\(\int x^3 (a+b \log (c x^n)) \, dx\) [43]

Optimal result

Integrand size = 14, antiderivative size = 27 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b n x^4+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/16*b*n*x^4+1/4*x^4*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2341} \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b n x^4 \]

[In]

Int[x^3*(a + b*Log[c*x^n]),x]

[Out]

-1/16*(b*n*x^4) + (x^4*(a + b*Log[c*x^n]))/4

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{16} b n x^4+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a x^4}{4}-\frac {1}{16} b n x^4+\frac {1}{4} b x^4 \log \left (c x^n\right ) \]

[In]

Integrate[x^3*(a + b*Log[c*x^n]),x]

[Out]

(a*x^4)/4 - (b*n*x^4)/16 + (b*x^4*Log[c*x^n])/4

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {x^{4} \ln \left (c \,x^{n}\right ) b}{4}-\frac {b n \,x^{4}}{16}+\frac {a \,x^{4}}{4}\)	\(27\)
risch	\(\frac {b \,x^{4} \ln \left (x^{n}\right )}{4}+\frac {x^{4} \left (-2 i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+2 i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+2 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-2 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 b \ln \left (c \right )-b n +4 a \right )}{16}\)	\(112\)

[In]

int(x^3*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)

[Out]

1/4*x^4*ln(c*x^n)*b-1/16*b*n*x^4+1/4*a*x^4

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} \, b n x^{4} \log \left (x\right ) + \frac {1}{4} \, b x^{4} \log \left (c\right ) - \frac {1}{16} \, {\left (b n - 4 \, a\right )} x^{4} \]

[In]

integrate(x^3*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

1/4*b*n*x^4*log(x) + 1/4*b*x^4*log(c) - 1/16*(b*n - 4*a)*x^4

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a x^{4}}{4} - \frac {b n x^{4}}{16} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4} \]

[In]

integrate(x**3*(a+b*ln(c*x**n)),x)

[Out]

a*x**4/4 - b*n*x**4/16 + b*x**4*log(c*x**n)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} \, b n x^{4} + \frac {1}{4} \, b x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a x^{4} \]

[In]

integrate(x^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-1/16*b*n*x^4 + 1/4*b*x^4*log(c*x^n) + 1/4*a*x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} \, b n x^{4} \log \left (x\right ) - \frac {1}{16} \, b n x^{4} + \frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} \]

[In]

integrate(x^3*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

1/4*b*n*x^4*log(x) - 1/16*b*n*x^4 + 1/4*b*x^4*log(c) + 1/4*a*x^4

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=x^4\,\left (\frac {a}{4}-\frac {b\,n}{16}\right )+\frac {b\,x^4\,\ln \left (c\,x^n\right )}{4} \]

[In]

int(x^3*(a + b*log(c*x^n)),x)

[Out]

x^4*(a/4 - (b*n)/16) + (b*x^4*log(c*x^n))/4