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

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 2.23, size = 155, normalized size = 2.04 \[ -\frac {2 \, {\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{2} x^{3} + 3 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{2} x^{2} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d^{2} x + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d^{2}} \]

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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giac [B]  time = 0.32, size = 266, normalized size = 3.50 \[ -\frac {4 \, {\left (\frac {c^{2} d^{4} {\left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} c}{b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)} + \frac {c^{2}}{{\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)\right )} \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}\right )} {\left | c \right |}}{b^{2} c^{3} d^{4} - 4 \, a c^{4} d^{4}} - \frac {2 \, \sqrt {c} {\left | c \right |} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{d^{2} {\left | c \right |}} \]

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]

-4*(c^2*d^4*(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)*c/(b^2*sgn(1/(2*c*d*x + b*
d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)) + c^2/((b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(
d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))*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)))*abs(c)/(b^2*c^3*d^4 - 4*a*c^4*d^4) - 2*sqrt(c)*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)/(b^4 -
8*a*b^2*c + 16*a^2*c^2))/(d^2*abs(c))

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maple [A]  time = 0.04, size = 68, normalized size = 0.89 \[ -\frac {2 \left (8 c^{2} x^{2}+8 b c x +4 a c +b^{2}\right )}{\left (2 c x +b \right ) \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 positive, negative or zero?

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mupad [B]  time = 0.72, size = 188, normalized size = 2.47 \[ -\frac {\left (\frac {2\,\left (4\,c^2+\frac {64\,b^2\,c^5}{32\,a\,c^4-8\,b^2\,c^3}\right )}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}+\frac {512\,c^7\,x^2}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )\,\left (32\,a\,c^4-8\,b^2\,c^3\right )}+\frac {512\,b\,c^6\,x}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )\,\left (32\,a\,c^4-8\,b^2\,c^3\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{x\,\left (b^2+2\,a\,c\right )+a\,b+2\,c^2\,x^3+3\,b\,c\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((2*(4*c^2 + (64*b^2*c^5)/(32*a*c^4 - 8*b^2*c^3)))/(d^2*(16*a*c^3 - 4*b^2*c^2)) + (512*c^7*x^2)/(d^2*(16*a*c
^3 - 4*b^2*c^2)*(32*a*c^4 - 8*b^2*c^3)) + (512*b*c^6*x)/(d^2*(16*a*c^3 - 4*b^2*c^2)*(32*a*c^4 - 8*b^2*c^3)))*(
a + b*x + c*x^2)^(1/2))/(x*(2*a*c + b^2) + a*b + 2*c^2*x^3 + 3*b*c*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{a b^{2} \sqrt {a + b x + c x^{2}} + 4 a b c x \sqrt {a + b x + c x^{2}} + 4 a c^{2} x^{2} \sqrt {a + b x + c x^{2}} + b^{3} x \sqrt {a + b x + c x^{2}} + 5 b^{2} c x^{2} \sqrt {a + b x + c x^{2}} + 8 b c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 4 c^{3} x^{4} \sqrt {a + b x + c x^{2}}}\, dx}{d^{2}} \]

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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