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

Optimal. Leaf size=113 \[ -\frac{b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}+\frac{b (b+2 c x) \sqrt{b x+c x^2} (5 b B-8 A c)}{64 c^3}-\frac{\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0496581, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {779, 612, 620, 206} \[ -\frac{b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}+\frac{b (b+2 c x) \sqrt{b x+c x^2} (5 b B-8 A c)}{64 c^3}-\frac{\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

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

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]

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

Mathematica [A]  time = 0.237667, size = 128, normalized size = 1.13 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-2 b^2 c (12 A+5 B x)+8 b c^2 x (2 A+B x)+16 c^3 x^2 (4 A+3 B x)+15 b^3 B\right )-\frac{3 b^{5/2} (5 b B-8 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B + 8*b*c^2*x*(2*A + B*x) + 16*c^3*x^2*(4*A + 3*B*x) - 2*b^2*c*(12*A + 5*B
*x)) - (3*b^(5/2)*(5*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/
2))

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Maple [B]  time = 0.007, size = 201, normalized size = 1.8 \begin{align*}{\frac{Bx}{4\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,bB}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}Bx}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{3}B}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{4}B}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{A}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{Abx}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{A{b}^{2}}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{A{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*x*(c*x^2+b*x)^(3/2)/c-5/24*B*b/c^2*(c*x^2+b*x)^(3/2)+5/32*B*b^2/c^2*x*(c*x^2+b*x)^(1/2)+5/64*B*b^3/c^3*(
c*x^2+b*x)^(1/2)-5/128*B*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*A*(c*x^2+b*x)^(3/2)/c-1/4*A
*b/c*x*(c*x^2+b*x)^(1/2)-1/8*A*b^2/c^2*(c*x^2+b*x)^(1/2)+1/16*A*b^3/c^(5/2)*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*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03651, size = 581, normalized size = 5.14 \begin{align*} \left [-\frac{3 \,{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \,{\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \,{\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{384 \, c^{4}}, \frac{3 \,{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \,{\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \,{\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{192 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 + 15*B
*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/1
92*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*B*c^4*x^3 + 15*B*b^3*c - 2
4*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16533, size = 178, normalized size = 1.58 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\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 - 8 \, A b c^{2}}{c^{3}}\right )} x + \frac{3 \,{\left (5 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{3}}\right )} + \frac{{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x + (B*b*c^2 + 8*A*c^3)/c^3)*x - (5*B*b^2*c - 8*A*b*c^2)/c^3)*x + 3*(5*B*b^
3 - 8*A*b^2*c)/c^3) + 1/128*(5*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(
7/2)