3.12.63 \(\int \frac {1}{(b d+2 c d x)^2 (a+b x+c x^2)} \, dx\) [1163]

Optimal. Leaf size=61 \[ \frac {2}{\left (b^2-4 a c\right ) d^2 (b+2 c x)}-\frac {2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} d^2} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {707, 632, 212} \begin {gather*} \frac {2}{d^2 \left (b^2-4 a c\right ) (b+2 c x)}-\frac {2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)),x]

[Out]

2/((b^2 - 4*a*c)*d^2*(b + 2*c*x)) - (2*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(3/2)*d^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 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 63, normalized size = 1.03 \begin {gather*} \frac {\frac {2}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)),x]

[Out]

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

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Maple [A]
time = 0.74, size = 62, normalized size = 1.02

method	result	size
default	\(\frac {-\frac {2}{\left (4 a c -b^{2}\right ) \left (2 c x +b \right )}-\frac {2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}}{d^{2}}\)	\(62\)
risch	\(-\frac {2}{\left (4 a c -b^{2}\right ) d^{2} \left (2 c x +b \right )}+\left (\munderset {\textit {\_R} =\RootOf \left (1+\left (64 d^{4} a^{3} c^{3}-48 b^{2} d^{4} c^{2} a^{2}+12 b^{4} d^{4} c a -b^{6} d^{4}\right ) \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (384 c^{4} d^{4} a^{3}-288 b^{2} c^{3} d^{4} a^{2}+72 a \,b^{4} c^{2} d^{4}-6 b^{6} c \,d^{4}\right ) \textit {\_R}^{2}+4 c \right ) x +\left (192 b \,c^{3} d^{4} a^{3}-144 b^{3} d^{4} c^{2} a^{2}+36 a \,b^{5} c \,d^{4}-3 b^{7} d^{4}\right ) \textit {\_R}^{2}+\left (16 a^{2} c^{2} d^{2}-8 a \,b^{2} c \,d^{2}+b^{4} d^{2}\right ) \textit {\_R} +2 b \right )\right )\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/d^2*(-2/(4*a*c-b^2)/(2*c*x+b)-2/(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(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a),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. 118 vs. \(2 (57) = 114\).
time = 1.79, size = 256, normalized size = 4.20 \begin {gather*} \left [-\frac {\sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )} \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 \, b^{2} + 8 \, a c}{2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{2} x + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d^{2}}, -\frac {2 \, {\left (\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - b^{2} + 4 \, a c\right )}}{2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{2} x + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (56) = 112\).
time = 0.48, size = 240, normalized size = 3.93 \begin {gather*} - \frac {2}{4 a b c d^{2} - b^{3} d^{2} + x \left (8 a c^{2} d^{2} - 2 b^{2} c d^{2}\right )} + \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {- 16 a^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b}{2 c} \right )}}{d^{2}} - \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {16 a^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b}{2 c} \right )}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**2/(c*x**2+b*x+a),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (57) = 114\).
time = 1.19, size = 117, normalized size = 1.92 \begin {gather*} \frac {2 \, c^{2} d^{3}}{{\left (b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4}\right )} {\left (2 \, c d x + b d\right )}} - \frac {2 \, \arctan \left (-\frac {\frac {b^{2} d}{2 \, c d x + b d} - \frac {4 \, a c d}{2 \, c d x + b d}}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.50, size = 115, normalized size = 1.89 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {b^3\,d^2-4\,a\,b\,c\,d^2}{d^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {2\,c\,x\,\left (b^2\,d^2-4\,a\,c\,d^2\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {2}{\left (4\,a\,c-b^2\right )\,\left (b\,d^2+2\,c\,d^2\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)),x)

[Out]

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

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