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

Optimal. Leaf size=99 \[ 6 a b^2 n^2 x-\frac {3 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{e}+\frac {6 b^3 n^2 (d+e x) \log \left (c (d+e x)^n\right )}{e}-6 b^3 n^3 x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2389, 2296, 2295} \[ 6 a b^2 n^2 x-\frac {3 b n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{e}+\frac {6 b^3 n^2 (d+e x) \log \left (c (d+e x)^n\right )}{e}-6 b^3 n^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^3,x]

[Out]

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 85, normalized size = 0.86 \[ \frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3-3 b n \left ((d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2-2 b n \left (e x (a-b n)+b (d+e x) \log \left (c (d+e x)^n\right )\right )\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 324, normalized size = 3.27 \[ \frac {b^{3} e x \log \relax (c)^{3} + {\left (b^{3} e n^{3} x + b^{3} d n^{3}\right )} \log \left (e x + d\right )^{3} - 3 \, {\left (b^{3} e n - a b^{2} e\right )} x \log \relax (c)^{2} - 3 \, {\left (b^{3} d n^{3} - a b^{2} d n^{2} + {\left (b^{3} e n^{3} - a b^{2} e n^{2}\right )} x - {\left (b^{3} e n^{2} x + b^{3} d n^{2}\right )} \log \relax (c)\right )} \log \left (e x + d\right )^{2} + 3 \, {\left (2 \, b^{3} e n^{2} - 2 \, a b^{2} e n + a^{2} b e\right )} x \log \relax (c) - {\left (6 \, b^{3} e n^{3} - 6 \, a b^{2} e n^{2} + 3 \, a^{2} b e n - a^{3} e\right )} x + 3 \, {\left (2 \, b^{3} d n^{3} - 2 \, a b^{2} d n^{2} + a^{2} b d n + {\left (b^{3} e n x + b^{3} d n\right )} \log \relax (c)^{2} + {\left (2 \, b^{3} e n^{3} - 2 \, a b^{2} e n^{2} + a^{2} b e n\right )} x - 2 \, {\left (b^{3} d n^{2} - a b^{2} d n + {\left (b^{3} e n^{2} - a b^{2} e n\right )} x\right )} \log \relax (c)\right )} \log \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 409, normalized size = 4.13 \[ {\left (x e + d\right )} b^{3} n^{3} e^{\left (-1\right )} \log \left (x e + d\right )^{3} - 3 \, {\left (x e + d\right )} b^{3} n^{3} e^{\left (-1\right )} \log \left (x e + d\right )^{2} + 3 \, {\left (x e + d\right )} b^{3} n^{2} e^{\left (-1\right )} \log \left (x e + d\right )^{2} \log \relax (c) + 6 \, {\left (x e + d\right )} b^{3} n^{3} e^{\left (-1\right )} \log \left (x e + d\right ) + 3 \, {\left (x e + d\right )} a b^{2} n^{2} e^{\left (-1\right )} \log \left (x e + d\right )^{2} - 6 \, {\left (x e + d\right )} b^{3} n^{2} e^{\left (-1\right )} \log \left (x e + d\right ) \log \relax (c) + 3 \, {\left (x e + d\right )} b^{3} n e^{\left (-1\right )} \log \left (x e + d\right ) \log \relax (c)^{2} - 6 \, {\left (x e + d\right )} b^{3} n^{3} e^{\left (-1\right )} - 6 \, {\left (x e + d\right )} a b^{2} n^{2} e^{\left (-1\right )} \log \left (x e + d\right ) + 6 \, {\left (x e + d\right )} b^{3} n^{2} e^{\left (-1\right )} \log \relax (c) + 6 \, {\left (x e + d\right )} a b^{2} n e^{\left (-1\right )} \log \left (x e + d\right ) \log \relax (c) - 3 \, {\left (x e + d\right )} b^{3} n e^{\left (-1\right )} \log \relax (c)^{2} + {\left (x e + d\right )} b^{3} e^{\left (-1\right )} \log \relax (c)^{3} + 6 \, {\left (x e + d\right )} a b^{2} n^{2} e^{\left (-1\right )} + 3 \, {\left (x e + d\right )} a^{2} b n e^{\left (-1\right )} \log \left (x e + d\right ) - 6 \, {\left (x e + d\right )} a b^{2} n e^{\left (-1\right )} \log \relax (c) + 3 \, {\left (x e + d\right )} a b^{2} e^{\left (-1\right )} \log \relax (c)^{2} - 3 \, {\left (x e + d\right )} a^{2} b n e^{\left (-1\right )} + 3 \, {\left (x e + d\right )} a^{2} b e^{\left (-1\right )} \log \relax (c) + {\left (x e + d\right )} a^{3} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.70, size = 4872, normalized size = 49.21 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x+x*b^3*ln((e*x+d)^n)^3-6*b^3*n^3*x+6*a*b^2*n^2*x-3*I/e*ln(c)*Pi*ln(e*x+d)*b^3*d*n*csgn(I*c)*csgn(I*(e*x+d
)^n)*csgn(I*c*(e*x+d)^n)-3*I/e*Pi*ln(e*x+d)*a*b^2*d*n*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+3/4*b*(8
*ln(e*x+d)*a*b*d*n-4*b^2*d*n^2*ln(e*x+d)^2+4*b^2*e*x*ln(c)^2-8*ln(e*x+d)*b^2*d*n^2+4*a^2*e*x+4*I*Pi*ln(e*x+d)*
b^2*d*n*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+4*I*ln(c)*Pi*b^2*e*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+4*I*ln(c)
*Pi*b^2*e*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-4*I*Pi*b^2*e*n*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-Pi^2*b^2*
e*x*csgn(I*c*(e*x+d)^n)^6+8*ln(c)*ln(e*x+d)*b^2*d*n+8*b^2*e*n^2*x-8*b^2*e*n*x*ln(c)+8*a*b*e*x*ln(c)-4*I*Pi*b^2
*e*n*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+4*I*Pi*a*b*e*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+4*I*Pi*a*b*e*x*c
sgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+4*I*Pi*ln(e*x+d)*b^2*d*n*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-4*I*ln(c)*Pi*b
^2*e*x*csgn(I*c*(e*x+d)^n)^3+4*I*Pi*b^2*e*n*x*csgn(I*c*(e*x+d)^n)^3-4*I*Pi*a*b*e*x*csgn(I*c*(e*x+d)^n)^3-4*I*l
n(c)*Pi*b^2*e*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+4*I*Pi*b^2*e*n*x*csgn(I*c)*csgn(I*(e*x+d)^n)*c
sgn(I*c*(e*x+d)^n)-4*I*Pi*ln(e*x+d)*b^2*d*n*csgn(I*c*(e*x+d)^n)^3+2*Pi^2*b^2*e*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)
*csgn(I*c*(e*x+d)^n)^3+2*Pi^2*b^2*e*x*csgn(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^3-4*Pi^2*b^2*e*x*csgn(
I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4-Pi^2*b^2*e*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^
2-8*a*b*e*n*x+2*Pi^2*b^2*e*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^5-Pi^2*b^2*e*x*csgn(I*c)^2*csgn(I*c*(e*x+d)
^n)^4+2*Pi^2*b^2*e*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5-Pi^2*b^2*e*x*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4-4*
I*Pi*a*b*e*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-4*I*Pi*ln(e*x+d)*b^2*d*n*csgn(I*c)*csgn(I*(e*x+d)
