\(\int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx\) [2376]

Optimal result

Integrand size = 20, antiderivative size = 68 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\frac {e \sqrt {a+b x+c x^2}}{c}+\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}} \]

[Out]

1/2*(-b*e+2*c*d)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)+e*(c*x^2+b*x+a)^(1/2)/c

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {654, 635, 212} \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}}+\frac {e \sqrt {a+b x+c x^2}}{c} \]

[In]

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

[Out]

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

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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 654

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

Rubi steps \begin{align*} \text {integral}& = \frac {e \sqrt {a+b x+c x^2}}{c}+\frac {(2 c d-b e) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c} \\ & = \frac {e \sqrt {a+b x+c x^2}}{c}+\frac {(2 c d-b e) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{c} \\ & = \frac {e \sqrt {a+b x+c x^2}}{c}+\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\frac {e \sqrt {a+x (b+c x)}}{c}+\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {e \sqrt {c \,x^{2}+b x +a}}{c}-\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\)	\(57\)
default	\(\frac {d \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+e \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(82\)

[In]

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

[Out]

e*(c*x^2+b*x+a)^(1/2)/c-1/2*(b*e-2*c*d)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.43 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {4 \, \sqrt {c x^{2} + b x + a} c e - {\left (2 \, c d - b e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right )}{4 \, c^{2}}, \frac {2 \, \sqrt {c x^{2} + b x + a} c e - {\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right )}{2 \, c^{2}}\right ] \]

[In]

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

[Out]

[1/4*(4*sqrt(c*x^2 + b*x + a)*c*e - (2*c*d - b*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x
+ a)*(2*c*x + b)*sqrt(c) - 4*a*c))/c^2, 1/2*(2*sqrt(c*x^2 + b*x + a)*c*e - (2*c*d - b*e)*sqrt(-c)*arctan(1/2*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)))/c^2]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (56) = 112\).

Time = 0.42 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.15 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (- \frac {b e}{2 c} + d\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \frac {e \sqrt {a + b x + c x^{2}}}{c} & \text {for}\: c \neq 0 \\\frac {2 d \sqrt {a + b x} + \frac {2 e \left (- a \sqrt {a + b x} + \frac {\left (a + b x\right )^{\frac {3}{2}}}{3}\right )}{b}}{b} & \text {for}\: b \neq 0 \\\frac {d x + \frac {e x^{2}}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((-b*e/(2*c) + d)*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/
(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + e*sqrt(a + b*x + c*x**2)/c, Ne(
c, 0)), ((2*d*sqrt(a + b*x) + 2*e*(-a*sqrt(a + b*x) + (a + b*x)**(3/2)/3)/b)/b, Ne(b, 0)), ((d*x + e*x**2/2)/s
qrt(a), True))

Maxima [F(-2)]

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

[In]

integrate((e*x+d)/(c*x^2+b*x+a)^(1/2),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 = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x + a} e}{c} - \frac {{\left (2 \, c d - b e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx=\frac {e\,\sqrt {c\,x^2+b\,x+a}}{c}+\frac {d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{\sqrt {c}}-\frac {b\,e\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{2\,c^{3/2}} \]

[In]

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

[Out]

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