3.10.84 \(\int \frac {(b d+2 c d x)^2}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac {4 c d^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {d^2 (b+2 c x)}{a+b x+c x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 62, 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 = {686, 618, 206} \begin {gather*} -\frac {4 c d^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {d^2 (b+2 c x)}{a+b x+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 618

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 686

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 65, normalized size = 1.05 \begin {gather*} d^2 \left (\frac {4 c \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {-b-2 c x}{a+b x+c x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 309, normalized size = 4.98 \begin {gather*} \left [-\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2} - 2 \, {\left (c^{2} d^{2} x^{2} + b c d^{2} x + a c d^{2}\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 )}{a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x}, -\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2} + 4 \, {\left (c^{2} d^{2} x^{2} + b c d^{2} x + a c d^{2}\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 )}{a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 66, normalized size = 1.06 \begin {gather*} \frac {4 \, c d^{2} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c d^{2} x + b d^{2}}{c x^{2} + b x + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 77, normalized size = 1.24 \begin {gather*} -\frac {2 c \,d^{2} x}{c \,x^{2}+b x +a}+\frac {4 c \,d^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {b \,d^{2}}{c \,x^{2}+b x +a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*d^2/(c*x^2+b*x+a)*c*x-d^2/(c*x^2+b*x+a)*b+4*d^2*c/(4*a*c-b^2)^(1/2)*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((2*c*d*x+b*d)^2/(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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.44, size = 88, normalized size = 1.42 \begin {gather*} \frac {4\,c\,d^2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}}-\frac {b\,d^2}{c\,x^2+b\,x+a}-\frac {2\,c\,d^2\,x}{c\,x^2+b\,x+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.65, size = 211, normalized size = 3.40 \begin {gather*} - 2 c d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {- 8 a c^{2} d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + 2 b^{2} c d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + 2 b c d^{2}}{4 c^{2} d^{2}} \right )} + 2 c d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {8 a c^{2} d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} - 2 b^{2} c d^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + 2 b c d^{2}}{4 c^{2} d^{2}} \right )} + \frac {- b d^{2} - 2 c d^{2} x}{a + b x + c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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