3.1240 \(\int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=95 \[ \frac{\sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}+\frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{2 \sqrt{c} d^3 \left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.055632, antiderivative size = 95, 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 = {693, 688, 205} \[ \frac{\sqrt{a+b x+c x^2}}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}+\frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{2 \sqrt{c} d^3 \left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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

Mathematica [A]  time = 0.284009, size = 107, normalized size = 1.13 \[ \frac{\sqrt{a+x (b+c x)} \left (\frac{2 \left (b^2-4 a c\right )}{(b+2 c x)^2}+\frac{\tanh ^{-1}\left (2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}\right )}{\sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}\right )}{2 d^3 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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Maple [B]  time = 0.192, size = 174, normalized size = 1.8 \begin{align*} -{\frac{1}{4\,{c}^{2}{d}^{3} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}}+{\frac{1}{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}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.18764, size = 936, normalized size = 9.85 \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}}{4 \,{\left (4 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{2} + 4 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x +{\left (b^{6} c - 8 \, a b^{4} c^{2} + 16 \, a^{2} 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}}{2 \,{\left (4 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{2} + 4 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x +{\left (b^{6} c - 8 \, a b^{4} c^{2} + 16 \, a^{2} b^{2} c^{3}\right )} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/4*((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^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^2 + 4*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x + (b^6*c - 8
*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^3), -1/2*((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^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^2 + 4*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x + (b^6*c - 8*a*b^
4*c^2 + 16*a^2*b^2*c^3)*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{b^{3} \sqrt{a + b x + c x^{2}} + 6 b^{2} c x \sqrt{a + b x + c x^{2}} + 12 b c^{2} x^{2} \sqrt{a + b x + c x^{2}} + 8 c^{3} x^{3} \sqrt{a + b x + c x^{2}}}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(b**3*sqrt(a + b*x + c*x**2) + 6*b**2*c*x*sqrt(a + b*x + c*x**2) + 12*b*c**2*x**2*sqrt(a + b*x + c*
x**2) + 8*c**3*x**3*sqrt(a + b*x + c*x**2)), 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(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError