\(\int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx\) [1195]

Optimal result

Integrand size = 26, antiderivative size = 83 \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=\frac {\sqrt {a+b x+c x^2}}{2 c d}-\frac {\sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{4 c^{3/2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {699, 702, 211} \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=\frac {\sqrt {a+b x+c x^2}}{2 c d}-\frac {\sqrt {b^2-4 a c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{4 c^{3/2} d} \]

[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 211

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

Rule 699

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 702

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)}+\sqrt {b^2-4 a c} \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{2 c^{3/2} d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {\sqrt {c \,x^{2}+b x +a}-\frac {\left (4 a c -b^{2}\right ) \operatorname {arctanh}\left (\frac {2 c \sqrt {c \,x^{2}+b x +a}}{\sqrt {c \left (4 a c -b^{2}\right )}}\right )}{2 \sqrt {c \left (4 a c -b^{2}\right )}}}{2 c d}\)	\(78\)
risch	\(\frac {\sqrt {c \,x^{2}+b x +a}}{2 c d}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{4 c^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, d}\)	\(132\)
default	\(\frac {\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}}{2 d c}\)	\(149\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.19 \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=\left [\frac {\sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, \sqrt {c x^{2} + b x + a}}{8 \, c d}, \frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) + 2 \, \sqrt {c x^{2} + b x + a}}{4 \, c d}\right ] \]

[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)]

Sympy [F]

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

[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

Maxima [F(-2)]

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

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d),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 [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=-\frac {{\left (b^{2} - 4 \, a c\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{2 \, \sqrt {b^{2} c - 4 \, a c^{2}} c d} + \frac {\sqrt {c x^{2} + b x + a}}{2 \, c d} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{b\,d+2\,c\,d\,x} \,d x \]

[In]

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

[Out]

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