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

Optimal. Leaf size=75 \[ \frac {d^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+d^2 (b+2 c x) \sqrt {a+b x+c x^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {692, 621, 206} \begin {gather*} \frac {d^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+d^2 (b+2 c x) \sqrt {a+b x+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2] + ((b^2 - 4*a*c)*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2
])])/(2*Sqrt[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 621

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

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 71, normalized size = 0.95 \begin {gather*} d^2 \left (\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 \sqrt {c}}+(b+2 c x) \sqrt {a+x (b+c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 82, normalized size = 1.09 \begin {gather*} \frac {\left (4 a c d^2-b^2 d^2\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{2 \sqrt {c}}+\sqrt {a+b x+c x^2} \left (b d^2+2 c d^2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 195, normalized size = 2.60 \begin {gather*} \left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} d^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, c^{2} d^{2} x + b c d^{2}\right )} \sqrt {c x^{2} + b x + a}}{4 \, c}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (2 \, c^{2} d^{2} x + b c d^{2}\right )} \sqrt {c x^{2} + b x + a}}{2 \, c}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((b^2 - 4*a*c)*sqrt(c)*d^2*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c)
- 4*a*c) - 4*(2*c^2*d^2*x + b*c*d^2)*sqrt(c*x^2 + b*x + a))/c, -1/2*((b^2 - 4*a*c)*sqrt(-c)*d^2*arctan(1/2*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^2*d^2*x + b*c*d^2)*sqrt(c*x^2 + b*x
+ a))/c]

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giac [A]  time = 0.25, size = 78, normalized size = 1.04 \begin {gather*} {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a} - \frac {{\left (b^{2} d^{2} - 4 \, a c d^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} \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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 108, normalized size = 1.44 \begin {gather*} -2 a \sqrt {c}\, d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+\frac {b^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}+2 \sqrt {c \,x^{2}+b x +a}\, c \,d^{2} x +\sqrt {c \,x^{2}+b x +a}\, b \,d^{2} \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)^(1/2),x)

[Out]

2*d^2*c*x*(c*x^2+b*x+a)^(1/2)+d^2*b*(c*x^2+b*x+a)^(1/2)+1/2*d^2*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)/c^(1/2)-2*d^2*c^(1/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(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)^(1/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \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)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int \frac {b^{2}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 c^{2} x^{2}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 b c x}{\sqrt {a + b x + c x^{2}}}\, dx\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)**(1/2),x)

[Out]

d**2*(Integral(b**2/sqrt(a + b*x + c*x**2), x) + Integral(4*c**2*x**2/sqrt(a + b*x + c*x**2), x) + Integral(4*
b*c*x/sqrt(a + b*x + c*x**2), x))

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