3.1.49 \(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^2} \, dx\) [49]

Optimal. Leaf size=132 \[ \frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(e f-d g) (f+g x)}-\frac {2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g (e f-d g)}-\frac {2 b^2 e n^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2444, 2441, 2440, 2438} \begin {gather*} -\frac {2 b^2 e n^2 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)}-\frac {2 b e n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x) (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2444

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[(d + e
*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f - d*g)*(f + g*x))), x] - Dist[b*e*n*(p/(e*f - d*g)), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 126, normalized size = 0.95 \begin {gather*} \frac {-\left (\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (a g (d+e x)+b g (d+e x) \log \left (c (d+e x)^n\right )-2 b e n (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right )+2 b^2 e n^2 (f+g x) \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )}{g (-e f+d g) (f+g x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 1092, normalized size = 8.27

method	result	size
risch	\(\text {Expression too large to display}\)	\(1092\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^2/g*n^2*e/(d*g-e*f)*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+b^2/g*n^2*e/(d*g-e*f)*ln(e*x+d)^2-1/4*(-I*b*Pi*c
sgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)
*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*a)^2/(g*x+f)/g+I/g*n*e/(d*g-e*f)*ln(g*x+f)*b^2
*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I/g*n*e/(d*g-e*f)*ln(e*x+d)*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2
+I/g*n*e/(d*g-e*f)*ln(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I/g*n*e/(d*g-e*f)*ln(g*x+f
)*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I/g*n*e/(d*g-e*f)*ln(e*x+d)*b^2*Pi*csgn(I*c*(e*x+d)^n
)^3-b^2/(g*x+f)/g*ln((e*x+d)^n)^2+I/(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*c*(e*x+d)^n)^3-I/(g*x+f)/g*ln((e*x+d
)^n)*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I/(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*c)*csgn(I*c*(e*x+d
)^n)^2-2*b/(g*x+f)/g*ln((e*x+d)^n)*a-2/(g*x+f)/g*ln((e*x+d)^n)*b^2*ln(c)+2*b/g*n*e/(d*g-e*f)*ln(g*x+f)*a-I/g*n
*e/(d*g-e*f)*ln(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-2*b^2/g*n^2*e/(d*g-e*f)*ln(g*x+f)*ln(((g*x+f)*e+
d*g-e*f)/(d*g-e*f))+I/(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+2*b^2/g*n
*e*ln((e*x+d)^n)/(d*g-e*f)*ln(g*x+f)-2*b^2/g*n*e*ln((e*x+d)^n)/(d*g-e*f)*ln(e*x+d)-2*b/g*n*e/(d*g-e*f)*ln(e*x+
d)*a+2/g*n*e/(d*g-e*f)*ln(g*x+f)*b^2*ln(c)-2/g*n*e/(d*g-e*f)*ln(e*x+d)*b^2*ln(c)-I/g*n*e/(d*g-e*f)*ln(g*x+f)*b
^2*Pi*csgn(I*c*(e*x+d)^n)^3+I/g*n*e/(d*g-e*f)*ln(g*x+f)*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2

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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*(e*x+d)^n))^2/(g*x+f)^2,x, algorithm="maxima")

[Out]

2*a*b*n*(log(g*x + f)/(d*g^2 - f*g*e) - log(x*e + d)/(d*g^2 - f*g*e))*e - b^2*(log((x*e + d)^n)^2/(g^2*x + f*g
) - integrate((g*x*e*log(c)^2 + d*g*log(c)^2 + 2*(f*n*e + (g*n + g*log(c))*x*e + d*g*log(c))*log((x*e + d)^n))
/(g^3*x^3*e + d*f^2*g + (d*g^3 + 2*f*g^2*e)*x^2 + (2*d*f*g^2 + f^2*g*e)*x), x)) - 2*a*b*log((x*e + d)^n*c)/(g^
2*x + f*g) - a^2/(g^2*x + f*g)

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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*(e*x+d)^n))^2/(g*x+f)^2,x, algorithm="fricas")

[Out]

integral((b^2*log((x*e + d)^n*c)^2 + 2*a*b*log((x*e + d)^n*c) + a^2)/(g^2*x^2 + 2*f*g*x + f^2), 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 \left (d + e x\right )^{n} \right )}\right )^{2}}{\left (f + g x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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