3.1.12 \(\int (d+e x)^2 (a+b \log (c x^n)) \, dx\) [12]

Optimal. Leaf size=70 \[ -b d^2 n x-\frac {1}{2} b d e n x^2-\frac {1}{9} b e^2 n x^3-\frac {b d^3 n \log (x)}{3 e}+\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e} \]

[Out]

-b*d^2*n*x-1/2*b*d*e*n*x^2-1/9*b*e^2*n*x^3-1/3*b*d^3*n*ln(x)/e+1/3*(e*x+d)^3*(a+b*ln(c*x^n))/e

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Rubi [A]
time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {32, 2350, 12, 45} \begin {gather*} \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b d^3 n \log (x)}{3 e}-b d^2 n x-\frac {1}{2} b d e n x^2-\frac {1}{9} b e^2 n x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^2*n*x) - (b*d*e*n*x^2)/2 - (b*e^2*n*x^3)/9 - (b*d^3*n*Log[x])/(3*e) + ((d + e*x)^3*(a + b*Log[c*x^n]))/(
3*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(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]

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-(b n) \int \frac {(d+e x)^3}{3 e x} \, dx\\ &=\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {(b n) \int \frac {(d+e x)^3}{x} \, dx}{3 e}\\ &=\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {(b n) \int \left (3 d^2 e+\frac {d^3}{x}+3 d e^2 x+e^3 x^2\right ) \, dx}{3 e}\\ &=-b d^2 n x-\frac {1}{2} b d e n x^2-\frac {1}{9} b e^2 n x^3-\frac {b d^3 n \log (x)}{3 e}+\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 77, normalized size = 1.10 \begin {gather*} \frac {1}{18} x \left (6 a \left (3 d^2+3 d e x+e^2 x^2\right )-b n \left (18 d^2+9 d e x+2 e^2 x^2\right )+6 b \left (3 d^2+3 d e x+e^2 x^2\right ) \log \left (c x^n\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(6*a*(3*d^2 + 3*d*e*x + e^2*x^2) - b*n*(18*d^2 + 9*d*e*x + 2*e^2*x^2) + 6*b*(3*d^2 + 3*d*e*x + e^2*x^2)*Log
[c*x^n]))/18

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.16, size = 414, normalized size = 5.91

method	result	size
risch	\(\frac {\left (e x +d \right )^{3} b \ln \left (x^{n}\right )}{3 e}-\frac {i e^{2} \pi b \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{6}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x}{2}-\frac {i e \pi b d \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i e^{2} \pi b \,x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2}-\frac {i e^{2} \pi b \,x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x}{2}+\frac {i e^{2} \pi b \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i e \pi b d \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i e \pi b d \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2}+\frac {i e \pi b d \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {\ln \left (c \right ) b \,e^{2} x^{3}}{3}-\frac {b \,e^{2} n \,x^{3}}{9}+\ln \left (c \right ) b d e \,x^{2}+\frac {a \,e^{2} x^{3}}{3}-\frac {b d e n \,x^{2}}{2}-\frac {b \,d^{3} n \ln \left (x \right )}{3 e}+\ln \left (c \right ) b \,d^{2} x +a d e \,x^{2}-b \,d^{2} n x +a \,d^{2} x\)	\(414\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(e*x+d)^3*b/e*ln(x^n)-1/6*I*e^2*Pi*b*x^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/2*I*Pi*b*d^2*csgn(I*c)*csgn
(I*x^n)*csgn(I*c*x^n)*x-1/2*I*e*Pi*b*d*x^2*csgn(I*c*x^n)^3+1/6*I*e^2*Pi*b*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*
I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2*x-1/6*I*e^2*Pi*b*x^3*csgn(I*c*x^n)^3-1/2*I*Pi*b*d^2*csgn(I*c*x^n)^3*x+1
/6*I*e^2*Pi*b*x^3*csgn(I*c)*csgn(I*c*x^n)^2-1/2*I*e*Pi*b*d*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*e*Pi*
b*d*x^2*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*Pi*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2*x+1/2*I*e*Pi*b*d*x^2*csgn(I*x^n)*cs
gn(I*c*x^n)^2+1/3*ln(c)*b*e^2*x^3-1/9*b*e^2*n*x^3+ln(c)*b*d*e*x^2+1/3*a*e^2*x^3-1/2*b*d*e*n*x^2-1/3*b*d^3*n*ln
(x)/e+ln(c)*b*d^2*x+a*d*e*x^2-b*d^2*n*x+a*d^2*x

