### 3.1196 $$\int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^2} \, dx$$

Optimal. Leaf size=75 $\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{3/2} d^2}-\frac{\sqrt{a+b x+c x^2}}{2 c d^2 (b+2 c x)}$

[Out]

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

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Rubi [A]  time = 0.0287825, antiderivative size = 75, 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, 621, 206} $\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{3/2} d^2}-\frac{\sqrt{a+b x+c x^2}}{2 c d^2 (b+2 c x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.299273, size = 114, normalized size = 1.52 $\frac{\sqrt{a+x (b+c x)} \left (\frac{\sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )}{\sqrt{4 a-\frac{b^2}{c}} \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}-\frac{2 \sqrt{c}}{b+2 c x}\right )}{4 c^{3/2} d^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.195, size = 291, normalized size = 3.9 \begin{align*} -{\frac{1}{c{d}^{2} \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 ) ^{-1}}+{\frac{x}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}+{\frac{b}{2\,c{d}^{2} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}+{\frac{a}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) }\ln \left ( \left ( x+{\frac{b}{2\,c}} \right ) \sqrt{c}+\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{2}}{4\,{d}^{2} \left ( 4\,ac-{b}^{2} \right ) }\ln \left ( \left ( x+{\frac{b}{2\,c}} \right ) \sqrt{c}+\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \right ){c}^{-{\frac{3}{2}}}} \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)^2,x)

[Out]

-1/c/d^2/(4*a*c-b^2)/(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+
1/4*(4*a*c-b^2)/c)^(1/2)*x+1/2/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b+1/c^(1/2)/d^2/(4*
a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*a-1/4/c^(3/2)/d^2/(4*a*c-b^2)*ln((x
+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.71201, size = 450, normalized size = 6. \begin{align*} \left [\frac{{\left (2 \, c x + b\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \, \sqrt{c x^{2} + b x + a} c}{8 \,{\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}}, -\frac{{\left (2 \, c x + b\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, \sqrt{c x^{2} + b x + a} c}{4 \,{\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\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)^2,x, algorithm="fricas")

[Out]

[1/8*((2*c*x + b)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c
) - 4*sqrt(c*x^2 + b*x + a)*c)/(2*c^3*d^2*x + b*c^2*d^2), -1/4*((2*c*x + b)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b
*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*sqrt(c*x^2 + b*x + a)*c)/(2*c^3*d^2*x + b*c^2*d^2)]

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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^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \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)**2,x)

[Out]

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

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Giac [B]  time = 1.19959, size = 261, normalized size = 3.48 \begin{align*} -\frac{1}{4} \, d^{2}{\left (\frac{{\left (\frac{c \arctan \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}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \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 )} \mathrm{sgn}\left (\frac{1}{2 \, c d x + b d}\right ) \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (d\right )}{c^{2} d^{4}{\left | c \right |}} - \frac{{\left (c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + \sqrt{-c} \sqrt{c}\right )} \mathrm{sgn}\left (\frac{1}{2 \, c d x + b d}\right ) \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (d\right )}{\sqrt{-c} c^{2} d^{4}{\left | c \right |}}\right )}{\left | c \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)^2,x, algorithm="giac")

[Out]

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