3.2.17 \(\int \frac {(a+b \log (c x^n))^2}{(d+e x)^4} \, dx\) [117]

Optimal. Leaf size=203 \[ -\frac {b^2 n^2}{3 d^2 e (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d^3 e}+\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)}-\frac {2 b n \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^3 e}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}+\frac {b^2 n^2 \log (d+e x)}{d^3 e}+\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {d}{e x}\right )}{3 d^3 e} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2356, 2389, 2379, 2438, 2351, 31, 46} \begin {gather*} \frac {2 b^2 n^2 \text {PolyLog}\left (2,-\frac {d}{e x}\right )}{3 d^3 e}-\frac {2 b n \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^3 e}-\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)}+\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}-\frac {b^2 n^2 \log (x)}{3 d^3 e}+\frac {b^2 n^2 \log (d+e x)}{d^3 e}-\frac {b^2 n^2}{3 d^2 e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b^2*n^2)/(d^2*e*(d + e*x)) - (b^2*n^2*Log[x])/(3*d^3*e) + (b*n*(a + b*Log[c*x^n]))/(3*d*e*(d + e*x)^2) -
(2*b*n*x*(a + b*Log[c*x^n]))/(3*d^3*(d + e*x)) - (2*b*n*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/(3*d^3*e) - (a +
b*Log[c*x^n])^2/(3*e*(d + e*x)^3) + (b^2*n^2*Log[d + e*x])/(d^3*e) + (2*b^2*n^2*PolyLog[2, -(d/(e*x))])/(3*d^
3*e)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx &=-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx}{3 e}\\ &=-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{3 d}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{3 d e}\\ &=\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{3 d^2}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{3 d^2 e}-\frac {\left (b^2 n^2\right ) \int \frac {1}{x (d+e x)^2} \, dx}{3 d e}\\ &=\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{3 d^3}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{3 d^3 e}+\frac {\left (2 b^2 n^2\right ) \int \frac {1}{d+e x} \, dx}{3 d^3}-\frac {\left (b^2 n^2\right ) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{3 d e}\\ &=-\frac {b^2 n^2}{3 d^2 e (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d^3 e}+\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 d^3 e}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}+\frac {b^2 n^2 \log (d+e x)}{d^3 e}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d^3 e}+\frac {\left (2 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{3 d^3 e}\\ &=-\frac {b^2 n^2}{3 d^2 e (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d^3 e}+\frac {b n \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 d^3 e}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}+\frac {b^2 n^2 \log (d+e x)}{d^3 e}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d^3 e}-\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{3 d^3 e}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 211, normalized size = 1.04 \begin {gather*} -\frac {\left (a+b \log \left (c x^n\right )\right )^2}{3 e (d+e x)^3}+\frac {2 b n \left (\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}+\frac {a+b \log \left (c x^n\right )}{d^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^3 n}-\frac {b n \left (\frac {1}{d (d+e x)}+\frac {\log (x)}{d^2}-\frac {\log (d+e x)}{d^2}\right )}{2 d}-\frac {b n \left (\frac {\log (x)}{d}-\frac {\log (d+e x)}{d}\right )}{d^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (\frac {d+e x}{d}\right )}{d^3}-\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^3}\right )}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a + b*Log[c*x^n])^2/(e*(d + e*x)^3) + (2*b*n*((a + b*Log[c*x^n])/(2*d*(d + e*x)^2) + (a + b*Log[c*x^n])/
(d^2*(d + e*x)) + (a + b*Log[c*x^n])^2/(2*b*d^3*n) - (b*n*(1/(d*(d + e*x)) + Log[x]/d^2 - Log[d + e*x]/d^2))/(
2*d) - (b*n*(Log[x]/d - Log[d + e*x]/d))/d^2 - ((a + b*Log[c*x^n])*Log[(d + e*x)/d])/d^3 - (b*n*PolyLog[2, -((
e*x)/d)])/d^3))/(3*e)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 1227, normalized size = 6.04

method	result	size
risch	\(\text {Expression too large to display}\)	\(1227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))^2/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

