### 3.1195 $$\int \frac{\sqrt{a+b x+c x^2}}{b d+2 c d x} \, dx$$

Optimal. Leaf size=83 $\frac{\sqrt{a+b x+c x^2}}{2 c d}-\frac{\sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{4 c^{3/2} d}$

[Out]

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

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Rubi [A]  time = 0.063362, antiderivative size = 83, 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 = {685, 688, 205} $\frac{\sqrt{a+b x+c x^2}}{2 c d}-\frac{\sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{4 c^{3/2} d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), 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] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

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

Mathematica [A]  time = 0.0540887, size = 80, normalized size = 0.96 $\frac{2 \sqrt{c} \sqrt{a+x (b+c x)}-\sqrt{b^2-4 a c} \tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+x (b+c x)}}{\sqrt{b^2-4 a c}}\right )}{4 c^{3/2} d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.212, size = 244, normalized size = 2.9 \begin{align*}{\frac{1}{4\,cd}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{a}{cd}\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}}}}}}+{\frac{{b}^{2}}{4\,{c}^{2}d}\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((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.61645, size = 410, normalized size = 4.94 \begin{align*} \left [\frac{\sqrt{-\frac{b^{2} - 4 \, a c}{c}} \log \left (-\frac{4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt{c x^{2} + b x + a} c \sqrt{-\frac{b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, \sqrt{c x^{2} + b x + a}}{8 \, c d}, \frac{\sqrt{\frac{b^{2} - 4 \, a c}{c}} \arctan \left (\frac{\sqrt{\frac{b^{2} - 4 \, a c}{c}}}{2 \, \sqrt{c x^{2} + b x + a}}\right ) + 2 \, \sqrt{c x^{2} + b x + a}}{4 \, c d}\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),x, algorithm="fricas")

[Out]

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

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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 c x}\, dx}{d} \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),x)

[Out]

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

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Giac [A]  time = 1.16911, size = 131, normalized size = 1.58 \begin{align*} -\frac{{\left (b^{2} - 4 \, a c\right )} \arctan \left (-\frac{2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} c + b \sqrt{c}}{\sqrt{b^{2} c - 4 \, a c^{2}}}\right )}{2 \, \sqrt{b^{2} c - 4 \, a c^{2}} c d} + \frac{\sqrt{c x^{2} + b x + a}}{2 \, c d} \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),x, algorithm="giac")

[Out]

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