3.9.38 \(\int x (A+B x) \sqrt {a+b x+c x^2} \, dx\)

Optimal. Leaf size=144 \[ -\frac {\left (b^2-4 a c\right ) \left (-4 a B c-8 A b c+5 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-8 A b c+5 b^2 B\right )}{64 c^3}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {779, 612, 621, 206} \begin {gather*} \frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-8 A b c+5 b^2 B\right )}{64 c^3}-\frac {\left (b^2-4 a c\right ) \left (-4 a B c-8 A b c+5 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*Sqrt[a + b*x + c*x^2],x]

[Out]

((5*b^2*B - 8*A*b*c - 4*a*B*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3) - ((5*b*B - 8*A*c - 6*B*c*x)*(a + b
*x + c*x^2)^(3/2))/(24*c^2) - ((b^2 - 4*a*c)*(5*b^2*B - 8*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt
[a + b*x + c*x^2])])/(128*c^(7/2))

Rule 206

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

Rule 612

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 621

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (A+B x) \sqrt {a+b x+c x^2} \, dx &=-\frac {(5 b B-8 A c-6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {\left (5 b^2 B-8 A b c-4 a B c\right ) \int \sqrt {a+b x+c x^2} \, dx}{16 c^2}\\ &=\frac {\left (5 b^2 B-8 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2 B-8 A b c-4 a B c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {\left (5 b^2 B-8 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2 B-8 A b c-4 a B c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{64 c^3}\\ &=\frac {\left (5 b^2 B-8 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^2-4 a c\right ) \left (5 b^2 B-8 A b c-4 a B c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 127, normalized size = 0.88 \begin {gather*} \frac {(a+x (b+c x))^{3/2} (8 A c-5 b B+6 B c x)-\frac {3 \left (-4 a B c-8 A b c+5 b^2 B\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{16 c^{3/2}}}{24 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(A + B*x)*Sqrt[a + b*x + c*x^2],x]

[Out]

((-5*b*B + 8*A*c + 6*B*c*x)*(a + x*(b + c*x))^(3/2) - (3*(5*b^2*B - 8*A*b*c - 4*a*B*c)*(-2*Sqrt[c]*(b + 2*c*x)
*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(16*c^(3/2)))/
(24*c^2)

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IntegrateAlgebraic [A]  time = 0.63, size = 177, normalized size = 1.23 \begin {gather*} \frac {\left (16 a^2 B c^2+32 a A b c^2-24 a b^2 B c-8 A b^3 c+5 b^4 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{128 c^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (64 a A c^2-52 a b B c+24 a B c^2 x-24 A b^2 c+16 A b c^2 x+64 A c^3 x^2+15 b^3 B-10 b^2 B c x+8 b B c^2 x^2+48 B c^3 x^3\right )}{192 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*(A + B*x)*Sqrt[a + b*x + c*x^2],x]

[Out]

(Sqrt[a + b*x + c*x^2]*(15*b^3*B - 24*A*b^2*c - 52*a*b*B*c + 64*a*A*c^2 - 10*b^2*B*c*x + 16*A*b*c^2*x + 24*a*B
*c^2*x + 8*b*B*c^2*x^2 + 64*A*c^3*x^2 + 48*B*c^3*x^3))/(192*c^3) + ((5*b^4*B - 8*A*b^3*c - 24*a*b^2*B*c + 32*a
*A*b*c^2 + 16*a^2*B*c^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(128*c^(7/2))

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fricas [A]  time = 0.48, size = 393, normalized size = 2.73 \begin {gather*} \left [\frac {3 \, {\left (5 \, B b^{4} + 16 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\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 (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 64 \, A a c^{3} - 4 \, {\left (13 \, B a b + 6 \, A b^{2}\right )} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 4 \, {\left (3 \, B a + 2 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} + 16 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\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 (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 64 \, A a c^{3} - 4 \, {\left (13 \, B a b + 6 \, A b^{2}\right )} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 4 \, {\left (3 \, B a + 2 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(5*B*b^4 + 16*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*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*(48*B*c^4*x^3 + 15*B*b^3*c + 64*A*a*c^3 - 4*(13*B
*a*b + 6*A*b^2)*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 4*(3*B*a + 2*A*b)*c^3)*x)*sqrt(c*x^2 + b*x
+ a))/c^4, 1/384*(3*(5*B*b^4 + 16*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*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*(48*B*c^4*x^3 + 15*B*b^3*c + 64*A*a*c^3 - 4*(
13*B*a*b + 6*A*b^2)*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 4*(3*B*a + 2*A*b)*c^3)*x)*sqrt(c*x^2 +
b*x + a))/c^4]

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giac [A]  time = 0.25, size = 178, normalized size = 1.24 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, B x + \frac {B b c^{2} + 8 \, A c^{3}}{c^{3}}\right )} x - \frac {5 \, B b^{2} c - 12 \, B a c^{2} - 8 \, A b c^{2}}{c^{3}}\right )} x + \frac {15 \, B b^{3} - 52 \, B a b c - 24 \, A b^{2} c + 64 \, A a c^{2}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{4} - 24 \, B a b^{2} c - 8 \, A b^{3} c + 16 \, B a^{2} c^{2} + 32 \, A a b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*x + (B*b*c^2 + 8*A*c^3)/c^3)*x - (5*B*b^2*c - 12*B*a*c^2 - 8*A*b*c^2)/c
^3)*x + (15*B*b^3 - 52*B*a*b*c - 24*A*b^2*c + 64*A*a*c^2)/c^3) + 1/128*(5*B*b^4 - 24*B*a*b^2*c - 8*A*b^3*c + 1
6*B*a^2*c^2 + 32*A*a*b*c^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.05, size = 352, normalized size = 2.44 \begin {gather*} -\frac {A a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\frac {A \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {B \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 B a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {5 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A b x}{4 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, B a x}{8 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{8 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B a b}{16 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3}}{64 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B x}{4 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{3 c}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{24 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*x*(c*x^2+b*x+a)^(3/2)/c-5/24*B*b/c^2*(c*x^2+b*x+a)^(3/2)+5/32*B*b^2/c^2*x*(c*x^2+b*x+a)^(1/2)+5/64*B*b^3
/c^3*(c*x^2+b*x+a)^(1/2)+3/16*B*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-5/128*B*b^4/c^(7/2)*
ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/8*B*a/c*x*(c*x^2+b*x+a)^(1/2)-1/16*B*a/c^2*(c*x^2+b*x+a)^(1/2)*b
-1/8*B*a^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*A*(c*x^2+b*x+a)^(3/2)/c-1/4*A*b/c*x*(c*x^2+
b*x+a)^(1/2)-1/8*A*b^2/c^2*(c*x^2+b*x+a)^(1/2)-1/4*A*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+1
/16*A*b^3/c^(5/2)*ln((c*x+1/2*b)/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(x*(B*x+A)*(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 details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 1.51, size = 256, normalized size = 1.78 \begin {gather*} \frac {A\,\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 {B\,a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\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}}\right )}{4\,c}-\frac {5\,B\,b\,\left (\frac {\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 {\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}\right )}{8\,c}+\frac {A\,\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}+\frac {B\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*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)) - (B*a*((x/2 + b/(4*c))*
(a + b*x + c*x^2)^(1/2) + (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))))/(4*
c) - (5*B*b*((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)) + ((8*c*(a +
c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) + (A*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(
a + b*x + c*x^2)^(1/2))/(24*c^2) + (B*x*(a + b*x + c*x^2)^(3/2))/(4*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(A + B*x)*sqrt(a + b*x + c*x**2), x)

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