\(\int (d+e x) \sqrt {a+b x+c x^2} \, dx\) [2335]

Optimal result

Integrand size = 20, antiderivative size = 115 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {(2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}} \]

[Out]

1/3*e*(c*x^2+b*x+a)^(3/2)/c-1/16*(-4*a*c+b^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^(5/2)+1/8*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 635, 212} \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e)}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c} \]

[In]

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

[Out]

((2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2) + (e*(a + b*x + c*x^2)^(3/2))/(3*c) - ((b^2 - 4*a*c)
*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2))

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 626

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

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 \left (a+b x+c x^2\right )^{3/2}}{3 c}+\frac {(2 c d-b e) \int \sqrt {a+b x+c x^2} \, dx}{2 c} \\ & = \frac {(2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2} \\ & = \frac {(2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e)\right ) \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 )}{8 c^2} \\ & = \frac {(2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (-3 b^2 e+2 b c (3 d+e x)+4 c (2 a e+c x (3 d+2 e x))\right )+3 \left (b^2-4 a c\right ) (-2 c d+b e) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{24 c^{5/2}} \]

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-3*b^2*e + 2*b*c*(3*d + e*x) + 4*c*(2*a*e + c*x*(3*d + 2*e*x))) + 3*(b^2 - 4*a
*c)*(-2*c*d + b*e)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/(24*c^(5/2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\left (8 e \,c^{2} x^{2}+2 b c e x +12 c^{2} d x +8 a c e -3 b^{2} e +6 b c d \right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{2}}-\frac {\left (4 a b c e -8 a \,c^{2} d -b^{3} e +2 b^{2} c d \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}\)	\(115\)
default	\(d \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\)	\(158\)

[In]

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

[Out]

1/24*(8*c^2*e*x^2+2*b*c*e*x+12*c^2*d*x+8*a*c*e-3*b^2*e+6*b*c*d)*(c*x^2+b*x+a)^(1/2)/c^2-1/16*(4*a*b*c*e-8*a*c^
2*d-b^3*e+2*b^2*c*d)/c^(5/2)*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 = 297, normalized size of antiderivative = 2.58 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=\left [\frac {3 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d - {\left (b^{3} - 4 \, a b c\right )} e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} e x^{2} + 6 \, b c^{2} d - {\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e + 2 \, {\left (6 \, c^{3} d + b c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{3}}, \frac {3 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d - {\left (b^{3} - 4 \, a b c\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} e x^{2} + 6 \, b c^{2} d - {\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e + 2 \, {\left (6 \, c^{3} d + b c^{2} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/96*(3*(2*(b^2*c - 4*a*c^2)*d - (b^3 - 4*a*b*c)*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) + 4*(8*c^3*e*x^2 + 6*b*c^2*d - (3*b^2*c - 8*a*c^2)*e + 2*(6*c^3*d + b*c^2
*e)*x)*sqrt(c*x^2 + b*x + a))/c^3, 1/48*(3*(2*(b^2*c - 4*a*c^2)*d - (b^3 - 4*a*b*c)*e)*sqrt(-c)*arctan(1/2*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(8*c^3*e*x^2 + 6*b*c^2*d - (3*b^2*c - 8*a
*c^2)*e + 2*(6*c^3*d + b*c^2*e)*x)*sqrt(c*x^2 + b*x + a))/c^3]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (102) = 204\).

Time = 0.52 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.95 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {e x^{2}}{3} + \frac {x \left (\frac {b e}{6} + c d\right )}{2 c} + \frac {\frac {a e}{3} + b d - \frac {3 b \left (\frac {b e}{6} + c d\right )}{4 c}}{c}\right ) + \left (a d - \frac {a \left (\frac {b e}{6} + c d\right )}{2 c} - \frac {b \left (\frac {a e}{3} + b d - \frac {3 b \left (\frac {b e}{6} + c d\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {e \left (a + b x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- a e + b d\right )}{3 b}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + b*x + c*x**2)*(e*x**2/3 + x*(b*e/6 + c*d)/(2*c) + (a*e/3 + b*d - 3*b*(b*e/6 + c*d)/(4*c))/
c) + (a*d - a*(b*e/6 + c*d)/(2*c) - b*(a*e/3 + b*d - 3*b*(b*e/6 + c*d)/(4*c))/(2*c))*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)), Ne(c, 0)), (2*(e*(a + b*x)**(5/2)/(5*b) + (a + b*x)**(3/2)*(-a*e + b*d)/(3*b))/b, Ne(
b, 0)), (sqrt(a)*(d*x + e*x**2/2), True))

Maxima [F(-2)]

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

[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 = 121, normalized size of antiderivative = 1.05 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, e x + \frac {6 \, c^{2} d + b c e}{c^{2}}\right )} x + \frac {6 \, b c d - 3 \, b^{2} e + 8 \, a c e}{c^{2}}\right )} + \frac {{\left (2 \, b^{2} c d - 8 \, a c^{2} d - b^{3} e + 4 \, a b c e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*e*x + (6*c^2*d + b*c*e)/c^2)*x + (6*b*c*d - 3*b^2*e + 8*a*c*e)/c^2) + 1/16*(2
*b^2*c*d - 8*a*c^2*d - b^3*e + 4*a*b*c*e)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.26 \[ \int (d+e x) \sqrt {a+b x+c x^2} \, dx=d\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2} \]

[In]

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

[Out]

d*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (d*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4
))/(2*c^(3/2)) + (e*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + (e*(8
*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)