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

Optimal. Leaf size=200 \[ \frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2} (3 b B-4 A c)}{512 c^5}-\frac{7 b^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{192 c^4}-\frac{7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{11/2}}+\frac{7 b x \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{160 c^3}-\frac{x^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c} \]

[Out]

(7*b^3*(3*b*B - 4*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^5) - (7*b^2*(3*b*B - 4*A*c)*(b*x + c*x^2)^(3/2))/
(192*c^4) + (7*b*(3*b*B - 4*A*c)*x*(b*x + c*x^2)^(3/2))/(160*c^3) - ((3*b*B - 4*A*c)*x^2*(b*x + c*x^2)^(3/2))/
(20*c^2) + (B*x^3*(b*x + c*x^2)^(3/2))/(6*c) - (7*b^5*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(512*c^(11/2))

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Rubi [A]  time = 0.191263, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {794, 670, 640, 612, 620, 206} \[ \frac{7 b^3 (b+2 c x) \sqrt{b x+c x^2} (3 b B-4 A c)}{512 c^5}-\frac{7 b^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{192 c^4}-\frac{7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{11/2}}+\frac{7 b x \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{160 c^3}-\frac{x^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(7*b^3*(3*b*B - 4*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^5) - (7*b^2*(3*b*B - 4*A*c)*(b*x + c*x^2)^(3/2))/
(192*c^4) + (7*b*(3*b*B - 4*A*c)*x*(b*x + c*x^2)^(3/2))/(160*c^3) - ((3*b*B - 4*A*c)*x^2*(b*x + c*x^2)^(3/2))/
(20*c^2) + (B*x^3*(b*x + c*x^2)^(3/2))/(6*c) - (7*b^5*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(512*c^(11/2))

Rule 794

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

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 (A+B x) \sqrt{b x+c x^2} \, dx &=\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac{\left (3 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right ) \int x^3 \sqrt{b x+c x^2} \, dx}{6 c}\\ &=-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac{(7 b (3 b B-4 A c)) \int x^2 \sqrt{b x+c x^2} \, dx}{40 c^2}\\ &=\frac{7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (7 b^2 (3 b B-4 A c)\right ) \int x \sqrt{b x+c x^2} \, dx}{64 c^3}\\ &=-\frac{7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac{7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac{\left (7 b^3 (3 b B-4 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{128 c^4}\\ &=\frac{7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt{b x+c x^2}}{512 c^5}-\frac{7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac{7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (7 b^5 (3 b B-4 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{1024 c^5}\\ &=\frac{7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt{b x+c x^2}}{512 c^5}-\frac{7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac{7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (7 b^5 (3 b B-4 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{512 c^5}\\ &=\frac{7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt{b x+c x^2}}{512 c^5}-\frac{7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac{7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac{(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac{7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.298467, size = 166, normalized size = 0.83 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-16 b^2 c^3 x^2 (14 A+9 B x)+56 b^3 c^2 x (5 A+3 B x)-210 b^4 c (2 A+B x)+64 b c^4 x^3 (3 A+2 B x)+256 c^5 x^4 (6 A+5 B x)+315 b^5 B\right )-\frac{105 b^{9/2} (3 b B-4 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{7680 c^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(315*b^5*B - 210*b^4*c*(2*A + B*x) + 64*b*c^4*x^3*(3*A + 2*B*x) + 56*b^3*c^2*x*(5*
A + 3*B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 16*b^2*c^3*x^2*(14*A + 9*B*x)) - (105*b^(9/2)*(3*b*B - 4*A*c)*ArcSinh
[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(7680*c^(11/2))

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Maple [A]  time = 0.012, size = 291, normalized size = 1.5 \begin{align*}{\frac{B{x}^{3}}{6\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,Bb{x}^{2}}{20\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{21\,{b}^{2}Bx}{160\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{3}B}{64\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{21\,{b}^{4}Bx}{256\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{21\,B{b}^{5}}{512\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{21\,B{b}^{6}}{1024}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{11}{2}}}}+{\frac{A{x}^{2}}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,Abx}{40\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{7\,A{b}^{2}}{48\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,A{b}^{3}x}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{7\,A{b}^{4}}{128\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{7\,A{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*(B*x+A)*(c*x^2+b*x)^(1/2),x)

[Out]

1/6*B*x^3*(c*x^2+b*x)^(3/2)/c-3/20*B*b/c^2*x^2*(c*x^2+b*x)^(3/2)+21/160*B*b^2/c^3*x*(c*x^2+b*x)^(3/2)-7/64*B*b
^3/c^4*(c*x^2+b*x)^(3/2)+21/256*B*b^4/c^4*(c*x^2+b*x)^(1/2)*x+21/512*B*b^5/c^5*(c*x^2+b*x)^(1/2)-21/1024*B*b^6
/c^(11/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/5*A*x^2*(c*x^2+b*x)^(3/2)/c-7/40*A*b/c^2*x*(c*x^2+b*x)^(
3/2)+7/48*A*b^2/c^3*(c*x^2+b*x)^(3/2)-7/64*A*b^3/c^3*(c*x^2+b*x)^(1/2)*x-7/128*A*b^4/c^4*(c*x^2+b*x)^(1/2)+7/2
56*A*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*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98418, size = 811, normalized size = 4.05 \begin{align*} \left [-\frac{105 \,{\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \,{\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \,{\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \,{\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \,{\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{15360 \, c^{6}}, \frac{105 \,{\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \,{\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \,{\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \,{\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \,{\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{7680 \, c^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15360*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5
+ 315*B*b^5*c - 420*A*b^4*c^2 + 128*(B*b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*
c^3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^6, 1/7680*(105*(3*B*b^6 - 4*A*
b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (1280*B*c^6*x^5 + 315*B*b^5*c - 420*A*b^4*c^2 + 128
*(B*b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*c^3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^
4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^6]

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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 )} \left (A + B x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*sqrt(x*(b + c*x))*(A + B*x), x)

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Giac [A]  time = 1.15305, size = 254, normalized size = 1.27 \begin{align*} \frac{1}{7680} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \, B x + \frac{B b c^{4} + 12 \, A c^{5}}{c^{5}}\right )} x - \frac{3 \,{\left (3 \, B b^{2} c^{3} - 4 \, A b c^{4}\right )}}{c^{5}}\right )} x + \frac{7 \,{\left (3 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )}}{c^{5}}\right )} x - \frac{35 \,{\left (3 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )}}{c^{5}}\right )} x + \frac{105 \,{\left (3 \, B b^{5} - 4 \, A b^{4} c\right )}}{c^{5}}\right )} + \frac{7 \,{\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*x + (B*b*c^4 + 12*A*c^5)/c^5)*x - 3*(3*B*b^2*c^3 - 4*A*b*c^4)/c^5)*
x + 7*(3*B*b^3*c^2 - 4*A*b^2*c^3)/c^5)*x - 35*(3*B*b^4*c - 4*A*b^3*c^2)/c^5)*x + 105*(3*B*b^5 - 4*A*b^4*c)/c^5
) + 7/1024*(3*B*b^6 - 4*A*b^5*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(11/2)