3.9.93 \(\int \frac {x (d+e x)}{(a+b x+c x^2)^2} \, dx\) [893]

Optimal. Leaf size=99 \[ \frac {a (2 c d-b e)+\left (b c d-b^2 e+2 a c e\right ) x}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 (b d-2 a e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

(a*(-b*e+2*c*d)+(2*a*c*e-b^2*e+b*c*d)*x)/c/(-4*a*c+b^2)/(c*x^2+b*x+a)-2*(-2*a*e+b*d)*arctanh((2*c*x+b)/(-4*a*c
+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)

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Rubi [A]
time = 0.03, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {791, 632, 212} \begin {gather*} \frac {x \left (2 a c e+b^2 (-e)+b c d\right )+a (2 c d-b e)}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 (b d-2 a e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*c*d - b*e) + (b*c*d - b^2*e + 2*a*c*e)*x)/(c*(b^2 - 4*a*c)*(a + b*x + c*x^2)) - (2*(b*d - 2*a*e)*ArcTanh
[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)

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 791

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x))*((a + b*x + c*x^2
)^(p + 1)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*
p + 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x (d+e x)}{\left (a+b x+c x^2\right )^2} \, dx &=\frac {a (2 c d-b e)+\left (b c d-b^2 e+2 a c e\right ) x}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {(b d-2 a e) \int \frac {1}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=\frac {a (2 c d-b e)+\left (b c d-b^2 e+2 a c e\right ) x}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(2 (b d-2 a e)) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{b^2-4 a c}\\ &=\frac {a (2 c d-b e)+\left (b c d-b^2 e+2 a c e\right ) x}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 (b d-2 a e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 99, normalized size = 1.00 \begin {gather*} \frac {a b e+b (-c d+b e) x-2 a c (d+e x)}{c \left (-b^2+4 a c\right ) (a+x (b+c x))}-\frac {2 (b d-2 a e) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*e + b*(-(c*d) + b*e)*x - 2*a*c*(d + e*x))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))) - (2*(b*d - 2*a*e)*ArcTan[
(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2)

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Maple [A]
time = 0.77, size = 118, normalized size = 1.19

