3.1290 \(\int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=101 \[ \frac{2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}} \]

[Out]

(2*Sqrt[d]*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (2*Sqrt[d]*ArcTanh
[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4)

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Rubi [A]  time = 0.0722369, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {694, 329, 298, 203, 206} \[ \frac{2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d]*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (2*Sqrt[d]*ArcTanh
[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4)

Rule 694

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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{b d+2 c d x}}{a+b x+c x^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{a-\frac{b^2}{4 c}+\frac{x^4}{4 c d^2}} \, dx,x,\sqrt{d (b+2 c x)}\right )}{c d}\\ &=-\left ((2 d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d-x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )\right )+(2 d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d+x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=\frac{2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}\\ \end{align*}

Mathematica [A]  time = 0.0386442, size = 87, normalized size = 0.86 \[ \frac{2 \sqrt{d (b+2 c x)} \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\sqrt [4]{b^2-4 a c} \sqrt{b+2 c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*(b + 2*c*x)]*(ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] - ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1
/4)]))/((b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x])

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Maple [B]  time = 0.193, size = 271, normalized size = 2.7 \begin{align*}{\frac{d\sqrt{2}}{2}\ln \left ({ \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+{d\sqrt{2}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}-{d\sqrt{2}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.00982, size = 518, normalized size = 5.13 \begin{align*} -4 \, \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2 \, c d x + b d} d \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} - \sqrt{2 \, c d^{3} x + b d^{3} +{\left (b^{2} - 4 \, a c\right )} d^{2} \sqrt{\frac{d^{2}}{b^{2} - 4 \, a c}}} \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}}{d^{2}}\right ) - \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} \log \left ({\left (b^{2} - 4 \, a c\right )} \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}} + \sqrt{2 \, c d x + b d} d\right ) + \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} \log \left (-{\left (b^{2} - 4 \, a c\right )} \left (\frac{d^{2}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}} + \sqrt{2 \, c d x + b d} d\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.54553, size = 65, normalized size = 0.64 \begin{align*} 4 d \operatorname{RootSum}{\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left ( t \mapsto t \log{\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt{b d + 2 c d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*d*RootSum(_t**4*(1024*a*c*d**2 - 256*b**2*d**2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d*
*2 + sqrt(b*d + 2*c*d*x))))

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Giac [B]  time = 1.14661, size = 531, normalized size = 5.26 \begin{align*} -\frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} + 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} d - 4 \, a c d} - \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} - 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} d - 4 \, a c d} + \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \log \left (2 \, c d x + b d + \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} d - 4 \, \sqrt{2} a c d} - \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} \log \left (2 \, c d x + b d - \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} d - 4 \, \sqrt{2} a c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*
x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2*d - 4*a*c*d) - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*arctan(-1/2*s
qrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2*d - 4
*a*c*d) + (-b^2*d^2 + 4*a*c*d^2)^(3/4)*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x +
b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2*d - 4*sqrt(2)*a*c*d) - (-b^2*d^2 + 4*a*c*d^2)^(3/4)*log(2*c*d
*x + b*d - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2
*d - 4*sqrt(2)*a*c*d)