3.12.97 \(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^3} \, dx\) [1197]

Optimal. Leaf size=91 \[ -\frac {\sqrt {a+b x+c x^2}}{4 c d^3 (b+2 c x)^2}+\frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{8 c^{3/2} \sqrt {b^2-4 a c} d^3} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 91, 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 = {698, 702, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{8 c^{3/2} d^3 \sqrt {b^2-4 a c}}-\frac {\sqrt {a+b x+c x^2}}{4 c d^3 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^3,x]

[Out]

-1/4*Sqrt[a + b*x + c*x^2]/(c*d^3*(b + 2*c*x)^2) + ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]]
/(8*c^(3/2)*Sqrt[b^2 - 4*a*c]*d^3)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 698

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]

Rule 702

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^3} \, dx &=-\frac {\sqrt {a+b x+c x^2}}{4 c d^3 (b+2 c x)^2}+\frac {\int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{8 c d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{4 c d^3 (b+2 c x)^2}+\frac {\text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{2 d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{4 c d^3 (b+2 c x)^2}+\frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{8 c^{3/2} \sqrt {b^2-4 a c} d^3}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 115, normalized size = 1.26 \begin {gather*} \frac {-\sqrt {c} \sqrt {b^2-4 a c} \sqrt {a+x (b+c x)}-(b+2 c x)^2 \tan ^{-1}\left (\frac {b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{4 c^{3/2} \sqrt {b^2-4 a c} d^3 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^3,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(221\) vs. \(2(75)=150\).
time = 0.80, size = 222, normalized size = 2.44

method	result	size
default	\(\frac {-\frac {2 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2}}+\frac {2 c^{2} \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}}{8 d^{3} c^{3}}\)	\(222\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

1/8/d^3/c^3*(-2/(4*a*c-b^2)*c/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+2*c^2/(4*a*c-b^2)*(1/2*(
4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)-1/2*(4*a*c-b^2)/c/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a
*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))))

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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)^3,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. 174 vs. \(2 (75) = 150\).
time = 1.94, size = 378, normalized size = 4.15 \begin {gather*} \left [-\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \sqrt {-b^{2} c + 4 \, a c^{2}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {-b^{2} c + 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{16 \, {\left (4 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} d^{3} x + {\left (b^{4} c^{2} - 4 \, a b^{2} c^{3}\right )} d^{3}\right )}}, -\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \sqrt {b^{2} c - 4 \, a c^{2}} \arctan \left (\frac {\sqrt {b^{2} c - 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{8 \, {\left (4 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} d^{3} x + {\left (b^{4} c^{2} - 4 \, a b^{2} c^{3}\right )} d^{3}\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)^3,x, algorithm="fricas")

[Out]

[-1/16*((4*c^2*x^2 + 4*b*c*x + b^2)*sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b
^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a
))/(4*(b^2*c^4 - 4*a*c^5)*d^3*x^2 + 4*(b^3*c^3 - 4*a*b*c^4)*d^3*x + (b^4*c^2 - 4*a*b^2*c^3)*d^3), -1/8*((4*c^2
*x^2 + 4*b*c*x + b^2)*sqrt(b^2*c - 4*a*c^2)*arctan(1/2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a)/(c^2*x^2 +
b*c*x + a*c)) + 2*(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(4*(b^2*c^4 - 4*a*c^5)*d^3*x^2 + 4*(b^3*c^3 - 4*a*b
*c^4)*d^3*x + (b^4*c^2 - 4*a*b^2*c^3)*d^3)]

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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^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx}{d^{3}} \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)**3,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x)/d**3

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{8,[3]%%%},[6,3,0,0]%%%}+%%%{%%{[%%%{-24,[2]%%%},0]:[
1,0,%%%{-1,

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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 )}^3} \,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)^3,x)

[Out]

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^3, x)

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