3.1.88 \(\int \frac {\log (d (a+b x+c x^2)^n)}{(d+e x)^2} \, dx\) [88]

Optimal. Leaf size=165 \[ \frac {\sqrt {b^2-4 a c} n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c d^2-b d e+a e^2}-\frac {(2 c d-b e) n \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}+\frac {(2 c d-b e) n \log \left (a+b x+c x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \]

[Out]

-(-b*e+2*c*d)*n*ln(e*x+d)/e/(a*e^2-b*d*e+c*d^2)+1/2*(-b*e+2*c*d)*n*ln(c*x^2+b*x+a)/e/(a*e^2-b*d*e+c*d^2)-ln(d*
(c*x^2+b*x+a)^n)/e/(e*x+d)+n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)

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Rubi [A]
time = 0.16, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2605, 814, 648, 632, 212, 642} \begin {gather*} \frac {n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {n (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac {n (2 c d-b e) \log (d+e x)}{e \left (a e^2-b d e+c d^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b^2 - 4*a*c]*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c*d^2 - b*d*e + a*e^2) - ((2*c*d - b*e)*n*Log[d
+ e*x])/(e*(c*d^2 - b*d*e + a*e^2)) + ((2*c*d - b*e)*n*Log[a + b*x + c*x^2])/(2*e*(c*d^2 - b*d*e + a*e^2)) - L
og[d*(a + b*x + c*x^2)^n]/(e*(d + e*x))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 166, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {-b^2+4 a c} n \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{-c d^2+e (b d-a e)}+\frac {(-2 c d+b e) n \log (d+e x)}{e \left (c d^2+e (-b d+a e)\right )}-\frac {(-2 c d+b e) n \log (a+x (b+c x))}{2 e \left (c d^2+e (-b d+a e)\right )}-\frac {\log \left (d (a+x (b+c x))^n\right )}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.12, size = 1785, normalized size = 10.82

