\(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x^2} \, dx\) [24]

Optimal result

Integrand size = 21, antiderivative size = 119 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d^3 n}{x}-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-b*d^3*n/x-3*b*d*e^2*n*x-1/4*b*e^3*n*x^2-3/2*b*d^2*e*n*ln(x)^2-d^3*(a+b*ln(c*x^n))/x+3*d*e^2*x*(a+b*ln(c*x^n))
+1/2*e^3*x^2*(a+b*ln(c*x^n))+3*d^2*e*ln(x)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {45, 2372, 2338} \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{x}-\frac {3}{2} b d^2 e n \log ^2(x)-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2 \]

[In]

Int[((d + e*x)^3*(a + b*Log[c*x^n]))/x^2,x]

[Out]

-((b*d^3*n)/x) - 3*b*d*e^2*n*x - (b*e^3*n*x^2)/4 - (3*b*d^2*e*n*Log[x]^2)/2 - (d^3*(a + b*Log[c*x^n]))/x + 3*d
*e^2*x*(a + b*Log[c*x^n]) + (e^3*x^2*(a + b*Log[c*x^n]))/2 + 3*d^2*e*Log[x]*(a + b*Log[c*x^n])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps \begin{align*} \text {integral}& = -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (3 d e^2-\frac {d^3}{x^2}+\frac {e^3 x}{2}+\frac {3 d^2 e \log (x)}{x}\right ) \, dx \\ & = -\frac {b d^3 n}{x}-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\left (3 b d^2 e n\right ) \int \frac {\log (x)}{x} \, dx \\ & = -\frac {b d^3 n}{x}-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d^3 n}{x}+3 a d e^2 x-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2+3 b d e^2 x \log \left (c x^n\right )-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \]

[In]

Integrate[((d + e*x)^3*(a + b*Log[c*x^n]))/x^2,x]

[Out]

-((b*d^3*n)/x) + 3*a*d*e^2*x - 3*b*d*e^2*n*x - (b*e^3*n*x^2)/4 + 3*b*d*e^2*x*Log[c*x^n] - (d^3*(a + b*Log[c*x^
n]))/x + (e^3*x^2*(a + b*Log[c*x^n]))/2 + (3*d^2*e*(a + b*Log[c*x^n])^2)/(2*b*n)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.22

method	result	size
parallelrisch	\(\frac {2 x^{3} \ln \left (c \,x^{n}\right ) b \,e^{3} n -x^{3} b \,e^{3} n^{2}+2 x^{3} a \,e^{3} n +12 x^{2} \ln \left (c \,x^{n}\right ) b d \,e^{2} n -12 x^{2} b d \,e^{2} n^{2}+12 \ln \left (x \right ) x a \,d^{2} e n +12 x^{2} a d \,e^{2} n +6 e \,d^{2} b \ln \left (c \,x^{n}\right )^{2} x -4 \ln \left (c \,x^{n}\right ) b \,d^{3} n -4 b \,d^{3} n^{2}-4 a \,d^{3} n}{4 x n}\)	\(145\)
risch	\(-\frac {b \left (-e^{3} x^{3}-6 e \,d^{2} \ln \left (x \right ) x -6 d \,e^{2} x^{2}+2 d^{3}\right ) \ln \left (x^{n}\right )}{2 x}-\frac {-2 \ln \left (c \right ) b \,e^{3} x^{3}+6 i \ln \left (x \right ) \pi b \,d^{2} e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x +6 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-6 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-6 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+4 a \,d^{3}-12 \ln \left (x \right ) a \,d^{2} e x +4 d^{3} b \ln \left (c \right )-2 a \,e^{3} x^{3}-i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+6 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 b \,d^{3} n -6 i \ln \left (x \right ) \pi b \,d^{2} e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x -6 i \ln \left (x \right ) \pi b \,d^{2} e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x -2 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-12 a d \,e^{2} x^{2}+6 i \ln \left (x \right ) \pi b \,d^{2} e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x -12 \ln \left (c \right ) b d \,e^{2} x^{2}+12 b d \,e^{2} n \,x^{2}-12 \ln \left (x \right ) \ln \left (c \right ) b \,d^{2} e x +6 e \,d^{2} b n \ln \left (x \right )^{2} x +b \,e^{3} n \,x^{3}+2 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4 x}\)	\(588\)

[In]

int((e*x+d)^3*(a+b*ln(c*x^n))/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4/x*(2*x^3*ln(c*x^n)*b*e^3*n-x^3*b*e^3*n^2+2*x^3*a*e^3*n+12*x^2*ln(c*x^n)*b*d*e^2*n-12*x^2*b*d*e^2*n^2+12*ln
(x)*x*a*d^2*e*n+12*x^2*a*d*e^2*n+6*e*d^2*b*ln(c*x^n)^2*x-4*ln(c*x^n)*b*d^3*n-4*b*d^3*n^2-4*a*d^3*n)/n

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.25 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {6 \, b d^{2} e n x \log \left (x\right )^{2} - 4 \, b d^{3} n - 4 \, a d^{3} - {\left (b e^{3} n - 2 \, a e^{3}\right )} x^{3} - 12 \, {\left (b d e^{2} n - a d e^{2}\right )} x^{2} + 2 \, {\left (b e^{3} x^{3} + 6 \, b d e^{2} x^{2} - 2 \, b d^{3}\right )} \log \left (c\right ) + 2 \, {\left (b e^{3} n x^{3} + 6 \, b d e^{2} n x^{2} + 6 \, b d^{2} e x \log \left (c\right ) - 2 \, b d^{3} n + 6 \, a d^{2} e x\right )} \log \left (x\right )}{4 \, x} \]

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="fricas")

[Out]

