3.82 \(\int f^{a+b x^2} x^{12} \, dx\)

Optimal. Leaf size=34 \[ -\frac {x^{13} f^a \Gamma \left (\frac {13}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{13/2}} \]

[Out]

-1/2*f^a*x^13*(524288/5621533568633696205238621875*GAMMA(51/2,-b*x^2*ln(f))-524288/562153356863369620523862187
5*(-b*x^2*ln(f))^(49/2)*exp(b*x^2*ln(f))-262144/114725174870075432759971875*(-b*x^2*ln(f))^(47/2)*exp(b*x^2*ln
(f))-131072/2440961167448413462978125*(-b*x^2*ln(f))^(45/2)*exp(b*x^2*ln(f))-65536/54243581498853632510625*(-b
*x^2*ln(f))^(43/2)*exp(b*x^2*ln(f))-32768/1261478639508224011875*(-b*x^2*ln(f))^(41/2)*exp(b*x^2*ln(f))-16384/
30767771695322536875*(-b*x^2*ln(f))^(39/2)*exp(b*x^2*ln(f))-8192/788917222956988125*(-b*x^2*ln(f))^(37/2)*exp(
b*x^2*ln(f))-4096/21322087106945625*(-b*x^2*ln(f))^(35/2)*exp(b*x^2*ln(f))-2048/609202488769875*(-b*x^2*ln(f))
^(33/2)*exp(b*x^2*ln(f))-1024/18460681477875*(-b*x^2*ln(f))^(31/2)*exp(b*x^2*ln(f))-512/595505854125*(-b*x^2*l
n(f))^(29/2)*exp(b*x^2*ln(f))-256/20534684625*(-b*x^2*ln(f))^(27/2)*exp(b*x^2*ln(f))-128/760543875*(-b*x^2*ln(
f))^(25/2)*exp(b*x^2*ln(f))-64/30421755*(-b*x^2*ln(f))^(23/2)*exp(b*x^2*ln(f))-32/1322685*(-b*x^2*ln(f))^(21/2
)*exp(b*x^2*ln(f))-16/62985*(-b*x^2*ln(f))^(19/2)*exp(b*x^2*ln(f))-8/3315*(-b*x^2*ln(f))^(17/2)*exp(b*x^2*ln(f
))-4/195*(-b*x^2*ln(f))^(15/2)*exp(b*x^2*ln(f))-2/13*(-b*x^2*ln(f))^(13/2)*exp(b*x^2*ln(f)))/(-b*x^2*ln(f))^(1
3/2)

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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} \[ -\frac {x^{13} f^a \text {Gamma}\left (\frac {13}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)*x^12,x]

[Out]

-(f^a*x^13*Gamma[13/2, -(b*x^2*Log[f])])/(2*(-(b*x^2*Log[f]))^(13/2))

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^{12} \, dx &=-\frac {f^a x^{13} \Gamma \left (\frac {13}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{13/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ -\frac {x^{13} f^a \Gamma \left (\frac {13}{2},-b x^2 \log (f)\right )}{2 \left (-b x^2 \log (f)\right )^{13/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)*x^12,x]

[Out]

-1/2*(f^a*x^13*Gamma[13/2, -(b*x^2*Log[f])])/(-(b*x^2*Log[f]))^(13/2)

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fricas [A]  time = 0.41, size = 113, normalized size = 3.32 \[ -\frac {10395 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) - 2 \, {\left (32 \, b^{6} x^{11} \log \relax (f)^{6} - 176 \, b^{5} x^{9} \log \relax (f)^{5} + 792 \, b^{4} x^{7} \log \relax (f)^{4} - 2772 \, b^{3} x^{5} \log \relax (f)^{3} + 6930 \, b^{2} x^{3} \log \relax (f)^{2} - 10395 \, b x \log \relax (f)\right )} f^{b x^{2} + a}}{128 \, b^{7} \log \relax (f)^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^12,x, algorithm="fricas")

[Out]

-1/128*(10395*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x) - 2*(32*b^6*x^11*log(f)^6 - 176*b^5*x^9*log(
f)^5 + 792*b^4*x^7*log(f)^4 - 2772*b^3*x^5*log(f)^3 + 6930*b^2*x^3*log(f)^2 - 10395*b*x*log(f))*f^(b*x^2 + a))
/(b^7*log(f)^7)

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giac [A]  time = 0.31, size = 116, normalized size = 3.41 \[ -\frac {10395 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (f)} x\right )}{128 \, \sqrt {-b \log \relax (f)} b^{6} \log \relax (f)^{6}} + \frac {{\left (32 \, b^{5} x^{11} \log \relax (f)^{5} - 176 \, b^{4} x^{9} \log \relax (f)^{4} + 792 \, b^{3} x^{7} \log \relax (f)^{3} - 2772 \, b^{2} x^{5} \log \relax (f)^{2} + 6930 \, b x^{3} \log \relax (f) - 10395 \, x\right )} e^{\left (b x^{2} \log \relax (f) + a \log \relax (f)\right )}}{64 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^12,x, algorithm="giac")

[Out]