^n)*csgn(I*c*(e*x+d)^n))/e*ln((e*x+d)^n)+ln(c)^3*b^3*x+3/2*b^2*(-I*Pi*b*e*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I
*c*(e*x+d)^n)+I*Pi*b*e*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*Pi*b*e*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*
Pi*b*e*x*csgn(I*c*(e*x+d)^n)^3+2*b*e*x*ln(c)+2*b*d*n*ln(e*x+d)-2*b*e*n*x+2*a*e*x)/e*ln((e*x+d)^n)^2-3*a^2*b*n*
x+3*ln(c)*a^2*b*x+3*ln(c)^2*a*b^2*x-3*ln(c)^2*b^3*n*x+6*ln(c)*b^3*n^2*x-3/4*Pi^2*a*b^2*x*csgn(I*c)^2*csgn(I*c*
(e*x+d)^n)^4-3/e*ln(c)*b^3*d*n^2*ln(e*x+d)^2+3/e*ln(c)^2*ln(e*x+d)*b^3*d*n-6/e*ln(c)*ln(e*x+d)*b^3*d*n^2-3/e*a
*b^2*d*n^2*ln(e*x+d)^2-6/e*ln(e*x+d)*a*b^2*d*n^2+3/e*ln(e*x+d)*a^2*b*d*n-3*I*Pi*b^3*n^2*x*csgn(I*c*(e*x+d)^n)^
3-3/2*I*Pi*a^2*b*x*csgn(I*c*(e*x+d)^n)^3-1/8*I*Pi^3*b^3*x*csgn(I*(e*x+d)^n)^3*csgn(I*c*(e*x+d)^n)^6+3/8*I*Pi^3
*b^3*x*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^7-3/8*I*Pi^3*b^3*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^8-3/8*
I*Pi^3*b^3*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^8+3/8*I*Pi^3*b^3*x*csgn(I*c)^2*csgn(I*c*(e*x+d)^n)^7-1/8*I*Pi^3*b^3
*x*csgn(I*c)^3*csgn(I*c*(e*x+d)^n)^6-3/2*I*ln(c)^2*Pi*b^3*x*csgn(I*c*(e*x+d)^n)^3+6/e*ln(c)*ln(e*x+d)*a*b^2*d*
n-3/4/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*c*(e*x+d)^n)^6+3/2*ln(c)*Pi^2*b^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I
*c*(e*x+d)^n)^3-3/4*ln(c)*Pi^2*b^3*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^2-3*ln(c)*Pi^2*b^3*x*
csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4+3/2*ln(c)*Pi^2*b^3*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)*csgn(I*c*
(e*x+d)^n)^3-3/2*Pi^2*b^3*n*x*csgn(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^3+3/4*Pi^2*b^3*n*x*csgn(I*c)^2
*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^2+3*Pi^2*b^3*n*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4-3/
2*Pi^2*b^3*n*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^3+3/2*Pi^2*a*b^2*x*csgn(I*c)*csgn(I*(e*x+d)^n
)^2*csgn(I*c*(e*x+d)^n)^3-3/4*Pi^2*a*b^2*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^2-3*Pi^2*a*b^2*
x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4+3/2*Pi^2*a*b^2*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e
*x+d)^n)^3+3/2*I/e*Pi*b^3*d*n^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*ln(e*x+d)^2+3*I/e*ln(c)*Pi*ln(
e*x+d)*b^3*d*n*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+3*I/e*ln(c)*Pi*ln(e*x+d)*b^3*d*n*csgn(I*c)*csgn(I*c*(e*
x+d)^n)^2+3*I/e*Pi*b^3*d*n^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*ln(e*x+d)+3*I/e*Pi*ln(e*x+d)*a*b^
2*d*n*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+3*I/e*Pi*ln(e*x+d)*a*b^2*d*n*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3/4
*ln(c)*Pi^2*b^3*x*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4+3/2*ln(c)*Pi^2*b^3*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e
