3.22 \(\int (d+e x)^3 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=85 \[ \frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{3}{4} b d^2 e n x^2-\frac{b d^4 n \log (x)}{4 e}-b d^3 n x-\frac{1}{3} b d e^2 n x^3-\frac{1}{16} b e^3 n x^4 \]

[Out]

-(b*d^3*n*x) - (3*b*d^2*e*n*x^2)/4 - (b*d*e^2*n*x^3)/3 - (b*e^3*n*x^4)/16 - (b*d^4*n*Log[x])/(4*e) + ((d + e*x
)^4*(a + b*Log[c*x^n]))/(4*e)

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Rubi [A]  time = 0.0434278, antiderivative size = 85, normalized size of antiderivative = 1., 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, 2313, 12, 43} \[ \frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{3}{4} b d^2 e n x^2-\frac{b d^4 n \log (x)}{4 e}-b d^3 n x-\frac{1}{3} b d e^2 n x^3-\frac{1}{16} b e^3 n x^4 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^3*n*x) - (3*b*d^2*e*n*x^2)/4 - (b*d*e^2*n*x^3)/3 - (b*e^3*n*x^4)/16 - (b*d^4*n*Log[x])/(4*e) + ((d + e*x
)^4*(a + b*Log[c*x^n]))/(4*e)

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 2313

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

Rule 12

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

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])

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}-(b n) \int \frac{(d+e x)^4}{4 e x} \, dx\\ &=\frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{(b n) \int \frac{(d+e x)^4}{x} \, dx}{4 e}\\ &=\frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{(b n) \int \left (4 d^3 e+\frac{d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx}{4 e}\\ &=-b d^3 n x-\frac{3}{4} b d^2 e n x^2-\frac{1}{3} b d e^2 n x^3-\frac{1}{16} b e^3 n x^4-\frac{b d^4 n \log (x)}{4 e}+\frac{(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}\\ \end{align*}

Mathematica [A]  time = 0.046026, size = 110, normalized size = 1.29 \[ \frac{1}{48} x \left (12 a \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+12 b \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right ) \log \left (c x^n\right )-b n \left (36 d^2 e x+48 d^3+16 d e^2 x^2+3 e^3 x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(12*a*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) - b*n*(48*d^3 + 36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3) +
12*b*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*Log[c*x^n]))/48

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Maple [C]  time = 0.227, size = 571, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*a*d^2*e*x^2+a*d*e^2*x^3+1/8*I*e^3*Pi*b*x^4*csgn(I*c*x^n)^2*csgn(I*c)+1/8*I*e^3*Pi*b*x^4*csgn(I*x^n)*csgn(I
*c*x^n)^2-1/2*I*e^2*Pi*b*d*x^3*csgn(I*c*x^n)^3-3/4*I*e*Pi*b*d^2*x^2*csgn(I*c*x^n)^3+1/2*I*Pi*b*d^3*csgn(I*c*x^
n)^2*csgn(I*c)*x+1/2*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2*x+3/2*ln(c)*b*d^2*e*x^2+ln(c)*b*d*e^2*x^3+ln(c)*b*
d^3*x+1/4*ln(c)*b*e^3*x^4-1/2*I*Pi*b*d^3*csgn(I*c*x^n)^3*x-1/8*I*e^3*Pi*b*x^4*csgn(I*c*x^n)^3+1/4*a*e^3*x^4+a*
d^3*x+1/4*b*(e*x+d)^4/e*ln(x^n)-1/8*I*e^3*Pi*b*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/2*I*e^2*Pi*b*d*x^3*cs
gn(I*c*x^n)^2*csgn(I*c)+1/2*I*e^2*Pi*b*d*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2+3/4*I*e*Pi*b*d^2*x^2*csgn(I*c*x^n)^2*
csgn(I*c)+3/4*I*e*Pi*b*d^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*
x-b*d^3*n*x-1/2*I*e^2*Pi*b*d*x^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3/4*I*e*Pi*b*d^2*x^2*csgn(I*x^n)*csgn(I*c
*x^n)*csgn(I*c)-1/4*b*d^4*n*ln(x)/e-1/16*b*e^3*n*x^4-3/4*b*d^2*e*n*x^2-1/3*b*d*e^2*n*x^3

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Maxima [A]  time = 1.13255, size = 180, normalized size = 2.12 \begin{align*} -\frac{1}{16} \, b e^{3} n x^{4} + \frac{1}{4} \, b e^{3} x^{4} \log \left (c x^{n}\right ) - \frac{1}{3} \, b d e^{2} n x^{3} + \frac{1}{4} \, a e^{3} x^{4} + b d e^{2} x^{3} \log \left (c x^{n}\right ) - \frac{3}{4} \, b d^{2} e n x^{2} + a d e^{2} x^{3} + \frac{3}{2} \, b d^{2} e x^{2} \log \left (c x^{n}\right ) - b d^{3} n x + \frac{3}{2} \, a d^{2} e x^{2} + b d^{3} x \log \left (c x^{n}\right ) + a d^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*e^3*n*x^4 + 1/4*b*e^3*x^4*log(c*x^n) - 1/3*b*d*e^2*n*x^3 + 1/4*a*e^3*x^4 + b*d*e^2*x^3*log(c*x^n) - 3/
4*b*d^2*e*n*x^2 + a*d*e^2*x^3 + 3/2*b*d^2*e*x^2*log(c*x^n) - b*d^3*n*x + 3/2*a*d^2*e*x^2 + b*d^3*x*log(c*x^n)
+ a*d^3*x

