3.1.79 \(\int \frac {f^{a+b x^2}}{x^7} \, dx\) [79]

Optimal. Leaf size=81 \[ -\frac {f^{a+b x^2}}{6 x^6}-\frac {b f^{a+b x^2} \log (f)}{12 x^4}-\frac {b^2 f^{a+b x^2} \log ^2(f)}{12 x^2}+\frac {1}{12} b^3 f^a \text {Ei}\left (b x^2 \log (f)\right ) \log ^3(f) \]

[Out]

-1/6*f^(b*x^2+a)/x^6-1/12*b*f^(b*x^2+a)*ln(f)/x^4-1/12*b^2*f^(b*x^2+a)*ln(f)^2/x^2+1/12*b^3*f^a*Ei(b*x^2*ln(f)
)*ln(f)^3

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \begin {gather*} \frac {1}{12} b^3 f^a \log ^3(f) \text {Ei}\left (b x^2 \log (f)\right )-\frac {b^2 \log ^2(f) f^{a+b x^2}}{12 x^2}-\frac {f^{a+b x^2}}{6 x^6}-\frac {b \log (f) f^{a+b x^2}}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^7,x]

[Out]

-1/6*f^(a + b*x^2)/x^6 - (b*f^(a + b*x^2)*Log[f])/(12*x^4) - (b^2*f^(a + b*x^2)*Log[f]^2)/(12*x^2) + (b^3*f^a*
ExpIntegralEi[b*x^2*Log[f]]*Log[f]^3)/12

Rule 2241

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

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

\begin {align*} \int \frac {f^{a+b x^2}}{x^7} \, dx &=-\frac {f^{a+b x^2}}{6 x^6}+\frac {1}{3} (b \log (f)) \int \frac {f^{a+b x^2}}{x^5} \, dx\\ &=-\frac {f^{a+b x^2}}{6 x^6}-\frac {b f^{a+b x^2} \log (f)}{12 x^4}+\frac {1}{6} \left (b^2 \log ^2(f)\right ) \int \frac {f^{a+b x^2}}{x^3} \, dx\\ &=-\frac {f^{a+b x^2}}{6 x^6}-\frac {b f^{a+b x^2} \log (f)}{12 x^4}-\frac {b^2 f^{a+b x^2} \log ^2(f)}{12 x^2}+\frac {1}{6} \left (b^3 \log ^3(f)\right ) \int \frac {f^{a+b x^2}}{x} \, dx\\ &=-\frac {f^{a+b x^2}}{6 x^6}-\frac {b f^{a+b x^2} \log (f)}{12 x^4}-\frac {b^2 f^{a+b x^2} \log ^2(f)}{12 x^2}+\frac {1}{12} b^3 f^a \text {Ei}\left (b x^2 \log (f)\right ) \log ^3(f)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 59, normalized size = 0.73 \begin {gather*} \frac {f^a \left (b^3 x^6 \text {Ei}\left (b x^2 \log (f)\right ) \log ^3(f)-f^{b x^2} \left (2+b x^2 \log (f)+b^2 x^4 \log ^2(f)\right )\right )}{12 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^7,x]

[Out]

(f^a*(b^3*x^6*ExpIntegralEi[b*x^2*Log[f]]*Log[f]^3 - f^(b*x^2)*(2 + b*x^2*Log[f] + b^2*x^4*Log[f]^2)))/(12*x^6
)

________________________________________________________________________________________

Maple [A]
time = 0.04, size = 79, normalized size = 0.98

method	result	size
risch	\(-\frac {f^{a} f^{b \,x^{2}}}{6 x^{6}}-\frac {f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{12 x^{4}}-\frac {f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{12 x^{2}}-\frac {f^{a} \ln \left (f \right )^{3} b^{3} \expIntegral \left (1, -b \,x^{2} \ln \left (f \right )\right )}{12}\)	\(79\)
meijerg	\(-\frac {f^{a} b^{3} \ln \left (f \right )^{3} \left (-\frac {22 b^{3} x^{6} \ln \left (f \right )^{3}+36 b^{2} x^{4} \ln \left (f \right )^{2}+36 b \,x^{2} \ln \left (f \right )+24}{72 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {\left (4 b^{2} x^{4} \ln \left (f \right )^{2}+4 b \,x^{2} \ln \left (f \right )+8\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{24 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {\ln \left (-b \,x^{2} \ln \left (f \right )\right )}{6}+\frac {\expIntegral \left (1, -b \,x^{2} \ln \left (f \right )\right )}{6}+\frac {11}{36}-\frac {\ln \left (x \right )}{3}-\frac {\ln \left (-b \right )}{6}-\frac {\ln \left (\ln \left (f \right )\right )}{6}+\frac {1}{3 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {1}{2 b^{2} x^{4} \ln \left (f \right )^{2}}+\frac {1}{2 b \,x^{2} \ln \left (f \right )}\right )}{2}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*f^a/x^6*f^(b*x^2)-1/12*f^a*ln(f)*b/x^4*f^(b*x^2)-1/12*f^a*ln(f)^2*b^2/x^2*f^(b*x^2)-1/12*f^a*ln(f)^3*b^3*
Ei(1,-b*x^2*ln(f))

________________________________________________________________________________________

Maxima [A]
time = 0.32, size = 22, normalized size = 0.27 \begin {gather*} \frac {1}{2} \, b^{3} f^{a} \Gamma \left (-3, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*f^a*gamma(-3, -b*x^2*log(f))*log(f)^3

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 59, normalized size = 0.73 \begin {gather*} \frac {b^{3} f^{a} x^{6} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{3} - {\left (b^{2} x^{4} \log \left (f\right )^{2} + b x^{2} \log \left (f\right ) + 2\right )} f^{b x^{2} + a}}{12 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(b^3*f^a*x^6*Ei(b*x^2*log(f))*log(f)^3 - (b^2*x^4*log(f)^2 + b*x^2*log(f) + 2)*f^(b*x^2 + a))/x^6

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f^{a + b x^{2}}}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**7,x)

[Out]

Integral(f**(a + b*x**2)/x**7, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^7, x)

________________________________________________________________________________________

Mupad [B]
time = 3.54, size = 69, normalized size = 0.85 \begin {gather*} -\frac {b^3\,f^a\,{\ln \left (f\right )}^3\,\left (f^{b\,x^2}\,\left (\frac {1}{6\,b\,x^2\,\ln \left (f\right )}+\frac {1}{6\,b^2\,x^4\,{\ln \left (f\right )}^2}+\frac {1}{3\,b^3\,x^6\,{\ln \left (f\right )}^3}\right )+\frac {\mathrm {expint}\left (-b\,x^2\,\ln \left (f\right )\right )}{6}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)/x^7,x)

[Out]

-(b^3*f^a*log(f)^3*(f^(b*x^2)*(1/(6*b*x^2*log(f)) + 1/(6*b^2*x^4*log(f)^2) + 1/(3*b^3*x^6*log(f)^3)) + expint(
-b*x^2*log(f))/6))/2

________________________________________________________________________________________