\(\int f^{a+b x^2} x^7 \, dx\) [72]

Optimal result

Integrand size = 13, antiderivative size = 86 \[ \int f^{a+b x^2} x^7 \, dx=-\frac {3 f^{a+b x^2}}{b^4 \log ^4(f)}+\frac {3 f^{a+b x^2} x^2}{b^3 \log ^3(f)}-\frac {3 f^{a+b x^2} x^4}{2 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^6}{2 b \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int f^{a+b x^2} x^7 \, dx=-\frac {3 f^{a+b x^2}}{b^4 \log ^4(f)}+\frac {3 x^2 f^{a+b x^2}}{b^3 \log ^3(f)}-\frac {3 x^4 f^{a+b x^2}}{2 b^2 \log ^2(f)}+\frac {x^6 f^{a+b x^2}}{2 b \log (f)} \]

[In]

Int[f^(a + b*x^2)*x^7,x]

[Out]

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

Rule 2240

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

Rule 2243

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2} x^6}{2 b \log (f)}-\frac {3 \int f^{a+b x^2} x^5 \, dx}{b \log (f)} \\ & = -\frac {3 f^{a+b x^2} x^4}{2 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^6}{2 b \log (f)}+\frac {6 \int f^{a+b x^2} x^3 \, dx}{b^2 \log ^2(f)} \\ & = \frac {3 f^{a+b x^2} x^2}{b^3 \log ^3(f)}-\frac {3 f^{a+b x^2} x^4}{2 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^6}{2 b \log (f)}-\frac {6 \int f^{a+b x^2} x \, dx}{b^3 \log ^3(f)} \\ & = -\frac {3 f^{a+b x^2}}{b^4 \log ^4(f)}+\frac {3 f^{a+b x^2} x^2}{b^3 \log ^3(f)}-\frac {3 f^{a+b x^2} x^4}{2 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^6}{2 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int f^{a+b x^2} x^7 \, dx=\frac {f^{a+b x^2} \left (-6+6 b x^2 \log (f)-3 b^2 x^4 \log ^2(f)+b^3 x^6 \log ^3(f)\right )}{2 b^4 \log ^4(f)} \]

[In]

Integrate[f^(a + b*x^2)*x^7,x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60

method	result	size
gosper	\(\frac {\left (b^{3} x^{6} \ln \left (f \right )^{3}-3 b^{2} x^{4} \ln \left (f \right )^{2}+6 b \,x^{2} \ln \left (f \right )-6\right ) f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{4} b^{4}}\)	\(52\)
risch	\(\frac {\left (b^{3} x^{6} \ln \left (f \right )^{3}-3 b^{2} x^{4} \ln \left (f \right )^{2}+6 b \,x^{2} \ln \left (f \right )-6\right ) f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{4} b^{4}}\)	\(52\)
meijerg	\(\frac {f^{a} \left (6-\frac {\left (-4 b^{3} x^{6} \ln \left (f \right )^{3}+12 b^{2} x^{4} \ln \left (f \right )^{2}-24 b \,x^{2} \ln \left (f \right )+24\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{4}\right )}{2 b^{4} \ln \left (f \right )^{4}}\)	\(59\)
parallelrisch	\(\frac {f^{b \,x^{2}+a} x^{6} \ln \left (f \right )^{3} b^{3}-3 f^{b \,x^{2}+a} x^{4} \ln \left (f \right )^{2} b^{2}+6 f^{b \,x^{2}+a} x^{2} \ln \left (f \right ) b -6 f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{4} b^{4}}\)	\(80\)
norman	\(-\frac {3 \,{\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {x^{6} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}+\frac {3 x^{2} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}-\frac {3 x^{4} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 \ln \left (f \right )^{2} b^{2}}\)	\(91\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.59 \[ \int f^{a+b x^2} x^7 \, dx=\frac {{\left (b^{3} x^{6} \log \left (f\right )^{3} - 3 \, b^{2} x^{4} \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) - 6\right )} f^{b x^{2} + a}}{2 \, b^{4} \log \left (f\right )^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.77 \[ \int f^{a+b x^2} x^7 \, dx=\begin {cases} \frac {f^{a + b x^{2}} \left (b^{3} x^{6} \log {\left (f \right )}^{3} - 3 b^{2} x^{4} \log {\left (f \right )}^{2} + 6 b x^{2} \log {\left (f \right )} - 6\right )}{2 b^{4} \log {\left (f \right )}^{4}} & \text {for}\: b^{4} \log {\left (f \right )}^{4} \neq 0 \\\frac {x^{8}}{8} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((f**(a + b*x**2)*(b**3*x**6*log(f)**3 - 3*b**2*x**4*log(f)**2 + 6*b*x**2*log(f) - 6)/(2*b**4*log(f)*
*4), Ne(b**4*log(f)**4, 0)), (x**8/8, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72 \[ \int f^{a+b x^2} x^7 \, dx=\frac {{\left (b^{3} f^{a} x^{6} \log \left (f\right )^{3} - 3 \, b^{2} f^{a} x^{4} \log \left (f\right )^{2} + 6 \, b f^{a} x^{2} \log \left (f\right ) - 6 \, f^{a}\right )} f^{b x^{2}}}{2 \, b^{4} \log \left (f\right )^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.64 \[ \int f^{a+b x^2} x^7 \, dx=\frac {{\left (b^{3} x^{6} \log \left (f\right )^{3} - 3 \, b^{2} x^{4} \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) - 6\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{2 \, b^{4} \log \left (f\right )^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int f^{a+b x^2} x^7 \, dx=-\frac {f^{b\,x^2+a}\,\left (-\frac {b^3\,x^6\,{\ln \left (f\right )}^3}{2}+\frac {3\,b^2\,x^4\,{\ln \left (f\right )}^2}{2}-3\,b\,x^2\,\ln \left (f\right )+3\right )}{b^4\,{\ln \left (f\right )}^4} \]

[In]

int(f^(a + b*x^2)*x^7,x)

[Out]

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