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Maxima [A]
time = 0.28, size = 90, normalized size = 1.29 \begin {gather*} -\frac {1}{9} \, b n x^{3} e^{2} - \frac {1}{2} \, b d n x^{2} e + \frac {1}{3} \, b x^{3} e^{2} \log \left (c x^{n}\right ) + b d x^{2} e \log \left (c x^{n}\right ) - b d^{2} n x + \frac {1}{3} \, a x^{3} e^{2} + a d x^{2} e + b d^{2} x \log \left (c x^{n}\right ) + a d^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*b*n*x^3*e^2 - 1/2*b*d*n*x^2*e + 1/3*b*x^3*e^2*log(c*x^n) + b*d*x^2*e*log(c*x^n) - b*d^2*n*x + 1/3*a*x^3*e
^2 + a*d*x^2*e + b*d^2*x*log(c*x^n) + a*d^2*x

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Fricas [A]
time = 0.35, size = 106, normalized size = 1.51 \begin {gather*} -\frac {1}{9} \, {\left (b n - 3 \, a\right )} x^{3} e^{2} - \frac {1}{2} \, {\left (b d n - 2 \, a d\right )} x^{2} e - {\left (b d^{2} n - a d^{2}\right )} x + \frac {1}{3} \, {\left (b x^{3} e^{2} + 3 \, b d x^{2} e + 3 \, b d^{2} x\right )} \log \left (c\right ) + \frac {1}{3} \, {\left (b n x^{3} e^{2} + 3 \, b d n x^{2} e + 3 \, b d^{2} n x\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(b*n - 3*a)*x^3*e^2 - 1/2*(b*d*n - 2*a*d)*x^2*e - (b*d^2*n - a*d^2)*x + 1/3*(b*x^3*e^2 + 3*b*d*x^2*e + 3*
b*d^2*x)*log(c) + 1/3*(b*n*x^3*e^2 + 3*b*d*n*x^2*e + 3*b*d^2*n*x)*log(x)

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Sympy [A]
time = 0.19, size = 102, normalized size = 1.46 \begin {gather*} a d^{2} x + a d e x^{2} + \frac {a e^{2} x^{3}}{3} - b d^{2} n x + b d^{2} x \log {\left (c x^{n} \right )} - \frac {b d e n x^{2}}{2} + b d e x^{2} \log {\left (c x^{n} \right )} - \frac {b e^{2} n x^{3}}{9} + \frac {b e^{2} x^{3} \log {\left (c x^{n} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*x + a*d*e*x**2 + a*e**2*x**3/3 - b*d**2*n*x + b*d**2*x*log(c*x**n) - b*d*e*n*x**2/2 + b*d*e*x**2*log(c*
x**n) - b*e**2*n*x**3/9 + b*e**2*x**3*log(c*x**n)/3

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Giac [A]
time = 2.24, size = 109, normalized size = 1.56 \begin {gather*} \frac {1}{3} \, b n x^{3} e^{2} \log \left (x\right ) + b d n x^{2} e \log \left (x\right ) - \frac {1}{9} \, b n x^{3} e^{2} - \frac {1}{2} \, b d n x^{2} e + \frac {1}{3} \, b x^{3} e^{2} \log \left (c\right ) + b d x^{2} e \log \left (c\right ) + b d^{2} n x \log \left (x\right ) - b d^{2} n x + \frac {1}{3} \, a x^{3} e^{2} + a d x^{2} e + b d^{2} x \log \left (c\right ) + a d^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*n*x^3*e^2*log(x) + b*d*n*x^2*e*log(x) - 1/9*b*n*x^3*e^2 - 1/2*b*d*n*x^2*e + 1/3*b*x^3*e^2*log(c) + b*d*x
^2*e*log(c) + b*d^2*n*x*log(x) - b*d^2*n*x + 1/3*a*x^3*e^2 + a*d*x^2*e + b*d^2*x*log(c) + a*d^2*x

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Mupad [B]
time = 3.61, size = 73, normalized size = 1.04 \begin {gather*} \ln \left (c\,x^n\right )\,\left (b\,d^2\,x+b\,d\,e\,x^2+\frac {b\,e^2\,x^3}{3}\right )+\frac {e^2\,x^3\,\left (3\,a-b\,n\right )}{9}+d^2\,x\,\left (a-b\,n\right )+\frac {d\,e\,x^2\,\left (2\,a-b\,n\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x^n))*(d + e*x)^2,x)

[Out]

log(c*x^n)*((b*e^2*x^3)/3 + b*d^2*x + b*d*e*x^2) + (e^2*x^3*(3*a - b*n))/9 + d^2*x*(a - b*n) + (d*e*x^2*(2*a -
b*n))/2

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