3.83 \(\int f^{a+b x^2} x^{10} \, dx\)

Optimal. Leaf size=34 \[ -\frac {x^{11} f^a \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}} \]

[Out]

-1/2*f^a*x^11*(1048576/61836869254970658257624840625*GAMMA(51/2,-b*x^2*ln(f))-1048576/618368692549706582576248
40625*(-b*x^2*ln(f))^(49/2)*exp(b*x^2*ln(f))-524288/1261976923570829760359690625*(-b*x^2*ln(f))^(47/2)*exp(b*x
^2*ln(f))-262144/26850572841932548092759375*(-b*x^2*ln(f))^(45/2)*exp(b*x^2*ln(f))-131072/59667939648738995761
6875*(-b*x^2*ln(f))^(43/2)*exp(b*x^2*ln(f))-65536/13876265034590464130625*(-b*x^2*ln(f))^(41/2)*exp(b*x^2*ln(f
))-32768/338445488648547905625*(-b*x^2*ln(f))^(39/2)*exp(b*x^2*ln(f))-16384/8678089452526869375*(-b*x^2*ln(f))
^(37/2)*exp(b*x^2*ln(f))-8192/234542958176401875*(-b*x^2*ln(f))^(35/2)*exp(b*x^2*ln(f))-4096/6701227376468625*
(-b*x^2*ln(f))^(33/2)*exp(b*x^2*ln(f))-2048/203067496256625*(-b*x^2*ln(f))^(31/2)*exp(b*x^2*ln(f))-1024/655056
4395375*(-b*x^2*ln(f))^(29/2)*exp(b*x^2*ln(f))-512/225881530875*(-b*x^2*ln(f))^(27/2)*exp(b*x^2*ln(f))-256/836
5982625*(-b*x^2*ln(f))^(25/2)*exp(b*x^2*ln(f))-128/334639305*(-b*x^2*ln(f))^(23/2)*exp(b*x^2*ln(f))-64/1454953
5*(-b*x^2*ln(f))^(21/2)*exp(b*x^2*ln(f))-32/692835*(-b*x^2*ln(f))^(19/2)*exp(b*x^2*ln(f))-16/36465*(-b*x^2*ln(
f))^(17/2)*exp(b*x^2*ln(f))-8/2145*(-b*x^2*ln(f))^(15/2)*exp(b*x^2*ln(f))-4/143*(-b*x^2*ln(f))^(13/2)*exp(b*x^
2*ln(f))-2/11*(-b*x^2*ln(f))^(11/2)*exp(b*x^2*ln(f)))/(-b*x^2*ln(f))^(11/2)

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {x^{11} f^a \text {Gamma}\left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)*x^10,x]

[Out]

-(f^a*x^11*Gamma[11/2, -(b*x^2*Log[f])])/(2*(-(b*x^2*Log[f]))^(11/2))

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int f^{a+b x^2} x^{10} \, dx &=-\frac {f^a x^{11} \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ -\frac {x^{11} f^a \Gamma \left (\frac {11}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)*x^10,x]

[Out]

-1/2*(f^a*x^11*Gamma[11/2, -(b*x^2*Log[f])])/(-(b*x^2*Log[f]))^(11/2)

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fricas [A]  time = 0.42, size = 101, normalized size = 2.97 \[ \frac {945 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) + 2 \, {\left (16 \, b^{5} x^{9} \log \relax (f)^{5} - 72 \, b^{4} x^{7} \log \relax (f)^{4} + 252 \, b^{3} x^{5} \log \relax (f)^{3} - 630 \, b^{2} x^{3} \log \relax (f)^{2} + 945 \, b x \log \relax (f)\right )} f^{b x^{2} + a}}{64 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^10,x, algorithm="fricas")

[Out]

1/64*(945*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x) + 2*(16*b^5*x^9*log(f)^5 - 72*b^4*x^7*log(f)^4 +
252*b^3*x^5*log(f)^3 - 630*b^2*x^3*log(f)^2 + 945*b*x*log(f))*f^(b*x^2 + a))/(b^6*log(f)^6)

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giac [A]  time = 0.33, size = 104, normalized size = 3.06 \[ \frac {945 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (f)} x\right )}{64 \, \sqrt {-b \log \relax (f)} b^{5} \log \relax (f)^{5}} + \frac {{\left (16 \, b^{4} x^{9} \log \relax (f)^{4} - 72 \, b^{3} x^{7} \log \relax (f)^{3} + 252 \, b^{2} x^{5} \log \relax (f)^{2} - 630 \, b x^{3} \log \relax (f) + 945 \, x\right )} e^{\left (b x^{2} \log \relax (f) + a \log \relax (f)\right )}}{32 \, b^{5} \log \relax (f)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^10,x, algorithm="giac")

