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

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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 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]

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 = 0.05, size = 85, normalized size = 0.84 \begin {gather*} -\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)}} \end {gather*}

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)])

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 0.28, size = 194, normalized size = 1.92 \begin {gather*} \frac {(1-i) \tan ^{-1}\left (\frac {-\frac {(1+i) c \sqrt {d} x}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {d}}{\sqrt [4]{b^2-4 a c}}+\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}}{\sqrt {b d+2 c d x}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{3/4}}-\frac {(1-i) \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{\sqrt {d} \left (\sqrt {b^2-4 a c}+i b+2 i c x\right )}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 447, normalized size = 4.43 \begin {gather*} 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 {gather*}

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))

________________________________________________________________________________________

giac [B]  time = 0.20, size = 393, normalized size = 3.89 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 271, normalized size = 2.68 \begin {gather*} -\frac {\sqrt {2}\, d \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, d \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}} \end {gather*}

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)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> 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 details)Is 4*a*c-b^2 positive, negative or zero?

________________________________________________________________________________________

mupad [B]  time = 0.54, size = 161, normalized size = 1.59 \begin {gather*} -\frac {2\,\mathrm {atan}\left (\frac {128\,d^{3/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {2\,\mathrm {atanh}\left (\frac {128\,d^{3/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {128\,b^2\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {512\,a\,c\,d^2}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*atan((128*d^(3/2)*(b*d + 2*c*d*x)^(1/2))/(((128*b^2*d^2)/(b^2 - 4*a*c)^(3/2) - (512*a*c*d^2)/(b^2 - 4*a*c
)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(d^(1/2)*(b^2 - 4*a*c)^(3/4)) - (2*atanh((128*d^(3/2)*(b*d + 2*c*d*x)^(1/2))/(
((128*b^2*d^2)/(b^2 - 4*a*c)^(3/2) - (512*a*c*d^2)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(d^(1/2)*(b^2 -
4*a*c)^(3/4))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )}\, dx \end {gather*}

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)

________________________________________________________________________________________