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

Optimal. Leaf size=75 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{3/2} d^2}-\frac {\sqrt {a+b x+c x^2}}{2 c d^2 (b+2 c x)} \]

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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 = {684, 621, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{3/2} d^2}-\frac {\sqrt {a+b x+c x^2}}{2 c d^2 (b+2 c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 684

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 114, normalized size = 1.52 \begin {gather*} \frac {\sqrt {a+x (b+c x)} \left (\frac {\sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )}{\sqrt {4 a-\frac {b^2}{c}} \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}-\frac {2 \sqrt {c}}{b+2 c x}\right )}{4 c^{3/2} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 86, normalized size = 1.15 \begin {gather*} -\frac {\log \left (-2 c^{3/2} d^2 \sqrt {a+b x+c x^2}+b c d^2+2 c^2 d^2 x\right )}{4 c^{3/2} d^2}-\frac {\sqrt {a+b x+c x^2}}{2 c d^2 (b+2 c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sqrt[a + b*x + c*x^2]/(c*d^2*(b + 2*c*x)) - Log[b*c*d^2 + 2*c^2*d^2*x - 2*c^(3/2)*d^2*Sqrt[a + b*x + c*x^
2]]/(4*c^(3/2)*d^2)

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fricas [A]  time = 0.46, size = 189, normalized size = 2.52 \begin {gather*} \left [\frac {{\left (2 \, c x + b\right )} \sqrt {c} \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 \, \sqrt {c x^{2} + b x + a} c}{8 \, {\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}}, -\frac {{\left (2 \, c x + b\right )} \sqrt {-c} \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 \, \sqrt {c x^{2} + b x + a} c}{4 \, {\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((2*c*x + b)*sqrt(c)*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*sqrt(c*x^2 + b*x + a)*c)/(2*c^3*d^2*x + b*c^2*d^2), -1/4*((2*c*x + b)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b
*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*sqrt(c*x^2 + b*x + a)*c)/(2*c^3*d^2*x + b*c^2*d^2)]

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giac [B]  time = 0.60, size = 215, normalized size = 2.87 \begin {gather*} -\frac {1}{4} \, d^{2} {\left (\frac {\arctan \left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)}{\sqrt {-c} c d^{4} {\left | c \right |}} - \frac {{\left (\sqrt {c} \arctan \left (\frac {\sqrt {c}}{\sqrt {-c}}\right ) + \sqrt {-c}\right )} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)}{\sqrt {-c} c^{\frac {3}{2}} d^{4} {\left | c \right |}} + \frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)}{c^{2} d^{4} {\left | c \right |}}\right )} {\left | c \right |} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 291, normalized size = 3.88 \begin {gather*} \frac {a \ln \left (\left (x +\frac {b}{2 c}\right ) \sqrt {c}+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {c}\, d^{2}}-\frac {b^{2} \ln \left (\left (x +\frac {b}{2 c}\right ) \sqrt {c}+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{4 \left (4 a c -b^{2}\right ) c^{\frac {3}{2}} d^{2}}+\frac {\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, x}{\left (4 a c -b^{2}\right ) d^{2}}+\frac {\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, b}{2 \left (4 a c -b^{2}\right ) c \,d^{2}}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right ) c \,d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c/d^2/(4*a*c-b^2)/(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+
1/4*(4*a*c-b^2)/c)^(1/2)*x+1/2/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b+1/c^(1/2)/d^2/(4*
a*c-b^2)*ln(c^(1/2)*(x+1/2*b/c)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*a-1/4/c^(3/2)/d^2/(4*a*c-b^2)*ln(c^
(1/2)*(x+1/2*b/c)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b^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((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^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 {\sqrt {c\,x^2+b\,x+a}}{{\left (b\,d+2\,c\,d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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