1/3*I/e*n/d^3*ln(x)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2/3*b/(e*x+d)^3/e*ln(x^n)*a+1/3*I/(e*x+d)^3/e*ln(x^n)*b
^2*Pi*csgn(I*c*x^n)^3+1/3*I/(e*x+d)^3/e*ln(x^n)*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2/3/e*n/d^2/(e*x+d)
*b^2*ln(c)+2/3/e*n/d^3*ln(x)*b^2*ln(c)-2/3/e*n/d^3*ln(e*x+d)*b^2*ln(c)-2/3*b^2/e*n/d^3*ln(x^n)*ln(e*x+d)+2/3*b
^2/e*n*ln(x^n)/d^2/(e*x+d)+1/3*b^2/e*n*ln(x^n)/d/(e*x+d)^2+2/3*b^2/e*n*ln(x^n)/d^3*ln(x)+1/3/e*n/d/(e*x+d)^2*b
^2*ln(c)+1/3*b/e*n/d/(e*x+d)^2*a+2/3*b/e*n/d^2/(e*x+d)*a+2/3*b/e*n/d^3*ln(x)*a-2/3*b/e*n/d^3*ln(e*x+d)*a-1/3*b
^2/e*n^2/d^3*ln(x)^2+2/3*b^2/e*n^2/d^3*dilog(-e*x/d)-1/6*I/e*n/d/(e*x+d)^2*b^2*Pi*csgn(I*c*x^n)^3-1/3*I/e*n/d^
3*ln(x)*b^2*Pi*csgn(I*c*x^n)^3-1/12*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n
)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)^2/(e*x+d)^3/e+2/3*b^2/e*n^2/d^3*l
n(e*x+d)*ln(-e*x/d)-1/3*I/(e*x+d)^3/e*ln(x^n)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/3*I/(e*x+d)^3/e*ln(x^n)*b^2
*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/3*b^2/(e*x+d)^3/e*ln(x^n)^2+1/3*I/e*n/d^3*ln(e*x+d)*b^2*Pi*csgn(I*c*x^n)^3-1/3
*I/e*n/d^2/(e*x+d)*b^2*Pi*csgn(I*c*x^n)^3+1/6*I/e*n/d/(e*x+d)^2*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/3*I/e*n/d
^3*ln(e*x+d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*I/e*n/d^2/(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2-2/3/(e*
x+d)^3/e*ln(x^n)*b^2*ln(c)-1/3*I/e*n/d^3*ln(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2-b^2*n^2*ln(x)/d^3/e+b^2*n^
2*ln(e*x+d)/d^3/e-1/3*b^2*n^2/d^2/e/(e*x+d)-1/6*I/e*n/d/(e*x+d)^2*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1
/6*I/e*n/d/(e*x+d)^2*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/3*I/e*n/d^3*ln(x)*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/3
*I/e*n/d^2/(e*x+d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*I/e*n/d^3*ln(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csg
n(I*c*x^n)-1/3*I/e*n/d^2/(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/3*I/e*n/d^3*ln(x)*b^2*Pi*csgn(I*
c)*csgn(I*x^n)*csgn(I*c*x^n)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/3*a*b*n*((2*x*e + 3*d)/(d^2*x^2*e^3 + 2*d^3*x*e^2 + d^4*e) - 2*e^(-1)*log(x*e + d)/d^3 + 2*e^(-1)*log(x)/d^3
) - 1/3*b^2*(log(x^n)^2/(x^3*e^4 + 3*d*x^2*e^3 + 3*d^2*x*e^2 + d^3*e) - 3*integrate(1/3*(3*x*e*log(c)^2 + 2*((
n + 3*log(c))*x*e + d*n)*log(x^n))/(x^5*e^5 + 4*d*x^4*e^4 + 6*d^2*x^3*e^3 + 4*d^3*x^2*e^2 + d^4*x*e), x)) - 2/
3*a*b*log(c*x^n)/(x^3*e^4 + 3*d*x^2*e^3 + 3*d^2*x*e^2 + d^3*e) - 1/3*a^2/(x^3*e^4 + 3*d*x^2*e^3 + 3*d^2*x*e^2
+ d^3*e)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))**2/(e*x+d)**4,x)

[Out]

Integral((a + b*log(c*x**n))**2/(d + e*x)**4, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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