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

Optimal. Leaf size=115 \[ \frac {(2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 c}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}} \]

[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

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Rubi [A]
time = 0.03, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 635, 212} \begin {gather*} -\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}}+\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int (d+e x) \sqrt {a+b x+c x^2} \, dx &=\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*}

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Mathematica [A]
time = 0.38, size = 112, normalized size = 0.97 \begin {gather*} \frac {2 \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) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{48 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*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)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(48*c^(5/2))

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Maple [A]
time = 0.84, size = 158, normalized size = 1.37

method	result	size
default	\(e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+d \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )\)	\(158\)
risch	\(\frac {\left (8 c^{2} e \,x^{2}+2 b e x c +12 c^{2} d x +8 a c e -3 b^{2} e +6 b c d \right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{2}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a b e}{4 c^{\frac {3}{2}}}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a d}{2 \sqrt {c}}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) e \,b^{3}}{16 c^{\frac {5}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{2} d}{8 c^{\frac {3}{2}}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((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

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Fricas [A]
time = 2.38, size = 293, normalized size = 2.55 \begin {gather*} \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 (12 \, c^{3} d x + 6 \, b c^{2} d + {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} e\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 (12 \, c^{3} d x + 6 \, b c^{2} d + {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} e\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*(12*c^3*d*x + 6*b*c^2*d + (8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c + 8*a*c^2)
*e)*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*sqrt(c
*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(12*c^3*d*x + 6*b*c^2*d + (8*c^3*x^2 + 2*b*c
^2*x - 3*b^2*c + 8*a*c^2)*e)*sqrt(c*x^2 + b*x + a))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.59, size = 129, normalized size = 1.12 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, x e + \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x*e + (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)

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Mupad [B]
time = 1.21, size = 145, normalized size = 1.26 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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