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Fricas [B]  time = 0.998267, size = 360, normalized size = 4.24 \begin{align*} -\frac{1}{16} \,{\left (b e^{3} n - 4 \, a e^{3}\right )} x^{4} - \frac{1}{3} \,{\left (b d e^{2} n - 3 \, a d e^{2}\right )} x^{3} - \frac{3}{4} \,{\left (b d^{2} e n - 2 \, a d^{2} e\right )} x^{2} -{\left (b d^{3} n - a d^{3}\right )} x + \frac{1}{4} \,{\left (b e^{3} x^{4} + 4 \, b d e^{2} x^{3} + 6 \, b d^{2} e x^{2} + 4 \, b d^{3} x\right )} \log \left (c\right ) + \frac{1}{4} \,{\left (b e^{3} n x^{4} + 4 \, b d e^{2} n x^{3} + 6 \, b d^{2} e n x^{2} + 4 \, b d^{3} n x\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b*e^3*n - 4*a*e^3)*x^4 - 1/3*(b*d*e^2*n - 3*a*d*e^2)*x^3 - 3/4*(b*d^2*e*n - 2*a*d^2*e)*x^2 - (b*d^3*n -
a*d^3)*x + 1/4*(b*e^3*x^4 + 4*b*d*e^2*x^3 + 6*b*d^2*e*x^2 + 4*b*d^3*x)*log(c) + 1/4*(b*e^3*n*x^4 + 4*b*d*e^2*
n*x^3 + 6*b*d^2*e*n*x^2 + 4*b*d^3*n*x)*log(x)

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Sympy [B]  time = 3.98225, size = 204, normalized size = 2.4 \begin{align*} a d^{3} x + \frac{3 a d^{2} e x^{2}}{2} + a d e^{2} x^{3} + \frac{a e^{3} x^{4}}{4} + b d^{3} n x \log{\left (x \right )} - b d^{3} n x + b d^{3} x \log{\left (c \right )} + \frac{3 b d^{2} e n x^{2} \log{\left (x \right )}}{2} - \frac{3 b d^{2} e n x^{2}}{4} + \frac{3 b d^{2} e x^{2} \log{\left (c \right )}}{2} + b d e^{2} n x^{3} \log{\left (x \right )} - \frac{b d e^{2} n x^{3}}{3} + b d e^{2} x^{3} \log{\left (c \right )} + \frac{b e^{3} n x^{4} \log{\left (x \right )}}{4} - \frac{b e^{3} n x^{4}}{16} + \frac{b e^{3} x^{4} \log{\left (c \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*x + 3*a*d**2*e*x**2/2 + a*d*e**2*x**3 + a*e**3*x**4/4 + b*d**3*n*x*log(x) - b*d**3*n*x + b*d**3*x*log(c
) + 3*b*d**2*e*n*x**2*log(x)/2 - 3*b*d**2*e*n*x**2/4 + 3*b*d**2*e*x**2*log(c)/2 + b*d*e**2*n*x**3*log(x) - b*d
*e**2*n*x**3/3 + b*d*e**2*x**3*log(c) + b*e**3*n*x**4*log(x)/4 - b*e**3*n*x**4/16 + b*e**3*x**4*log(c)/4

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Giac [B]  time = 1.27096, size = 215, normalized size = 2.53 \begin{align*} \frac{1}{4} \, b n x^{4} e^{3} \log \left (x\right ) + b d n x^{3} e^{2} \log \left (x\right ) + \frac{3}{2} \, b d^{2} n x^{2} e \log \left (x\right ) - \frac{1}{16} \, b n x^{4} e^{3} - \frac{1}{3} \, b d n x^{3} e^{2} - \frac{3}{4} \, b d^{2} n x^{2} e + \frac{1}{4} \, b x^{4} e^{3} \log \left (c\right ) + b d x^{3} e^{2} \log \left (c\right ) + \frac{3}{2} \, b d^{2} x^{2} e \log \left (c\right ) + b d^{3} n x \log \left (x\right ) - b d^{3} n x + \frac{1}{4} \, a x^{4} e^{3} + a d x^{3} e^{2} + \frac{3}{2} \, a d^{2} x^{2} e + b d^{3} x \log \left (c\right ) + a d^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*n*x^4*e^3*log(x) + b*d*n*x^3*e^2*log(x) + 3/2*b*d^2*n*x^2*e*log(x) - 1/16*b*n*x^4*e^3 - 1/3*b*d*n*x^3*e^
2 - 3/4*b*d^2*n*x^2*e + 1/4*b*x^4*e^3*log(c) + b*d*x^3*e^2*log(c) + 3/2*b*d^2*x^2*e*log(c) + b*d^3*n*x*log(x)
- b*d^3*n*x + 1/4*a*x^4*e^3 + a*d*x^3*e^2 + 3/2*a*d^2*x^2*e + b*d^3*x*log(c) + a*d^3*x