method	result	size
default	\(\frac {-\frac {\left (2 a c e -b^{2} e +b c d \right ) x}{c \left (4 a c -b^{2}\right )}+\frac {a \left (b e -2 c d \right )}{c \left (4 a c -b^{2}\right )}}{c \,x^{2}+b x +a}+\frac {2 \left (2 a e -b d \right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\)	\(118\)
risch	\(\frac {-\frac {\left (2 a c e -b^{2} e +b c d \right ) x}{c \left (4 a c -b^{2}\right )}+\frac {a \left (b e -2 c d \right )}{c \left (4 a c -b^{2}\right )}}{c \,x^{2}+b x +a}+\frac {2 \ln \left (\left (-8 c^{2} a +2 b^{2} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) a e}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left (\left (-8 c^{2} a +2 b^{2} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) b d}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {2 \ln \left (\left (8 c^{2} a -2 b^{2} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) a e}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}+\frac {\ln \left (\left (8 c^{2} a -2 b^{2} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) b d}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x*(e*x+d)/(c*x^2+b*x+a)^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 [B] Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (99) = 198\).
time = 1.09, size = 514, normalized size = 5.19 \begin {gather*} \left [\frac {{\left (b^{3} c - 4 \, a b c^{2}\right )} d x + {\left (b c^{2} d x^{2} + b^{2} c d x + a b c d - 2 \, {\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} 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 ) + 2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x\right )} e}{a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}, \frac {{\left (b^{3} c - 4 \, a b c^{2}\right )} d x - 2 \, {\left (b c^{2} d x^{2} + b^{2} c d x + a b c d - 2 \, {\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} 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 ) + 2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x\right )} e}{a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((b^3*c - 4*a*b*c^2)*d*x + (b*c^2*d*x^2 + b^2*c*d*x + a*b*c*d - 2*(a*c^2*x^2 + a*b*c*x + a^2*c)*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)) + 2*(a*b^2*
c - 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c + (b^4 - 6*a*b^2*c + 8*a^2*c^2)*x)*e)/(a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*
c^3 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x), ((b^3*c - 4*a*b*c^2)
*d*x - 2*(b*c^2*d*x^2 + b^2*c*d*x + a*b*c*d - 2*(a*c^2*x^2 + a*b*c*x + a^2*c)*e)*sqrt(-b^2 + 4*a*c)*arctan(-sq
rt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(a*b^2*c - 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c + (b^4 - 6*a*b^2*
c + 8*a^2*c^2)*x)*e)/(a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^2 + (b^5*c
- 8*a*b^3*c^2 + 16*a^2*b*c^3)*x)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (90) = 180\).
time = 0.58, size = 379, normalized size = 3.83 \begin {gather*} - \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) \log {\left (x + \frac {- 16 a^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) + 8 a b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) + 2 a b e - b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) - b^{2} d}{4 a c e - 2 b c d} \right )} + \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) \log {\left (x + \frac {16 a^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) - 8 a b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) + 2 a b e + b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \cdot \left (2 a e - b d\right ) - b^{2} d}{4 a c e - 2 b c d} \right )} + \frac {a b e - 2 a c d + x \left (- 2 a c e + b^{2} e - b c d\right )}{4 a^{2} c^{2} - a b^{2} c + x^{2} \cdot \left (4 a c^{3} - b^{2} c^{2}\right ) + x \left (4 a b c^{2} - b^{3} c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d)*log(x + (-16*a**2*c**2*sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d) + 8*
a*b**2*c*sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d) + 2*a*b*e - b**4*sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d) -
b**2*d)/(4*a*c*e - 2*b*c*d)) + sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d)*log(x + (16*a**2*c**2*sqrt(-1/(4*a*c -
b**2)**3)*(2*a*e - b*d) - 8*a*b**2*c*sqrt(-1/(4*a*c - b**2)**3)*(2*a*e - b*d) + 2*a*b*e + b**4*sqrt(-1/(4*a*c
- b**2)**3)*(2*a*e - b*d) - b**2*d)/(4*a*c*e - 2*b*c*d)) + (a*b*e - 2*a*c*d + x*(-2*a*c*e + b**2*e - b*c*d))/
(4*a**2*c**2 - a*b**2*c + x**2*(4*a*c**3 - b**2*c**2) + x*(4*a*b*c**2 - b**3*c))

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Giac [A]
time = 0.81, size = 113, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (b d - 2 \, a e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {b c d x - b^{2} x e + 2 \, a c x e + 2 \, a c d - a b e}{{\left (b^{2} c - 4 \, a c^{2}\right )} {\left (c x^{2} + b x + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.28, size = 177, normalized size = 1.79 \begin {gather*} \frac {\frac {a\,\left (b\,e-2\,c\,d\right )}{c\,\left (4\,a\,c-b^2\right )}-\frac {x\,\left (-e\,b^2+c\,d\,b+2\,a\,c\,e\right )}{c\,\left (4\,a\,c-b^2\right )}}{c\,x^2+b\,x+a}-\frac {2\,\mathrm {atan}\left (\frac {\left (4\,a\,c-b^2\right )\,\left (\frac {\left (b^3-4\,a\,b\,c\right )\,\left (2\,a\,e-b\,d\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {2\,c\,x\,\left (2\,a\,e-b\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{2\,a\,e-b\,d}\right )\,\left (2\,a\,e-b\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*(b*e - 2*c*d))/(c*(4*a*c - b^2)) - (x*(2*a*c*e - b^2*e + b*c*d))/(c*(4*a*c - b^2)))/(a + b*x + c*x^2) - (2
*atan(((4*a*c - b^2)*(((b^3 - 4*a*b*c)*(2*a*e - b*d))/(4*a*c - b^2)^(5/2) - (2*c*x*(2*a*e - b*d))/(4*a*c - b^2
)^(3/2)))/(2*a*e - b*d))*(2*a*e - b*d))/(4*a*c - b^2)^(3/2)

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