[Out]

945/64*sqrt(pi)*f^a*erf(-sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^5*log(f)^5) + 1/32*(16*b^4*x^9*log(f)^4 - 72*b^
3*x^7*log(f)^3 + 252*b^2*x^5*log(f)^2 - 630*b*x^3*log(f) + 945*x)*e^(b*x^2*log(f) + a*log(f))/(b^5*log(f)^5)

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maple [A]  time = 0.08, size = 142, normalized size = 4.18 \[ \frac {x^{9} f^{a} f^{b \,x^{2}}}{2 b \ln \relax (f )}-\frac {9 x^{7} f^{a} f^{b \,x^{2}}}{4 b^{2} \ln \relax (f )^{2}}+\frac {63 x^{5} f^{a} f^{b \,x^{2}}}{8 b^{3} \ln \relax (f )^{3}}-\frac {315 x^{3} f^{a} f^{b \,x^{2}}}{16 b^{4} \ln \relax (f )^{4}}+\frac {945 x \,f^{a} f^{b \,x^{2}}}{32 b^{5} \ln \relax (f )^{5}}-\frac {945 \sqrt {\pi }\, f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right )}{64 \sqrt {-b \ln \relax (f )}\, b^{5} \ln \relax (f )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)*x^10,x)

[Out]

1/2*f^a/ln(f)/b*x^9*f^(b*x^2)-9/4*f^a/ln(f)^2/b^2*x^7*f^(b*x^2)+63/8*f^a/ln(f)^3/b^3*x^5*f^(b*x^2)-315/16*f^a/
ln(f)^4/b^4*x^3*f^(b*x^2)+945/32*f^a/ln(f)^5/b^5*x*f^(b*x^2)-945/64*f^a/ln(f)^5/b^5*Pi^(1/2)/(-b*ln(f))^(1/2)*
erf((-b*ln(f))^(1/2)*x)

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maxima [A]  time = 0.75, size = 112, normalized size = 3.29 \[ \frac {{\left (16 \, b^{4} f^{a} x^{9} \log \relax (f)^{4} - 72 \, b^{3} f^{a} x^{7} \log \relax (f)^{3} + 252 \, b^{2} f^{a} x^{5} \log \relax (f)^{2} - 630 \, b f^{a} x^{3} \log \relax (f) + 945 \, f^{a} x\right )} f^{b x^{2}}}{32 \, b^{5} \log \relax (f)^{5}} - \frac {945 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right )}{64 \, \sqrt {-b \log \relax (f)} b^{5} \log \relax (f)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^10,x, algorithm="maxima")

[Out]

1/32*(16*b^4*f^a*x^9*log(f)^4 - 72*b^3*f^a*x^7*log(f)^3 + 252*b^2*f^a*x^5*log(f)^2 - 630*b*f^a*x^3*log(f) + 94
5*f^a*x)*f^(b*x^2)/(b^5*log(f)^5) - 945/64*sqrt(pi)*f^a*erf(sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^5*log(f)^5)

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mupad [B]  time = 3.57, size = 139, normalized size = 4.09 \[ -\frac {\frac {f^a\,\left (945\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \relax (f)}{\sqrt {b\,\ln \relax (f)}}\right )-1890\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \relax (f)}\right )}{64\,\sqrt {b\,\ln \relax (f)}}-\frac {63\,b^2\,f^a\,f^{b\,x^2}\,x^5\,{\ln \relax (f)}^2}{8}+\frac {9\,b^3\,f^a\,f^{b\,x^2}\,x^7\,{\ln \relax (f)}^3}{4}-\frac {b^4\,f^a\,f^{b\,x^2}\,x^9\,{\ln \relax (f)}^4}{2}+\frac {315\,b\,f^a\,f^{b\,x^2}\,x^3\,\ln \relax (f)}{16}}{b^5\,{\ln \relax (f)}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)*x^10,x)

[Out]

-((f^a*(945*pi^(1/2)*erfi((b*x*log(f))/(b*log(f))^(1/2)) - 1890*f^(b*x^2)*x*(b*log(f))^(1/2)))/(64*(b*log(f))^
(1/2)) - (63*b^2*f^a*f^(b*x^2)*x^5*log(f)^2)/8 + (9*b^3*f^a*f^(b*x^2)*x^7*log(f)^3)/4 - (b^4*f^a*f^(b*x^2)*x^9
*log(f)^4)/2 + (315*b*f^a*f^(b*x^2)*x^3*log(f))/16)/(b^5*log(f)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x^{2}} x^{10}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)*x**10,x)

[Out]

Integral(f**(a + b*x**2)*x**10, x)

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