\(\int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx\) [1239]

Optimal result

Integrand size = 26, antiderivative size = 37 \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d^2 (b+2 c x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {696} \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {a+b x+c x^2}}{d^2 \left (b^2-4 a c\right ) (b+2 c x)} \]

[In]

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

[Out]

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

Rule 696

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*} \text {integral}& = \frac {2 \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) d^2 (b+2 c x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 2 in optimal.

Time = 0.74 (sec) , antiderivative size = 236, normalized size of antiderivative = 6.38 \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {\frac {2 a+x (b+2 c x)-2 \sqrt {a} \sqrt {a+x (b+c x)}}{2 \sqrt {a} c (b+2 c x) \left (2 a+b x-2 \sqrt {a} \sqrt {a+x (b+c x)}\right )}+\frac {2 \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b \sqrt {c} \sqrt {b^2-4 a c}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b \sqrt {c} \sqrt {-b^2+4 a c}}}{d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03

method	result	size
gosper	\(-\frac {2 \sqrt {c \,x^{2}+b x +a}}{\left (2 c x +b \right ) d^{2} \left (4 a c -b^{2}\right )}\)	\(38\)
trager	\(-\frac {2 \sqrt {c \,x^{2}+b x +a}}{\left (2 c x +b \right ) d^{2} \left (4 a c -b^{2}\right )}\)	\(38\)
default	\(-\frac {\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{d^{2} c \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}\)	\(61\)

[In]

int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {2 \, \sqrt {c x^{2} + b x + a}}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} x + {\left (b^{3} - 4 \, a b c\right )} d^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\frac {\int \frac {1}{b^{2} \sqrt {a + b x + c x^{2}} + 4 b c x \sqrt {a + b x + c x^{2}} + 4 c^{2} x^{2} \sqrt {a + b x + c x^{2}}}\, dx}{d^{2}} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (35) = 70\).

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=-\frac {\sqrt {c} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{b^{2} c d^{2} - 4 \, a c^{2} d^{2}} + \frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}{b^{2} c d^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c^{2} d^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(b d+2 c d x)^2 \sqrt {a+b x+c x^2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x+a}}{d^2\,\left (4\,a\,c-b^2\right )\,\left (b+2\,c\,x\right )} \]

[In]

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

[Out]

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