method	result	size
risch	\(\text {Expression too large to display}\)	\(1785\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e/(e*x+d)*ln((c*x^2+b*x+a)^n)+1/2*(-2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*c*d*e^3-2*c^2*d^2*e^2)*_R^2+(b*c*
e^2*n-2*c^2*d*e*n)*_R+4*c^2*n^2)*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^2)*_R^2+(-2*a*c*e^2*n+b^2*e^2*n-b
*c*d*e*n)*_R+2*b*c*n^2),_R=RootOf((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c*d*e*n)*_Z+c*n^2))*b*d*e^3*x-4*ln
(e*x+d)*c*d*e*n*x-I*Pi*b*e*csgn(I*d*(c*x^2+b*x+a)^n)^3*d+2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*c*d*e^3-2*c^2*d
^2*e^2)*_R^2+(b*c*e^2*n-2*c^2*d*e*n)*_R+4*c^2*n^2)*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^2)*_R^2+(-2*a*c
*e^2*n+b^2*e^2*n-b*c*d*e*n)*_R+2*b*c*n^2),_R=RootOf((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c*d*e*n)*_Z+c*n^
2))*c*d^3*e-4*ln(e*x+d)*c*d^2*n+2*ln(d)*b*e*d-I*Pi*b*e*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a
)^n)*d+2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*c*d*e^3-2*c^2*d^2*e^2)*_R^2+(b*c*e^2*n-2*c^2*d*e*n)*_R+4*c^2*n^2)
*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^2)*_R^2+(-2*a*c*e^2*n+b^2*e^2*n-b*c*d*e*n)*_R+2*b*c*n^2),_R=RootO
f((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c*d*e*n)*_Z+c*n^2))*a*e^4*x+2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*
c*d*e^3-2*c^2*d^2*e^2)*_R^2+(b*c*e^2*n-2*c^2*d*e*n)*_R+4*c^2*n^2)*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^
2)*_R^2+(-2*a*c*e^2*n+b^2*e^2*n-b*c*d*e*n)*_R+2*b*c*n^2),_R=RootOf((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c
*d*e*n)*_Z+c*n^2))*a*d*e^3-2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*c*d*e^3-2*c^2*d^2*e^2)*_R^2+(b*c*e^2*n-2*c^2*
d*e*n)*_R+4*c^2*n^2)*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^2)*_R^2+(-2*a*c*e^2*n+b^2*e^2*n-b*c*d*e*n)*_R
+2*b*c*n^2),_R=RootOf((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c*d*e*n)*_Z+c*n^2))*b*d^2*e^2-2*ln(d)*a*e^2-2*
ln(d)*c*d^2+2*ln(e*x+d)*b*d*e*n-I*Pi*c*d^2*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*c*d^2*csgn(I*(c*x^2+b*x+
a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*a*e^2*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*a*e^2*csgn(I*(c*x^2+b*
x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2+I*Pi*a*e^2*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+I*P
i*b*e*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2*d+I*Pi*b*e*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2*d+I
*Pi*c*d^2*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+I*Pi*a*e^2*csgn(I*d*(c*x^2+b*x+a)^n)^3+I
*Pi*c*d^2*csgn(I*d*(c*x^2+b*x+a)^n)^3+2*sum(_R*ln(((6*a*c*e^4-2*b^2*e^4+2*b*c*d*e^3-2*c^2*d^2*e^2)*_R^2+(b*c*e
^2*n-2*c^2*d*e*n)*_R+4*c^2*n^2)*x+(-a*b*e^4+8*a*c*d*e^3-b^2*d*e^3-b*c*d^2*e^2)*_R^2+(-2*a*c*e^2*n+b^2*e^2*n-b*
c*d*e*n)*_R+2*b*c*n^2),_R=RootOf((a*e^4-b*d*e^3+c*d^2*e^2)*_Z^2+(b*e^2*n-2*c*d*e*n)*_Z+c*n^2))*c*d^2*e^2*x+2*l
n(e*x+d)*b*e^2*n*x)/e/(a*e^2-b*d*e+c*d^2)/(e*x+d)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.42, size = 425, normalized size = 2.58 \begin {gather*} \left [\frac {{\left (n x e^{2} + d n e\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - {\left ({\left (b n x + 2 \, a n\right )} e^{2} - {\left (2 \, c d n x + b d n\right )} e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d^{2} n - b n x e^{2} + {\left (2 \, c d n x - b d n\right )} e\right )} \log \left (x e + d\right ) - 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \log \left (d\right )}{2 \, {\left (c d^{3} e + a x e^{4} - {\left (b d x - a d\right )} e^{3} + {\left (c d^{2} x - b d^{2}\right )} e^{2}\right )}}, \frac {2 \, {\left (n x e^{2} + d n e\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b n x + 2 \, a n\right )} e^{2} - {\left (2 \, c d n x + b d n\right )} e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d^{2} n - b n x e^{2} + {\left (2 \, c d n x - b d n\right )} e\right )} \log \left (x e + d\right ) - 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \log \left (d\right )}{2 \, {\left (c d^{3} e + a x e^{4} - {\left (b d x - a d\right )} e^{3} + {\left (c d^{2} x - b d^{2}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((n*x*e^2 + d*n*e)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x +
