### 3.1291 $$\int \frac{1}{\sqrt{b d+2 c d x} (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=101 $-\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{3/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{3/4}}$

[Out]

(-2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(3/4)*Sqrt[d]) - (2*ArcTanh[Sqrt
[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(3/4)*Sqrt[d])

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Rubi [A]  time = 0.0774185, 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, 212, 206, 203} $-\frac{2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{3/4}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt{d} \left (b^2-4 a c\right )^{3/4}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(3/4)*Sqrt[d]) - (2*ArcTanh[Sqrt
[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(3/4)*Sqrt[d])

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 212

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0503212, size = 85, normalized size = 0.84 $-\frac{2 \sqrt{b+2 c x} \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt{d (b+2 c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*Sqrt[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)^(3/4)*Sqrt[d*(b + 2*c*x)])

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Maple [B]  time = 0.19, 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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}+{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(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)^(3/4)*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)^(3/4)*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)^(3/4)*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(1/(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 = 2.26901, size = 1033, normalized size = 10.23 \begin{align*} 4 \, \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{1}{4}} \arctan \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{2} \sqrt{\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}} + 2 \, c d x + b d} d \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{3}{4}} -{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{2 \, c d x + b d} d \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{3}{4}}\right ) - \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{1}{4}} \log \left ({\left (b^{2} - 4 \, a c\right )} d \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{1}{4}} + \sqrt{2 \, c d x + b d}\right ) + \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{1}{4}} \log \left (-{\left (b^{2} - 4 \, a c\right )} d \left (\frac{1}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}}\right )^{\frac{1}{4}} + \sqrt{2 \, c d x + b d}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4)*arctan((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(
(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^2*sqrt(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2)) + 2*c*d*x +
b*d)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(3/4) - (b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(
2*c*d*x + b*d)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(3/4)) - (1/((b^6 - 12*a*b^4*c + 4
8*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4)*log((b^2 - 4*a*c)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^
3)*d^2))^(1/4) + sqrt(2*c*d*x + b*d)) + (1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4)*log(-
(b^2 - 4*a*c)*d*(1/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2))^(1/4) + sqrt(2*c*d*x + b*d))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13272, size = 531, normalized size = 5.26 \begin{align*} -\frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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{1}{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{1}{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{1}{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(1/(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)^(1/4)*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)^(1/4)*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)^(1/4)*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)^(1/4)*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)