3.1 \(\int x^3 \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}+\frac{7 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \]

[Out]

(-7*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (7*b^2*(b*x + c*x^2)^(3/2))/(48*c^3) - (7*b*x*(b*x + c*x^2)
^(3/2))/(40*c^2) + (x^2*(b*x + c*x^2)^(3/2))/(5*c) + (7*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(9/
2))

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Rubi [A]  time = 0.0597608, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {670, 640, 612, 620, 206} \[ -\frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}+\frac{7 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(-7*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (7*b^2*(b*x + c*x^2)^(3/2))/(48*c^3) - (7*b*x*(b*x + c*x^2)
^(3/2))/(40*c^2) + (x^2*(b*x + c*x^2)^(3/2))/(5*c) + (7*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(9/
2))

Rule 670

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

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]

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 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])

Rubi steps

\begin{align*} \int x^3 \sqrt{b x+c x^2} \, dx &=\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac{(7 b) \int x^2 \sqrt{b x+c x^2} \, dx}{10 c}\\ &=-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (7 b^2\right ) \int x \sqrt{b x+c x^2} \, dx}{16 c^2}\\ &=\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac{\left (7 b^3\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^3}\\ &=-\frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (7 b^5\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^4}\\ &=-\frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (7 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^4}\\ &=-\frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac{7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac{x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{7 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.154078, size = 109, normalized size = 0.83 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-56 b^2 c^2 x^2+70 b^3 c x-105 b^4+48 b c^3 x^3+384 c^4 x^4\right )+\frac{105 b^{9/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{1920 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4 + 70*b^3*c*x - 56*b^2*c^2*x^2 + 48*b*c^3*x^3 + 384*c^4*x^4) + (105*b^(9/
2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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Maple [A]  time = 0.047, size = 129, normalized size = 1. \begin{align*}{\frac{{x}^{2}}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,bx}{40\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{2}}{48\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{3}x}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{7\,{b}^{4}}{128\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{7\,{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*x^2+b*x)^(1/2),x)

[Out]

1/5*x^2*(c*x^2+b*x)^(3/2)/c-7/40*b*x*(c*x^2+b*x)^(3/2)/c^2+7/48*b^2*(c*x^2+b*x)^(3/2)/c^3-7/64*b^3/c^3*(c*x^2+
b*x)^(1/2)*x-7/128*b^4/c^4*(c*x^2+b*x)^(1/2)+7/256*b^5/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07286, size = 459, normalized size = 3.5 \begin{align*} \left [\frac{105 \, b^{5} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt{c x^{2} + b x}}{3840 \, c^{5}}, -\frac{105 \, b^{5} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt{c x^{2} + b x}}{1920 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/3840*(105*b^5*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2
*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(105*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x
)*sqrt(-c)/(c*x)) - (384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x)
)/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{x \left (b + c x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**3*sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.30265, size = 131, normalized size = 1. \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, x + \frac{b}{c}\right )} x - \frac{7 \, b^{2}}{c^{2}}\right )} x + \frac{35 \, b^{3}}{c^{3}}\right )} x - \frac{105 \, b^{4}}{c^{4}}\right )} - \frac{7 \, b^{5} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*x + b/c)*x - 7*b^2/c^2)*x + 35*b^3/c^3)*x - 105*b^4/c^4) - 7/256*b^5*log(
abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)