1/4*(6*b*d^2*e*n*x*log(x)^2 - 4*b*d^3*n - 4*a*d^3 - (b*e^3*n - 2*a*e^3)*x^3 - 12*(b*d*e^2*n - a*d*e^2)*x^2 + 2
*(b*e^3*x^3 + 6*b*d*e^2*x^2 - 2*b*d^3)*log(c) + 2*(b*e^3*n*x^3 + 6*b*d*e^2*n*x^2 + 6*b*d^2*e*x*log(c) - 2*b*d^
3*n + 6*a*d^2*e*x)*log(x))/x

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.53 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\begin {cases} - \frac {a d^{3}}{x} + \frac {3 a d^{2} e \log {\left (c x^{n} \right )}}{n} + 3 a d e^{2} x + \frac {a e^{3} x^{2}}{2} - \frac {b d^{3} n}{x} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{x} + \frac {3 b d^{2} e \log {\left (c x^{n} \right )}^{2}}{2 n} - 3 b d e^{2} n x + 3 b d e^{2} x \log {\left (c x^{n} \right )} - \frac {b e^{3} n x^{2}}{4} + \frac {b e^{3} x^{2} \log {\left (c x^{n} \right )}}{2} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (- \frac {d^{3}}{x} + 3 d^{2} e \log {\left (x \right )} + 3 d e^{2} x + \frac {e^{3} x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**2,x)

[Out]

Piecewise((-a*d**3/x + 3*a*d**2*e*log(c*x**n)/n + 3*a*d*e**2*x + a*e**3*x**2/2 - b*d**3*n/x - b*d**3*log(c*x**
n)/x + 3*b*d**2*e*log(c*x**n)**2/(2*n) - 3*b*d*e**2*n*x + 3*b*d*e**2*x*log(c*x**n) - b*e**3*n*x**2/4 + b*e**3*
x**2*log(c*x**n)/2, Ne(n, 0)), ((a + b*log(c))*(-d**3/x + 3*d**2*e*log(x) + 3*d*e**2*x + e**3*x**2/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {1}{4} \, b e^{3} n x^{2} + \frac {1}{2} \, b e^{3} x^{2} \log \left (c x^{n}\right ) - 3 \, b d e^{2} n x + \frac {1}{2} \, a e^{3} x^{2} + 3 \, b d e^{2} x \log \left (c x^{n}\right ) + 3 \, a d e^{2} x + \frac {3 \, b d^{2} e \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d^{2} e \log \left (x\right ) - \frac {b d^{3} n}{x} - \frac {b d^{3} \log \left (c x^{n}\right )}{x} - \frac {a d^{3}}{x} \]

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="maxima")

[Out]

-1/4*b*e^3*n*x^2 + 1/2*b*e^3*x^2*log(c*x^n) - 3*b*d*e^2*n*x + 1/2*a*e^3*x^2 + 3*b*d*e^2*x*log(c*x^n) + 3*a*d*e
^2*x + 3/2*b*d^2*e*log(c*x^n)^2/n + 3*a*d^2*e*log(x) - b*d^3*n/x - b*d^3*log(c*x^n)/x - a*d^3/x

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {1}{2} \, b e^{3} x^{2} \log \left (c\right ) + \frac {3}{2} \, b d^{2} e n \log \left (x\right )^{2} + 3 \, {\left (x \log \left (x\right ) - x\right )} b d e^{2} n + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} b e^{3} n + \frac {1}{2} \, a e^{3} x^{2} - b d^{3} n {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} + 3 \, b d e^{2} x \log \left (c\right ) + 3 \, b d^{2} e \log \left (c\right ) \log \left ({\left | x \right |}\right ) + 3 \, a d e^{2} x + 3 \, a d^{2} e \log \left ({\left | x \right |}\right ) - \frac {b d^{3} \log \left (c\right )}{x} - \frac {a d^{3}}{x} \]

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="giac")

[Out]

1/2*b*e^3*x^2*log(c) + 3/2*b*d^2*e*n*log(x)^2 + 3*(x*log(x) - x)*b*d*e^2*n + 1/4*(2*x^2*log(x) - x^2)*b*e^3*n
+ 1/2*a*e^3*x^2 - b*d^3*n*(log(x)/x + 1/x) + 3*b*d*e^2*x*log(c) + 3*b*d^2*e*log(c)*log(abs(x)) + 3*a*d*e^2*x +
3*a*d^2*e*log(abs(x)) - b*d^3*log(c)/x - a*d^3/x

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.29 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\ln \left (x\right )\,\left (3\,a\,d^2\,e+3\,b\,d^2\,e\,n\right )-\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3+3\,b\,d^2\,e\,x+3\,b\,d\,e^2\,x^2+b\,e^3\,x^3}{x}-\frac {\frac {3\,b\,e^3\,x^3}{2}+6\,b\,d\,e^2\,x^2}{x}\right )-\frac {a\,d^3+b\,d^3\,n}{x}+\frac {e^3\,x^2\,\left (2\,a-b\,n\right )}{4}+3\,d\,e^2\,x\,\left (a-b\,n\right )+\frac {3\,b\,d^2\,e\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

[In]

int(((a + b*log(c*x^n))*(d + e*x)^3)/x^2,x)

[Out]

log(x)*(3*a*d^2*e + 3*b*d^2*e*n) - log(c*x^n)*((b*d^3 + b*e^3*x^3 + 3*b*d^2*e*x + 3*b*d*e^2*x^2)/x - ((3*b*e^3
*x^3)/2 + 6*b*d*e^2*x^2)/x) - (a*d^3 + b*d^3*n)/x + (e^3*x^2*(2*a - b*n))/4 + 3*d*e^2*x*(a - b*n) + (3*b*d^2*e
*log(c*x^n)^2)/(2*n)