*x+d)^n)^5+3/2*ln(c)*Pi^2*b^3*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5-3/4*ln(c)*Pi^2*b^3*x*csgn(I*c)^2*csgn(I*c*(e*x
+d)^n)^4+3/4*Pi^2*b^3*n*x*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4-3/2*Pi^2*b^3*n*x*csgn(I*(e*x+d)^n)*csgn(I*
c*(e*x+d)^n)^5-3/2*Pi^2*b^3*n*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5+3/4*Pi^2*b^3*n*x*csgn(I*c)^2*csgn(I*c*(e*x+d)^
n)^4-3/4*Pi^2*a*b^2*x*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4+3/2*Pi^2*a*b^2*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e
*x+d)^n)^5+3/2*Pi^2*a*b^2*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5-6*ln(c)*a*b^2*n*x+1/e*b^3*d*n^3*ln(e*x+d)^3+3/e*b^
3*d*n^3*ln(e*x+d)^2-3/4*ln(c)*Pi^2*b^3*x*csgn(I*c*(e*x+d)^n)^6+3/4*Pi^2*b^3*n*x*csgn(I*c*(e*x+d)^n)^6-3/4*Pi^2
*a*b^2*x*csgn(I*c*(e*x+d)^n)^6+1/8*I*Pi^3*b^3*x*csgn(I*c*(e*x+d)^n)^9+6*b^3*d*n^3/e*ln(e*x+d)-3/4/e*Pi^2*ln(e*
x+d)*b^3*d*n*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^2-3/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*c)*csgn(I
*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4+3/2/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*c)^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^
n)^3+3*I*ln(c)*Pi*b^3*n*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-3*I*ln(c)*Pi*a*b^2*x*csgn(I*c)*csgn(
I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+3*I*Pi*a*b^2*n*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-3/2*I/e*Pi*b
^3*d*n^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*ln(e*x+d)^2-3/2*I/e*Pi*b^3*d*n^2*csgn(I*c)*csgn(I*c*(e*x+d)^n
)^2*ln(e*x+d)^2-3*I/e*ln(c)*Pi*ln(e*x+d)*b^3*d*n*csgn(I*c*(e*x+d)^n)^3-3*I/e*Pi*b^3*d*n^2*csgn(I*(e*x+d)^n)*cs
gn(I*c*(e*x+d)^n)^2*ln(e*x+d)-3*I/e*Pi*b^3*d*n^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*ln(e*x+d)-3*I/e*Pi*ln(e*x+d)*
a*b^2*d*n*csgn(I*c*(e*x+d)^n)^3+3/2/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)
^3-3/4/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4+3/2/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*
(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^5+3/2/e*Pi^2*ln(e*x+d)*b^3*d*n*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5-3/4/e*Pi^2*ln(e*
x+d)*b^3*d*n*csgn(I*c)^2*csgn(I*c*(e*x+d)^n)^4+3/2*I/e*Pi*b^3*d*n^2*csgn(I*c*(e*x+d)^n)^3*ln(e*x+d)^2+3*I/e*Pi
*b^3*d*n^2*csgn(I*c*(e*x+d)^n)^3*ln(e*x+d)-3/2*I*ln(c)^2*Pi*b^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)
^n)-3*I*ln(c)*Pi*b^3*n*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-3*I*ln(c)*Pi*b^3*n*x*csgn(I*c)*csgn(I*c*(e*x+
d)^n)^2-3*I*Pi*b^3*n^2*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+3*I*ln(c)*Pi*a*b^2*x*csgn(I*(e*x+d)^n
)*csgn(I*c*(e*x+d)^n)^2+3*I*ln(c)*Pi*a*b^2*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3*I*Pi*a*b^2*n*x*csgn(I*(e*x+d)^n
)*csgn(I*c*(e*x+d)^n)^2-3*I*Pi*a*b^2*n*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3/2*I*Pi*a^2*b*x*csgn(I*c)*csgn(I*(e*
