\(\int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\) [1290]

Optimal result

Integrand size = 26, antiderivative size = 101 \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\sqrt [4]{b^2-4 a c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {708, 335, 304, 209, 212} \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}} \]

[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 209

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

Rule 212

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

Rule 304

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 335

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 708

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*} \text {integral}& = \frac {\text {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 {\text {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) \text {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) \text {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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=-\frac {(1+i) \sqrt {d (b+2 c x)} \left (\arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(227\) vs. \(2(81)=162\).

Time = 2.84 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.26

method	result	size
derivativedivides	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\)	\(228\)
default	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\)	\(228\)
pseudoelliptic	\(\frac {d \sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\)	\(242\)

[In]

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

[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)))+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)-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))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=-\left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \log \left ({\left (b^{2} - 4 \, a c\right )} \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d} d\right ) + \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \log \left (-{\left (b^{2} - 4 \, a c\right )} \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d} d\right ) + i \, \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \log \left ({\left (i \, b^{2} - 4 i \, a c\right )} \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d} d\right ) - i \, \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \log \left ({\left (-i \, b^{2} + 4 i \, a c\right )} \left (\frac {d^{2}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} + \sqrt {2 \, c d x + b d} d\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=\int \frac {\sqrt {d \left (b + 2 c x\right )}}{a + b x + c x^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((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 deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (81) = 162\).

Time = 0.29 (sec) , antiderivative size = 393, normalized size of antiderivative = 3.89 \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} d - 4 \, a c d} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{2} d - 4 \, a c d} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} d - 4 \, \sqrt {2} a c d} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{2} d - 4 \, \sqrt {2} a c d} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx=\frac {2\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}}-\frac {2\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}} \]

[In]

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

[Out]

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