3.1.3 \(\int x \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=81 \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}}-\frac {b (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {\left (b x+c x^2\right )^{3/2}}{3 c} \]

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {640, 612, 620, 206} \begin {gather*} \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}}-\frac {b (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {\left (b x+c x^2\right )^{3/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[b*x + c*x^2],x]

[Out]

-(b*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (b*x + c*x^2)^(3/2)/(3*c) + (b^3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c
*x^2]])/(8*c^(5/2))

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 620

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

Rule 640

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

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Mathematica [A]  time = 0.11, size = 87, normalized size = 1.07 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {3 b^{5/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (-3 b^2+2 b c x+8 c^2 x^2\right )\right )}{24 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^2 + 2*b*c*x + 8*c^2*x^2) + (3*b^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(S
qrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 85, normalized size = 1.05 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-3 b^2+2 b c x+8 c^2 x^2\right )}{24 c^2}-\frac {b^3 \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[b*x + c*x^2]*(-3*b^2 + 2*b*c*x + 8*c^2*x^2))/(24*c^2) - (b^3*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[b*x +
c*x^2]])/(16*c^(5/2))

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fricas [A]  time = 0.42, size = 148, normalized size = 1.83 \begin {gather*} \left [\frac {3 \, b^{3} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{48 \, c^{3}}, -\frac {3 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{24 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*b^3*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c)*sqrt(c
*x^2 + b*x))/c^3, -1/24*(3*b^3*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*c^3*x^2 + 2*b*c^2*x - 3*
b^2*c)*sqrt(c*x^2 + b*x))/c^3]

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giac [A]  time = 0.20, size = 73, normalized size = 0.90 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, x + \frac {b}{c}\right )} x - \frac {3 \, b^{2}}{c^{2}}\right )} - \frac {b^{3} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\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(x*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*x + b/c)*x - 3*b^2/c^2) - 1/16*b^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqr
t(c) - b))/c^(5/2)

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maple [A]  time = 0.04, size = 87, normalized size = 1.07 \begin {gather*} \frac {b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}-\frac {\sqrt {c \,x^{2}+b x}\, b x}{4 c}-\frac {\sqrt {c \,x^{2}+b x}\, b^{2}}{8 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.19, size = 85, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {c x^{2} + b x} b x}{4 \, c} + \frac {b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {\sqrt {c x^{2} + b x} b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(c*x^2 + b*x)*b*x/c + 1/16*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 1/8*sqrt(c*x^2
+ b*x)*b^2/c^2 + 1/3*(c*x^2 + b*x)^(3/2)/c

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mupad [B]  time = 0.09, size = 69, normalized size = 0.85 \begin {gather*} \frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x + c*x^2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(x*(b + c*x)), x)

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