x+d)^n)*csgn(I*c*(e*x+d)^n)+3/8*I*Pi^3*b^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)^3*csgn(I*c*(e*x+d)^n)^5-3/8*I*Pi^3*b^
3*x*csgn(I*c)^2*csgn(I*(e*x+d)^n)^3*csgn(I*c*(e*x+d)^n)^4+1/8*I*Pi^3*b^3*x*csgn(I*c)^3*csgn(I*(e*x+d)^n)^3*csg
n(I*c*(e*x+d)^n)^3-9/8*I*Pi^3*b^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^6+9/8*I*Pi^3*b^3*x*csgn(
I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^5-3/8*I*Pi^3*b^3*x*csgn(I*c)^3*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*
x+d)^n)^4+9/8*I*Pi^3*b^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^7-9/8*I*Pi^3*b^3*x*csgn(I*c)^2*csgn
(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^6+3/8*I*Pi^3*b^3*x*csgn(I*c)^3*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^5+3/2*I
*ln(c)^2*Pi*b^3*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+3/2*I*ln(c)^2*Pi*b^3*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)
^2+3*I*ln(c)*Pi*b^3*n*x*csgn(I*c*(e*x+d)^n)^3+3*I*Pi*b^3*n^2*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+3*I*Pi*
b^3*n^2*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3*I*ln(c)*Pi*a*b^2*x*csgn(I*c*(e*x+d)^n)^3+3*I*Pi*a*b^2*n*x*csgn(I*c
*(e*x+d)^n)^3+3/2*I*Pi*a^2*b*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+3/2*I*Pi*a^2*b*x*csgn(I*c)*csgn(I*c*(e*
x+d)^n)^2

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maxima [B]  time = 0.71, size = 282, normalized size = 2.85 \[ b^{3} x \log \left ({\left (e x + d\right )}^{n} c\right )^{3} - 3 \, a^{2} b e n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + 3 \, a b^{2} x \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 3 \, a^{2} b x \log \left ({\left (e x + d\right )}^{n} c\right ) - 3 \, {\left (2 \, e n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {{\left (d \log \left (e x + d\right )^{2} - 2 \, e x + 2 \, d \log \left (e x + d\right )\right )} n^{2}}{e}\right )} a b^{2} - {\left (3 \, e n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} - e n {\left (\frac {{\left (d \log \left (e x + d\right )^{3} + 3 \, d \log \left (e x + d\right )^{2} - 6 \, e x + 6 \, d \log \left (e x + d\right )\right )} n^{2}}{e^{2}} - \frac {3 \, {\left (d \log \left (e x + d\right )^{2} - 2 \, e x + 2 \, d \log \left (e x + d\right )\right )} n \log \left ({\left (e x + d\right )}^{n} c\right )}{e^{2}}\right )}\right )} b^{3} + a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 172, normalized size = 1.74 \[ x\,\left (a^3-3\,a^2\,b\,n+6\,a\,b^2\,n^2-6\,b^3\,n^3\right )+{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^3\,\left (b^3\,x+\frac {b^3\,d}{e}\right )+{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2\,\left (\frac {3\,\left (a\,b^2\,d-b^3\,d\,n\right )}{e}+3\,b^2\,x\,\left (a-b\,n\right )\right )+\frac {\ln \left (d+e\,x\right )\,\left (3\,d\,a^2\,b\,n-6\,d\,a\,b^2\,n^2+6\,d\,b^3\,n^3\right )}{e}+3\,b\,x\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (a^2-2\,a\,b\,n+2\,b^2\,n^2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x)^n))^3,x)

