{7/2} \log ^{\frac {7}{2}}(F)}+\frac {15 \sqrt {x} F^{a+b x}}{4 b^3 \log ^3(F)}-\frac {5 x^{3/2} F^{a+b x}}{2 b^2 \log ^2(F)}+\frac {x^{5/2} F^{a+b x}}{b \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2176, 2180, 2204} \[ -\frac {15 \sqrt {\pi } F^a \text {Erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{8 b^{7/2} \log ^{\frac {7}{2}}(F)}-\frac {5 x^{3/2} F^{a+b x}}{2 b^2 \log ^2(F)}+\frac {15 \sqrt {x} F^{a+b x}}{4 b^3 \log ^3(F)}+\frac {x^{5/2} F^{a+b x}}{b \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*x)*x^(5/2),x]

[Out]

(-15*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqrt[x]*Sqrt[Log[F]]])/(8*b^(7/2)*Log[F]^(7/2)) + (15*F^(a + b*x)*Sqrt[x])/(4*b
^3*Log[F]^3) - (5*F^(a + b*x)*x^(3/2))/(2*b^2*Log[F]^2) + (F^(a + b*x)*x^(5/2))/(b*Log[F])

Rule 2176

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] - Dist[(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] &&  !$UseGamma === True

Rule 2180

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] &&  !$UseGamma === True

Rule 2204

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*} \int F^{a+b x} x^{5/2} \, dx &=\frac {F^{a+b x} x^{5/2}}{b \log (F)}-\frac {5 \int F^{a+b x} x^{3/2} \, dx}{2 b \log (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)}+\frac {15 \int F^{a+b x} \sqrt {x} \, dx}{4 b^2 \log ^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)}-\frac {15 \int \frac {F^{a+b x}}{\sqrt {x}} \, dx}{8 b^3 \log ^3(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)}-\frac {15 \operatorname {Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^3 \log ^3(F)}\\ &=-\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)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.33 \[ \frac {\sqrt {x} F^a \Gamma \left (\frac {7}{2},-b x \log (F)\right )}{b^3 \log ^3(F) \sqrt {-b x \log (F)}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*x)*x^(5/2),x]

[Out]

(F^a*Sqrt[x]*Gamma[7/2, -(b*x*Log[F])])/(b^3*Log[F]^3*Sqrt[-(b*x*Log[F])])

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fricas [A]  time = 0.42, size = 77, normalized size = 0.71 \[ \frac {15 \, \sqrt {\pi } \sqrt {-b \log \relax (F)} F^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (F)} \sqrt {x}\right ) + 2 \, {\left (4 \, b^{3} x^{2} \log \relax (F)^{3} - 10 \, b^{2} x \log \relax (F)^{2} + 15 \, b \log \relax (F)\right )} F^{b x + a} \sqrt {x}}{8 \, b^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(5/2),x, algorithm="fricas")

[Out]

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)

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giac [A]  time = 0.48, size = 82, normalized size = 0.76 \[ \frac {15 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (F)} \sqrt {x}\right )}{8 \, \sqrt {-b \log \relax (F)} b^{3} \log \relax (F)^{3}} + \frac {{\left (4 \, b^{2} x^{\frac {5}{2}} \log \relax (F)^{2} - 10 \, b x^{\frac {3}{2}} \log \relax (F) + 15 \, \sqrt {x}\right )} e^{\left (b x \log \relax (F) + a \log \relax (F)\right )}}{4 \, b^{3} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(5/2),x, algorithm="giac")

[Out]

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)

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maple [A]  time = 0.02, size = 87, normalized size = 0.81 \[ -\frac {\left (\frac {\left (-b \right )^{\frac {7}{2}} \left (28 b^{2} x^{2} \ln \relax (F )^{2}-70 b x \ln \relax (F )+105\right ) \sqrt {x}\, {\mathrm e}^{b x \ln \relax (F )} \sqrt {\ln \relax (F )}}{28 b^{3}}-\frac {15 \left (-b \right )^{\frac {7}{2}} \sqrt {\pi }\, \erfi \left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \relax (F )}\right )}{8 b^{\frac {7}{2}}}\right ) F^{a}}{\left (-b \right )^{\frac {5}{2}} b \ln \relax (F )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*x+a)*x^(5/2),x)

[Out]

-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*b*x*ln(F)+105)/b^3*e
xp(b*x*ln(F))-15/8*(-b)^(7/2)/b^(7/2)*Pi^(1/2)*erfi(b^(1/2)*x^(1/2)*ln(F)^(1/2)))

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maxima [A]  time = 0.70, size = 24, normalized size = 0.22 \[ -\frac {F^{a} x^{\frac {7}{2}} \Gamma \left (\frac {7}{2}, -b x \log \relax (F)\right )}{\left (-b x \log \relax (F)\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*x^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.46, size = 72, normalized size = 0.67 \[ \frac {F^a\,x^{5/2}\,\left (F^{b\,x}\,\left (\frac {15\,\sqrt {-b\,x\,\ln \relax (F)}}{4}+\frac {5\,{\left (-b\,x\,\ln \relax (F)\right )}^{3/2}}{2}+{\left (-b\,x\,\ln \relax (F)\right )}^{5/2}\right )+\frac {15\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \relax (F)}\right )}{8}\right )}{b\,\ln \relax (F)\,{\left (-b\,x\,\ln \relax (F)\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*x)*x^(5/2),x)

[Out]

(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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(b*x+a)*x**(5/2),x)

[Out]

Timed out

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