\(\int F^{a+b x} x^{7/2} \, dx\) [30]

Optimal result

Integrand size = 13, antiderivative size = 131 \[ \int F^{a+b x} x^{7/2} \, dx=\frac {105 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{16 b^{9/2} \log ^{\frac {9}{2}}(F)}-\frac {105 F^{a+b x} \sqrt {x}}{8 b^4 \log ^4(F)}+\frac {35 F^{a+b x} x^{3/2}}{4 b^3 \log ^3(F)}-\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)} \]

[Out]

35/4*F^(b*x+a)*x^(3/2)/b^3/ln(F)^3-7/2*F^(b*x+a)*x^(5/2)/b^2/ln(F)^2+F^(b*x+a)*x^(7/2)/b/ln(F)+105/16*F^a*erfi
(b^(1/2)*x^(1/2)*ln(F)^(1/2))*Pi^(1/2)/b^(9/2)/ln(F)^(9/2)-105/8*F^(b*x+a)*x^(1/2)/b^4/ln(F)^4

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 131, 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 = {2207, 2211, 2235} \[ \int F^{a+b x} x^{7/2} \, dx=\frac {105 \sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{16 b^{9/2} \log ^{\frac {9}{2}}(F)}-\frac {105 \sqrt {x} F^{a+b x}}{8 b^4 \log ^4(F)}+\frac {35 x^{3/2} F^{a+b x}}{4 b^3 \log ^3(F)}-\frac {7 x^{5/2} F^{a+b x}}{2 b^2 \log ^2(F)}+\frac {x^{7/2} F^{a+b x}}{b \log (F)} \]

[In]

Int[F^(a + b*x)*x^(7/2),x]

[Out]

(105*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*Sqrt[x]*Sqrt[Log[F]]])/(16*b^(9/2)*Log[F]^(9/2)) - (105*F^(a + b*x)*Sqrt[x])/(8
*b^4*Log[F]^4) + (35*F^(a + b*x)*x^(3/2))/(4*b^3*Log[F]^3) - (7*F^(a + b*x)*x^(5/2))/(2*b^2*Log[F]^2) + (F^(a
+ b*x)*x^(7/2))/(b*Log[F])

Rule 2207

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] &&  !TrueQ[$UseGamma]

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 {F^{a+b x} x^{7/2}}{b \log (F)}-\frac {7 \int F^{a+b x} x^{5/2} \, dx}{2 b \log (F)} \\ & = -\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)}+\frac {35 \int F^{a+b x} x^{3/2} \, dx}{4 b^2 \log ^2(F)} \\ & = \frac {35 F^{a+b x} x^{3/2}}{4 b^3 \log ^3(F)}-\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)}-\frac {105 \int F^{a+b x} \sqrt {x} \, dx}{8 b^3 \log ^3(F)} \\ & = -\frac {105 F^{a+b x} \sqrt {x}}{8 b^4 \log ^4(F)}+\frac {35 F^{a+b x} x^{3/2}}{4 b^3 \log ^3(F)}-\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)}+\frac {105 \int \frac {F^{a+b x}}{\sqrt {x}} \, dx}{16 b^4 \log ^4(F)} \\ & = -\frac {105 F^{a+b x} \sqrt {x}}{8 b^4 \log ^4(F)}+\frac {35 F^{a+b x} x^{3/2}}{4 b^3 \log ^3(F)}-\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)}+\frac {105 \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt {x}\right )}{8 b^4 \log ^4(F)} \\ & = \frac {105 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )}{16 b^{9/2} \log ^{\frac {9}{2}}(F)}-\frac {105 F^{a+b x} \sqrt {x}}{8 b^4 \log ^4(F)}+\frac {35 F^{a+b x} x^{3/2}}{4 b^3 \log ^3(F)}-\frac {7 F^{a+b x} x^{5/2}}{2 b^2 \log ^2(F)}+\frac {F^{a+b x} x^{7/2}}{b \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.27 \[ \int F^{a+b x} x^{7/2} \, dx=\frac {F^a \Gamma \left (\frac {9}{2},-b x \log (F)\right ) \sqrt {-b x \log (F)}}{b^5 \sqrt {x} \log ^5(F)} \]

[In]

Integrate[F^(a + b*x)*x^(7/2),x]

[Out]

