\(\int (d+e x) \log (d (a+b x+c x^2)^n) \, dx\) [85]

Optimal result

Integrand size = 21, antiderivative size = 154 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {1}{2} \left (4 d-\frac {b e}{c}\right ) n x-\frac {1}{2} e n x^2+\frac {\sqrt {b^2-4 a c} (2 c d-b e) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 c^2}-\frac {\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n \log \left (a+b x+c x^2\right )}{4 c^2 e}+\frac {(d+e x)^2 \log \left (d \left (a+b x+c x^2\right )^n\right )}{2 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2605, 814, 648, 632, 212, 642} \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\frac {n \sqrt {b^2-4 a c} (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 c^2}-\frac {n \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{4 c^2 e}+\frac {(d+e x)^2 \log \left (d \left (a+b x+c x^2\right )^n\right )}{2 e}-\frac {1}{2} n x \left (4 d-\frac {b e}{c}\right )-\frac {1}{2} e n x^2 \]

[In]

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

[Out]

-1/2*((4*d - (b*e)/c)*n*x) - (e*n*x^2)/2 + (Sqrt[b^2 - 4*a*c]*(2*c*d - b*e)*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4
*a*c]])/(2*c^2) - ((2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*n*Log[a + b*x + c*x^2])/(4*c^2*e) + ((d + e*x)^2*
Log[d*(a + b*x + c*x^2)^n])/(2*e)

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.80 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\frac {-2 \sqrt {b^2-4 a c} (-2 c d+b e) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+\left (2 b c d-b^2 e+2 a c e\right ) n \log (a+x (b+c x))+2 c x \left (b e n-c n (4 d+e x)+c (2 d+e x) \log \left (d (a+x (b+c x))^n\right )\right )}{4 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.10

method	result	size
parts	\(\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) e \,x^{2}}{2}+\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) d x -\frac {n \left (-\frac {-c e \,x^{2}+b e x -4 c d x}{c}+\frac {\frac {\left (-2 a c e +e \,b^{2}-2 b c d \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (a b e -4 a c d -\frac {\left (-2 a c e +e \,b^{2}-2 b c d \right ) b}{2 c}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right )}{\sqrt {4 c a -b^{2}}}}{c}\right )}{2}\)	\(170\)
risch	\(\text {Expression too large to display}\)	\(1706\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.18 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\left [-\frac {2 \, c^{2} e n x^{2} + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} n \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (4 \, c^{2} d - b c e\right )} n x - {\left (2 \, c^{2} e n x^{2} + 4 \, c^{2} d n x + {\left (2 \, b c d - {\left (b^{2} - 2 \, a c\right )} e\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (c^{2} e x^{2} + 2 \, c^{2} d x\right )} \log \left (d\right )}{4 \, c^{2}}, -\frac {2 \, c^{2} e n x^{2} - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} n \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (4 \, c^{2} d - b c e\right )} n x - {\left (2 \, c^{2} e n x^{2} + 4 \, c^{2} d n x + {\left (2 \, b c d - {\left (b^{2} - 2 \, a c\right )} e\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (c^{2} e x^{2} + 2 \, c^{2} d x\right )} \log \left (d\right )}{4 \, c^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (139) = 278\).

Time = 88.72 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.46 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\begin {cases} \frac {a e \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{2 c} - \frac {b^{2} e \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{4 c^{2}} + \frac {b d \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{2 c} + \frac {b e n x}{2 c} - \frac {b e n \sqrt {- 4 a c + b^{2}} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{2 c^{2}} + \frac {b e \sqrt {- 4 a c + b^{2}} \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{4 c^{2}} - 2 d n x + d x \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )} - \frac {e n x^{2}}{2} + \frac {e x^{2} \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{2} + \frac {d n \sqrt {- 4 a c + b^{2}} \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{c} - \frac {d \sqrt {- 4 a c + b^{2}} \log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{2 c} & \text {for}\: c \neq 0 \\- \frac {a^{2} e \log {\left (d \left (a + b x\right )^{n} \right )}}{2 b^{2}} + \frac {a d \log {\left (d \left (a + b x\right )^{n} \right )}}{b} + \frac {a e n x}{2 b} - d n x + d x \log {\left (d \left (a + b x\right )^{n} \right )} - \frac {e n x^{2}}{4} + \frac {e x^{2} \log {\left (d \left (a + b x\right )^{n} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*e*log(d*(a + b*x + c*x**2)**n)/(2*c) - b**2*e*log(d*(a + b*x + c*x**2)**n)/(4*c**2) + b*d*log(d*(
a + b*x + c*x**2)**n)/(2*c) + b*e*n*x/(2*c) - b*e*n*sqrt(-4*a*c + b**2)*log(b/(2*c) + x + sqrt(-4*a*c + b**2)/
(2*c))/(2*c**2) + b*e*sqrt(-4*a*c + b**2)*log(d*(a + b*x + c*x**2)**n)/(4*c**2) - 2*d*n*x + d*x*log(d*(a + b*x
+ c*x**2)**n) - e*n*x**2/2 + e*x**2*log(d*(a + b*x + c*x**2)**n)/2 + d*n*sqrt(-4*a*c + b**2)*log(b/(2*c) + x
+ sqrt(-4*a*c + b**2)/(2*c))/c - d*sqrt(-4*a*c + b**2)*log(d*(a + b*x + c*x**2)**n)/(2*c), Ne(c, 0)), (-a**2*e
*log(d*(a + b*x)**n)/(2*b**2) + a*d*log(d*(a + b*x)**n)/b + a*e*n*x/(2*b) - d*n*x + d*x*log(d*(a + b*x)**n) -
e*n*x**2/4 + e*x**2*log(d*(a + b*x)**n)/2, True))

Maxima [F(-2)]

Exception generated. \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {1}{2} \, {\left (e n - e \log \left (d\right )\right )} x^{2} + \frac {1}{2} \, {\left (e n x^{2} + 2 \, d n x\right )} \log \left (c x^{2} + b x + a\right ) - \frac {{\left (4 \, c d n - b e n - 2 \, c d \log \left (d\right )\right )} x}{2 \, c} + \frac {{\left (2 \, b c d n - b^{2} e n + 2 \, a c e n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, c^{2}} - \frac {{\left (2 \, b^{2} c d n - 8 \, a c^{2} d n - b^{3} e n + 4 \, a b c e n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.57 \[ \int (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )\,\left (\frac {e\,x^2}{2}+d\,x\right )-x\,\left (\frac {n\,\left (b\,e+4\,c\,d\right )}{2\,c}-\frac {b\,e\,n}{c}\right )-\frac {e\,n\,x^2}{2}+\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (c\,\left (\frac {a\,e\,n}{2}+\frac {b\,d\,n}{2}-\frac {d\,n\,\sqrt {b^2-4\,a\,c}}{2}\right )-\frac {b^2\,e\,n}{4}+\frac {b\,e\,n\,\sqrt {b^2-4\,a\,c}}{4}\right )}{c^2}-\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^2\,e\,n}{4}-c\,\left (\frac {a\,e\,n}{2}+\frac {b\,d\,n}{2}+\frac {d\,n\,\sqrt {b^2-4\,a\,c}}{2}\right )+\frac {b\,e\,n\,\sqrt {b^2-4\,a\,c}}{4}\right )}{c^2} \]

[In]

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

[Out]

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