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

Optimal. Leaf size=84 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{x}+e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {b n (d+4 e x)^2}{4 x^2}-\frac {1}{2} b e^2 n \log ^2(x) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 67, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {43, 2334, 37, 2301} \[ -\frac {1}{2} \left (\frac {d^2}{x^2}+\frac {4 d e}{x}-2 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n (d+4 e x)^2}{4 x^2}-\frac {1}{2} b e^2 n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 43

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 2301

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 2334

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]}, Simp[u*(a + b*Log[c*x^n]), 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*} \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx &=-\frac {1}{2} \left (\frac {d^2}{x^2}+\frac {4 d e}{x}-2 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (-\frac {d (d+4 e x)}{2 x^3}+\frac {e^2 \log (x)}{x}\right ) \, dx\\ &=-\frac {1}{2} \left (\frac {d^2}{x^2}+\frac {4 d e}{x}-2 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} (b d n) \int \frac {d+4 e x}{x^3} \, dx-\left (b e^2 n\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {b n (d+4 e x)^2}{4 x^2}-\frac {1}{2} b e^2 n \log ^2(x)-\frac {1}{2} \left (\frac {d^2}{x^2}+\frac {4 d e}{x}-2 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 84, normalized size = 1.00 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}-\frac {b d^2 n}{4 x^2}-\frac {2 b d e n}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*d^2*n)/x^2 - (2*b*d*e*n)/x - (d^2*(a + b*Log[c*x^n]))/(2*x^2) - (2*d*e*(a + b*Log[c*x^n]))/x + (e^2*(a
+ b*Log[c*x^n])^2)/(2*b*n)

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fricas [A]  time = 0.51, size = 101, normalized size = 1.20 \[ \frac {2 \, b e^{2} n x^{2} \log \relax (x)^{2} - b d^{2} n - 2 \, a d^{2} - 8 \, {\left (b d e n + a d e\right )} x - 2 \, {\left (4 \, b d e x + b d^{2}\right )} \log \relax (c) + 2 \, {\left (2 \, b e^{2} x^{2} \log \relax (c) - 4 \, b d e n x + 2 \, a e^{2} x^{2} - b d^{2} n\right )} \log \relax (x)}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*e^2*n*x^2*log(x)^2 - b*d^2*n - 2*a*d^2 - 8*(b*d*e*n + a*d*e)*x - 2*(4*b*d*e*x + b*d^2)*log(c) + 2*(2*
b*e^2*x^2*log(c) - 4*b*d*e*n*x + 2*a*e^2*x^2 - b*d^2*n)*log(x))/x^2

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giac [A]  time = 0.29, size = 105, normalized size = 1.25 \[ \frac {2 \, b n x^{2} e^{2} \log \relax (x)^{2} - 8 \, b d n x e \log \relax (x) + 4 \, b x^{2} e^{2} \log \relax (c) \log \relax (x) - 8 \, b d n x e - 8 \, b d x e \log \relax (c) - 2 \, b d^{2} n \log \relax (x) + 4 \, a x^{2} e^{2} \log \relax (x) - b d^{2} n - 8 \, a d x e - 2 \, b d^{2} \log \relax (c) - 2 \, a d^{2}}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*n*x^2*e^2*log(x)^2 - 8*b*d*n*x*e*log(x) + 4*b*x^2*e^2*log(c)*log(x) - 8*b*d*n*x*e - 8*b*d*x*e*log(c)
- 2*b*d^2*n*log(x) + 4*a*x^2*e^2*log(x) - b*d^2*n - 8*a*d*x*e - 2*b*d^2*log(c) - 2*a*d^2)/x^2

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maple [C]  time = 0.24, size = 418, normalized size = 4.98 \[ -\frac {\left (-2 e^{2} x^{2} \ln \relax (x )+4 d e x +d^{2}\right ) b \ln \left (x^{n}\right )}{2 x^{2}}-\frac {2 i \pi b \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )-2 i \pi b \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )-2 i \pi b \,e^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )+2 i \pi b \,e^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )+2 b \,e^{2} n \,x^{2} \ln \relax (x )^{2}-4 i \pi b d e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+4 i \pi b d e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+4 i \pi b d e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-4 i \pi b d e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-4 b \,e^{2} x^{2} \ln \relax (c ) \ln \relax (x )-4 a \,e^{2} x^{2} \ln \relax (x )-i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+8 b d e n x +8 b d e x \ln \relax (c )+8 a d e x +b \,d^{2} n +2 b \,d^{2} \ln \relax (c )+2 a \,d^{2}}{4 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*(-2*e^2*ln(x)*x^2+4*d*e*x+d^2)/x^2*ln(x^n)-1/4*(I*Pi*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2+4*I*Pi*b*d*e*x*csg
n(I*x^n)*csgn(I*c*x^n)^2-2*I*ln(x)*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2*x^2+2*I*ln(x)*Pi*b*e^2*csgn(I*c*x^n)^3
*x^2-2*I*ln(x)*Pi*b*e^2*csgn(I*c*x^n)^2*csgn(I*c)*x^2+4*I*Pi*b*d*e*x*csgn(I*c*x^n)^2*csgn(I*c)+2*I*ln(x)*Pi*b*
e^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^2-I*Pi*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-4*I*Pi*b*d*e*x*csgn
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-4*I*Pi*b*d*e*x*csgn(I*c*x^n)^3+I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*d
^2*csgn(I*c*x^n)^3+2*b*e^2*n*ln(x)^2*x^2-4*ln(x)*ln(c)*b*e^2*x^2-4*ln(x)*a*e^2*x^2+8*b*d*e*x*ln(c)+8*b*d*e*n*x
+2*b*d^2*ln(c)+8*a*d*e*x+b*d^2*n+2*a*d^2)/x^2

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maxima [A]  time = 0.51, size = 90, normalized size = 1.07 \[ \frac {b e^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + a e^{2} \log \relax (x) - \frac {2 \, b d e n}{x} - \frac {2 \, b d e \log \left (c x^{n}\right )}{x} - \frac {b d^{2} n}{4 \, x^{2}} - \frac {2 \, a d e}{x} - \frac {b d^{2} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {a d^{2}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.69, size = 99, normalized size = 1.18 \[ \ln \relax (x)\,\left (a\,e^2+\frac {3\,b\,e^2\,n}{2}\right )-\frac {a\,d^2+x\,\left (4\,a\,d\,e+4\,b\,d\,e\,n\right )+\frac {b\,d^2\,n}{2}}{2\,x^2}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{2}+2\,b\,d\,e\,x+\frac {3\,b\,e^2\,x^2}{2}\right )}{x^2}+\frac {b\,e^2\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.47, size = 99, normalized size = 1.18 \[ - \frac {a d^{2}}{2 x^{2}} - \frac {2 a d e}{x} + a e^{2} \log {\relax (x )} + b d^{2} \left (- \frac {n}{4 x^{2}} - \frac {\log {\left (c x^{n} \right )}}{2 x^{2}}\right ) + 2 b d e \left (- \frac {n}{x} - \frac {\log {\left (c x^{n} \right )}}{x}\right ) - b e^{2} \left (\begin {cases} - \log {\relax (c )} \log {\relax (x )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{2}}{2 n} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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