-10395/128*sqrt(pi)*f^a*erf(-sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^6*log(f)^6) + 1/64*(32*b^5*x^11*log(f)^5 -
176*b^4*x^9*log(f)^4 + 792*b^3*x^7*log(f)^3 - 2772*b^2*x^5*log(f)^2 + 6930*b*x^3*log(f) - 10395*x)*e^(b*x^2*lo
g(f) + a*log(f))/(b^6*log(f)^6)

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maple [A]  time = 0.15, size = 164, normalized size = 4.82 \[ \frac {x^{11} f^{a} f^{b \,x^{2}}}{2 b \ln \relax (f )}-\frac {11 x^{9} f^{a} f^{b \,x^{2}}}{4 b^{2} \ln \relax (f )^{2}}+\frac {99 x^{7} f^{a} f^{b \,x^{2}}}{8 b^{3} \ln \relax (f )^{3}}-\frac {693 x^{5} f^{a} f^{b \,x^{2}}}{16 b^{4} \ln \relax (f )^{4}}+\frac {3465 x^{3} f^{a} f^{b \,x^{2}}}{32 b^{5} \ln \relax (f )^{5}}-\frac {10395 x \,f^{a} f^{b \,x^{2}}}{64 b^{6} \ln \relax (f )^{6}}+\frac {10395 \sqrt {\pi }\, f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right )}{128 \sqrt {-b \ln \relax (f )}\, b^{6} \ln \relax (f )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)*x^12,x)

[Out]

1/2*f^a*f^(b*x^2)*x^11/ln(f)/b-11/4*f^a/ln(f)^2/b^2*x^9*f^(b*x^2)+99/8*f^a/ln(f)^3/b^3*x^7*f^(b*x^2)-693/16*f^
a/ln(f)^4/b^4*x^5*f^(b*x^2)+3465/32*f^a/ln(f)^5/b^5*x^3*f^(b*x^2)-10395/64*f^a/ln(f)^6/b^6*x*f^(b*x^2)+10395/1
28*f^a/ln(f)^6/b^6*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)*x)

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maxima [A]  time = 0.76, size = 127, normalized size = 3.74 \[ \frac {{\left (32 \, b^{5} f^{a} x^{11} \log \relax (f)^{5} - 176 \, b^{4} f^{a} x^{9} \log \relax (f)^{4} + 792 \, b^{3} f^{a} x^{7} \log \relax (f)^{3} - 2772 \, b^{2} f^{a} x^{5} \log \relax (f)^{2} + 6930 \, b f^{a} x^{3} \log \relax (f) - 10395 \, f^{a} x\right )} f^{b x^{2}}}{64 \, b^{6} \log \relax (f)^{6}} + \frac {10395 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right )}{128 \, \sqrt {-b \log \relax (f)} b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^12,x, algorithm="maxima")

[Out]

1/64*(32*b^5*f^a*x^11*log(f)^5 - 176*b^4*f^a*x^9*log(f)^4 + 792*b^3*f^a*x^7*log(f)^3 - 2772*b^2*f^a*x^5*log(f)
^2 + 6930*b*f^a*x^3*log(f) - 10395*f^a*x)*f^(b*x^2)/(b^6*log(f)^6) + 10395/128*sqrt(pi)*f^a*erf(sqrt(-b*log(f)
)*x)/(sqrt(-b*log(f))*b^6*log(f)^6)

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mupad [B]  time = 3.63, size = 154, normalized size = 4.53 \[ \frac {\frac {f^a\,\left (\frac {10395\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \relax (f)}{\sqrt {b\,\ln \relax (f)}}\right )}{128}-\frac {10395\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \relax (f)}}{64}\right )}{\sqrt {b\,\ln \relax (f)}}-\frac {693\,b^2\,f^{b\,x^2+a}\,x^5\,{\ln \relax (f)}^2}{16}+\frac {99\,b^3\,f^{b\,x^2+a}\,x^7\,{\ln \relax (f)}^3}{8}-\frac {11\,b^4\,f^{b\,x^2+a}\,x^9\,{\ln \relax (f)}^4}{4}+\frac {b^5\,f^{b\,x^2+a}\,x^{11}\,{\ln \relax (f)}^5}{2}+\frac {3465\,b\,f^{b\,x^2+a}\,x^3\,\ln \relax (f)}{32}}{b^6\,{\ln \relax (f)}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)*x^12,x)

[Out]

((f^a*((10395*pi^(1/2)*erfi((b*x*log(f))/(b*log(f))^(1/2)))/128 - (10395*f^(b*x^2)*x*(b*log(f))^(1/2))/64))/(b
*log(f))^(1/2) - (693*b^2*f^(a + b*x^2)*x^5*log(f)^2)/16 + (99*b^3*f^(a + b*x^2)*x^7*log(f)^3)/8 - (11*b^4*f^(
a + b*x^2)*x^9*log(f)^4)/4 + (b^5*f^(a + b*x^2)*x^11*log(f)^5)/2 + (3465*b*f^(a + b*x^2)*x^3*log(f))/32)/(b^6*
log(f)^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x^{2}} x^{12}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)*x**12,x)

[Out]

Integral(f**(a + b*x**2)*x**12, x)

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