b))/(c*x^2 + b*x + a)) - ((b*n*x + 2*a*n)*e^2 - (2*c*d*n*x + b*d*n)*e)*log(c*x^2 + b*x + a) - 2*(2*c*d^2*n - b
*n*x*e^2 + (2*c*d*n*x - b*d*n)*e)*log(x*e + d) - 2*(c*d^2 - b*d*e + a*e^2)*log(d))/(c*d^3*e + a*x*e^4 - (b*d*x
- a*d)*e^3 + (c*d^2*x - b*d^2)*e^2), 1/2*(2*(n*x*e^2 + d*n*e)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(
2*c*x + b)/(b^2 - 4*a*c)) - ((b*n*x + 2*a*n)*e^2 - (2*c*d*n*x + b*d*n)*e)*log(c*x^2 + b*x + a) - 2*(2*c*d^2*n
- b*n*x*e^2 + (2*c*d*n*x - b*d*n)*e)*log(x*e + d) - 2*(c*d^2 - b*d*e + a*e^2)*log(d))/(c*d^3*e + a*x*e^4 - (b*
d*x - a*d)*e^3 + (c*d^2*x - b*d^2)*e^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 6.37, size = 284, normalized size = 1.72 \begin {gather*} \frac {{\left (2 \, c d n - b n e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac {{\left (b^{2} n - 4 \, a c n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c d n x e \log \left (x e + d\right ) + c d^{2} n \log \left (c x^{2} + b x + a\right ) - b d n e \log \left (c x^{2} + b x + a\right ) + 2 \, c d^{2} n \log \left (x e + d\right ) - b n x e^{2} \log \left (x e + d\right ) - b d n e \log \left (x e + d\right ) + a n e^{2} \log \left (c x^{2} + b x + a\right ) + c d^{2} \log \left (d\right ) - b d e \log \left (d\right ) + a e^{2} \log \left (d\right )}{c d^{2} x e^{2} + c d^{3} e - b d x e^{3} - b d^{2} e^{2} + a x e^{4} + a d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.34, size = 590, normalized size = 3.58 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (b\,e\,n-2\,c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{e\,\left (d+e\,x\right )}-\frac {\ln \left (\frac {2\,b\,c^2\,n^2}{e}+\frac {4\,c^3\,n^2\,x}{e}-\frac {n\,\left (b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (c^2\,n\,x\,\left (b\,e-2\,c\,d\right )-c\,n\,\left (-e\,b^2+c\,d\,b+2\,a\,c\,e\right )+\frac {c\,e\,n\,\left (b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )\,\left (e\,\left (\frac {b\,n}{2}+\frac {n\,\sqrt {b^2-4\,a\,c}}{2}\right )-c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3}-\frac {\ln \left (\frac {2\,b\,c^2\,n^2}{e}+\frac {4\,c^3\,n^2\,x}{e}-\frac {n\,\left (2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (c\,n\,\left (-e\,b^2+c\,d\,b+2\,a\,c\,e\right )-c^2\,n\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,n\,\left (2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )\,\left (e\,\left (\frac {b\,n}{2}-\frac {n\,\sqrt {b^2-4\,a\,c}}{2}\right )-c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(d + e*x)*(b*e*n - 2*c*d*n))/(a*e^3 - b*d*e^2 + c*d^2*e) - log(d*(a + b*x + c*x^2)^n)/(e*(d + e*x)) - (log
((2*b*c^2*n^2)/e + (4*c^3*n^2*x)/e - (n*(b*e - 2*c*d + e*(b^2 - 4*a*c)^(1/2))*(c^2*n*x*(b*e - 2*c*d) - c*n*(2*
a*c*e - b^2*e + b*c*d) + (c*e*n*(b*e - 2*c*d + e*(b^2 - 4*a*c)^(1/2))*(2*b^2*e^2*x + 2*c^2*d^2*x + a*b*e^2 + b
*c*d^2 + b^2*d*e - 6*a*c*e^2*x - 8*a*c*d*e - 2*b*c*d*e*x))/(2*(a*e^3 - b*d*e^2 + c*d^2*e))))/(2*(a*e^3 - b*d*e
^2 + c*d^2*e)))*(e*((b*n)/2 + (n*(b^2 - 4*a*c)^(1/2))/2) - c*d*n))/(a*e^3 - b*d*e^2 + c*d^2*e) - (log((2*b*c^2
*n^2)/e + (4*c^3*n^2*x)/e - (n*(2*c*d - b*e + e*(b^2 - 4*a*c)^(1/2))*(c*n*(2*a*c*e - b^2*e + b*c*d) - c^2*n*x*
(b*e - 2*c*d) + (c*e*n*(2*c*d - b*e + e*(b^2 - 4*a*c)^(1/2))*(2*b^2*e^2*x + 2*c^2*d^2*x + a*b*e^2 + b*c*d^2 +
b^2*d*e - 6*a*c*e^2*x - 8*a*c*d*e - 2*b*c*d*e*x))/(2*(a*e^3 - b*d*e^2 + c*d^2*e))))/(2*(a*e^3 - b*d*e^2 + c*d^
2*e)))*(e*((b*n)/2 - (n*(b^2 - 4*a*c)^(1/2))/2) - c*d*n))/(a*e^3 - b*d*e^2 + c*d^2*e)

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