3.1197 \(\int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^3} \, dx\)

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

[Out]

-Sqrt[a + b*x + c*x^2]/(4*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)

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Rubi [A]  time = 0.051964, antiderivative size = 91, normalized size of antiderivative = 1., 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, 688, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{8 c^{3/2} d^3 \sqrt{b^2-4 a c}}-\frac{\sqrt{a+b x+c x^2}}{4 c d^3 (b+2 c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[a + b*x + c*x^2]/(4*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 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]

Rule 688

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]

Rule 205

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

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{\operatorname{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*}

Mathematica [A]  time = 0.201273, size = 103, normalized size = 1.13 \[ \frac{-\sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \tanh ^{-1}\left (2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}\right )-\frac{2 c (a+x (b+c x))}{(b+2 c x)^2}}{8 c^2 d^3 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.196, size = 340, normalized size = 3.7 \begin{align*} -{\frac{1}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}}+{\frac{1}{8\,c{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{a}{2\,c{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}}+{\frac{{b}^{2}}{8\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 3.07557, size = 806, normalized size = 8.86 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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