[Out]

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

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sympy [A]  time = 3.32, size = 527, normalized size = 5.32 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b d n \log {\left (d + e x \right )}}{e} + 3 a^{2} b n x \log {\left (d + e x \right )} - 3 a^{2} b n x + 3 a^{2} b x \log {\relax (c )} + \frac {3 a b^{2} d n^{2} \log {\left (d + e x \right )}^{2}}{e} - \frac {6 a b^{2} d n^{2} \log {\left (d + e x \right )}}{e} + \frac {6 a b^{2} d n \log {\relax (c )} \log {\left (d + e x \right )}}{e} + 3 a b^{2} n^{2} x \log {\left (d + e x \right )}^{2} - 6 a b^{2} n^{2} x \log {\left (d + e x \right )} + 6 a b^{2} n^{2} x + 6 a b^{2} n x \log {\relax (c )} \log {\left (d + e x \right )} - 6 a b^{2} n x \log {\relax (c )} + 3 a b^{2} x \log {\relax (c )}^{2} + \frac {b^{3} d n^{3} \log {\left (d + e x \right )}^{3}}{e} - \frac {3 b^{3} d n^{3} \log {\left (d + e x \right )}^{2}}{e} + \frac {6 b^{3} d n^{3} \log {\left (d + e x \right )}}{e} + \frac {3 b^{3} d n^{2} \log {\relax (c )} \log {\left (d + e x \right )}^{2}}{e} - \frac {6 b^{3} d n^{2} \log {\relax (c )} \log {\left (d + e x \right )}}{e} + \frac {3 b^{3} d n \log {\relax (c )}^{2} \log {\left (d + e x \right )}}{e} + b^{3} n^{3} x \log {\left (d + e x \right )}^{3} - 3 b^{3} n^{3} x \log {\left (d + e x \right )}^{2} + 6 b^{3} n^{3} x \log {\left (d + e x \right )} - 6 b^{3} n^{3} x + 3 b^{3} n^{2} x \log {\relax (c )} \log {\left (d + e x \right )}^{2} - 6 b^{3} n^{2} x \log {\relax (c )} \log {\left (d + e x \right )} + 6 b^{3} n^{2} x \log {\relax (c )} + 3 b^{3} n x \log {\relax (c )}^{2} \log {\left (d + e x \right )} - 3 b^{3} n x \log {\relax (c )}^{2} + b^{3} x \log {\relax (c )}^{3} & \text {for}\: e \neq 0 \\x \left (a + b \log {\left (c d^{n} \right )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*d*n*log(d + e*x)/e + 3*a**2*b*n*x*log(d + e*x) - 3*a**2*b*n*x + 3*a**2*b*x*log(c)
+ 3*a*b**2*d*n**2*log(d + e*x)**2/e - 6*a*b**2*d*n**2*log(d + e*x)/e + 6*a*b**2*d*n*log(c)*log(d + e*x)/e + 3
*a*b**2*n**2*x*log(d + e*x)**2 - 6*a*b**2*n**2*x*log(d + e*x) + 6*a*b**2*n**2*x + 6*a*b**2*n*x*log(c)*log(d +
e*x) - 6*a*b**2*n*x*log(c) + 3*a*b**2*x*log(c)**2 + b**3*d*n**3*log(d + e*x)**3/e - 3*b**3*d*n**3*log(d + e*x)
**2/e + 6*b**3*d*n**3*log(d + e*x)/e + 3*b**3*d*n**2*log(c)*log(d + e*x)**2/e - 6*b**3*d*n**2*log(c)*log(d + e
*x)/e + 3*b**3*d*n*log(c)**2*log(d + e*x)/e + b**3*n**3*x*log(d + e*x)**3 - 3*b**3*n**3*x*log(d + e*x)**2 + 6*
b**3*n**3*x*log(d + e*x) - 6*b**3*n**3*x + 3*b**3*n**2*x*log(c)*log(d + e*x)**2 - 6*b**3*n**2*x*log(c)*log(d +
e*x) + 6*b**3*n**2*x*log(c) + 3*b**3*n*x*log(c)**2*log(d + e*x) - 3*b**3*n*x*log(c)**2 + b**3*x*log(c)**3, Ne
(e, 0)), (x*(a + b*log(c*d**n))**3, True))

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