(F^a*Gamma[9/2, -(b*x*Log[F])]*Sqrt[-(b*x*Log[F])])/(b^5*Sqrt[x]*Log[F]^5)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.76

method	result	size
meijerg	\(-\frac {F^{a} \left (-\frac {\sqrt {x}\, \left (-b \right )^{\frac {9}{2}} \sqrt {\ln \left (F \right )}\, \left (-72 b^{3} x^{3} \ln \left (F \right )^{3}+252 b^{2} x^{2} \ln \left (F \right )^{2}-630 x b \ln \left (F \right )+945\right ) {\mathrm e}^{x b \ln \left (F \right )}}{72 b^{4}}+\frac {105 \left (-b \right )^{\frac {9}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{16 b^{\frac {9}{2}}}\right )}{\left (-b \right )^{\frac {7}{2}} \ln \left (F \right )^{\frac {9}{2}} b}\)	\(99\)

[In]

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

[Out]

-F^a/(-b)^(7/2)/ln(F)^(9/2)/b*(-1/72*x^(1/2)*(-b)^(9/2)*ln(F)^(1/2)*(-72*b^3*x^3*ln(F)^3+252*b^2*x^2*ln(F)^2-6
30*x*b*ln(F)+945)/b^4*exp(x*b*ln(F))+105/16*(-b)^(9/2)/b^(9/2)*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 = 89, normalized size of antiderivative = 0.68 \[ \int F^{a+b x} x^{7/2} \, dx=-\frac {105 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) - 2 \, {\left (8 \, b^{4} x^{3} \log \left (F\right )^{4} - 28 \, b^{3} x^{2} \log \left (F\right )^{3} + 70 \, b^{2} x \log \left (F\right )^{2} - 105 \, b \log \left (F\right )\right )} F^{b x + a} \sqrt {x}}{16 \, b^{5} \log \left (F\right )^{5}} \]

[In]

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

[Out]

-1/16*(105*sqrt(pi)*sqrt(-b*log(F))*F^a*erf(sqrt(-b*log(F))*sqrt(x)) - 2*(8*b^4*x^3*log(F)^4 - 28*b^3*x^2*log(
F)^3 + 70*b^2*x*log(F)^2 - 105*b*log(F))*F^(b*x + a)*sqrt(x))/(b^5*log(F)^5)

Sympy [F]

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

[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 = 0.18 \[ \int F^{a+b x} x^{7/2} \, dx=-\frac {F^{a} x^{\frac {9}{2}} \Gamma \left (\frac {9}{2}, -b x \log \left (F\right )\right )}{\left (-b x \log \left (F\right )\right )^{\frac {9}{2}}} \]

[In]

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

[Out]

-F^a*x^(9/2)*gamma(9/2, -b*x*log(F))/(-b*x*log(F))^(9/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int F^{a+b x} x^{7/2} \, dx=-\frac {105 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} \sqrt {x}\right )}{16 \, \sqrt {-b \log \left (F\right )} b^{4} \log \left (F\right )^{4}} + \frac {{\left (8 \, b^{3} x^{\frac {7}{2}} \log \left (F\right )^{3} - 28 \, b^{2} x^{\frac {5}{2}} \log \left (F\right )^{2} + 70 \, b x^{\frac {3}{2}} \log \left (F\right ) - 105 \, \sqrt {x}\right )} e^{\left (b x \log \left (F\right ) + a \log \left (F\right )\right )}}{8 \, b^{4} \log \left (F\right )^{4}} \]

[In]

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

[Out]

-105/16*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*sqrt(x))/(sqrt(-b*log(F))*b^4*log(F)^4) + 1/8*(8*b^3*x^(7/2)*log(F)^
3 - 28*b^2*x^(5/2)*log(F)^2 + 70*b*x^(3/2)*log(F) - 105*sqrt(x))*e^(b*x*log(F) + a*log(F))/(b^4*log(F)^4)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.63 \[ \int F^{a+b x} x^{7/2} \, dx=\frac {F^a\,x^{7/2}\,\left (\frac {105\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )}{16}+F^{b\,x}\,\left (\frac {105\,\sqrt {-b\,x\,\ln \left (F\right )}}{8}+\frac {35\,{\left (-b\,x\,\ln \left (F\right )\right )}^{3/2}}{4}+\frac {7\,{\left (-b\,x\,\ln \left (F\right )\right )}^{5/2}}{2}+{\left (-b\,x\,\ln \left (F\right )\right )}^{7/2}\right )\right )}{b\,\ln \left (F\right )\,{\left (-b\,x\,\ln \left (F\right )\right )}^{7/2}} \]

[In]

int(F^(a + b*x)*x^(7/2),x)

[Out]

(F^a*x^(7/2)*((105*pi^(1/2)*erfc((-b*x*log(F))^(1/2)))/16 + F^(b*x)*((105*(-b*x*log(F))^(1/2))/8 + (35*(-b*x*l
og(F))^(3/2))/4 + (7*(-b*x*log(F))^(5/2))/2 + (-b*x*log(F))^(7/2))))/(b*log(F)*(-b*x*log(F))^(7/2))