### 3.1247 $$\int \frac{1}{(b d+2 c d x)^2 (a+b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=76 $-\frac{16 c \sqrt{a+b x+c x^2}}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x)}-\frac{2}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}$

[Out]

-2/((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2]) - (16*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^2*(
b + 2*c*x))

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Rubi [A]  time = 0.0340286, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {687, 682} $-\frac{16 c \sqrt{a+b x+c x^2}}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x)}-\frac{2}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-2/((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2]) - (16*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^2*(
b + 2*c*x))

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 682

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt{a+b x+c x^2}}-\frac{(8 c) \int \frac{1}{(b d+2 c d x)^2 \sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt{a+b x+c x^2}}-\frac{16 c \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 0.0280347, size = 56, normalized size = 0.74 $-\frac{2 \left (4 c \left (a+2 c x^2\right )+b^2+8 b c x\right )}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 68, normalized size = 0.9 \begin{align*} -2\,{\frac{8\,{c}^{2}{x}^{2}+8\,bcx+4\,ac+{b}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ){d}^{2} \left ( 2\,cx+b \right ) \sqrt{c{x}^{2}+bx+a}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(8*c^2*x^2+8*b*c*x+4*a*c+b^2)/(2*c*x+b)/d^2/(16*a^2*c^2-8*a*b^2*c+b^4)/(c*x^2+b*x+a)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.74985, size = 327, normalized size = 4.3 \begin{align*} -\frac{2 \,{\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt{c x^{2} + b x + a}}{2 \,{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{2} x^{3} + 3 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{2} x^{2} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d^{2} x +{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)*sqrt(c*x^2 + b*x + a)/(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 +
3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c
+ 16*a^3*b*c^2)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a b^{2} \sqrt{a + b x + c x^{2}} + 4 a b c x \sqrt{a + b x + c x^{2}} + 4 a c^{2} x^{2} \sqrt{a + b x + c x^{2}} + b^{3} x \sqrt{a + b x + c x^{2}} + 5 b^{2} c x^{2} \sqrt{a + b x + c x^{2}} + 8 b c^{2} x^{3} \sqrt{a + b x + c x^{2}} + 4 c^{3} x^{4} \sqrt{a + b x + c x^{2}}}\, dx}{d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a*b**2*sqrt(a + b*x + c*x**2) + 4*a*b*c*x*sqrt(a + b*x + c*x**2) + 4*a*c**2*x**2*sqrt(a + b*x + c*
x**2) + b**3*x*sqrt(a + b*x + c*x**2) + 5*b**2*c*x**2*sqrt(a + b*x + c*x**2) + 8*b*c**2*x**3*sqrt(a + b*x + c*
x**2) + 4*c**3*x**4*sqrt(a + b*x + c*x**2)), x)/d**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (2 \, c d x + b d\right )}^{2}{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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