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

Optimal. Leaf size=165 \[ \frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}-\frac {b^2 (b+2 c x) \sqrt {b x+c x^2} (7 b B-10 A c)}{128 c^4}+\frac {b \left (b x+c x^2\right )^{3/2} (7 b B-10 A c)}{48 c^3}-\frac {x \left (b x+c x^2\right )^{3/2} (7 b B-10 A c)}{40 c^2}+\frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \]

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Rubi [A]  time = 0.16, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \begin {gather*} -\frac {b^2 (b+2 c x) \sqrt {b x+c x^2} (7 b B-10 A c)}{128 c^4}+\frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}+\frac {b \left (b x+c x^2\right )^{3/2} (7 b B-10 A c)}{48 c^3}-\frac {x \left (b x+c x^2\right )^{3/2} (7 b B-10 A c)}{40 c^2}+\frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(7*b*B - 10*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (b*(7*b*B - 10*A*c)*(b*x + c*x^2)^(3/2))/(48
*c^3) - ((7*b*B - 10*A*c)*x*(b*x + c*x^2)^(3/2))/(40*c^2) + (B*x^2*(b*x + c*x^2)^(3/2))/(5*c) + (b^4*(7*b*B -
10*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(9/2))

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]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 148, normalized size = 0.90 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{7/2} (7 b B-10 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (10 b^3 c (15 A+7 B x)-4 b^2 c^2 x (25 A+14 B x)+16 b c^3 x^2 (5 A+3 B x)+96 c^4 x^3 (5 A+4 B x)-105 b^4 B\right )\right )}{1920 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*B + 16*b*c^3*x^2*(5*A + 3*B*x) + 96*c^4*x^3*(5*A + 4*B*x) + 10*b^3*c*(15
*A + 7*B*x) - 4*b^2*c^2*x*(25*A + 14*B*x)) + (15*b^(7/2)*(7*b*B - 10*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/
(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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IntegrateAlgebraic [A]  time = 0.54, size = 153, normalized size = 0.93 \begin {gather*} \frac {\left (10 A b^4 c-7 b^5 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{9/2}}+\frac {\sqrt {b x+c x^2} \left (150 A b^3 c-100 A b^2 c^2 x+80 A b c^3 x^2+480 A c^4 x^3-105 b^4 B+70 b^3 B c x-56 b^2 B c^2 x^2+48 b B c^3 x^3+384 B c^4 x^4\right )}{1920 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-105*b^4*B + 150*A*b^3*c + 70*b^3*B*c*x - 100*A*b^2*c^2*x - 56*b^2*B*c^2*x^2 + 80*A*b*c^3*
x^2 + 48*b*B*c^3*x^3 + 480*A*c^4*x^3 + 384*B*c^4*x^4))/(1920*c^4) + ((-7*b^5*B + 10*A*b^4*c)*Log[b + 2*c*x - 2
*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(9/2))

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fricas [A]  time = 0.42, size = 302, normalized size = 1.83 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*x^4 -
105*B*b^4*c + 150*A*b^3*c^2 + 48*(B*b*c^4 + 10*A*c^5)*x^3 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^2 + 10*(7*B*b^3*c^2
- 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*
x)*sqrt(-c)/(c*x)) - (384*B*c^5*x^4 - 105*B*b^4*c + 150*A*b^3*c^2 + 48*(B*b*c^4 + 10*A*c^5)*x^3 - 8*(7*B*b^2*c
^3 - 10*A*b*c^4)*x^2 + 10*(7*B*b^3*c^2 - 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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giac [A]  time = 0.24, size = 160, normalized size = 0.97 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x + \frac {B b c^{3} + 10 \, A c^{4}}{c^{4}}\right )} x - \frac {7 \, B b^{2} c^{2} - 10 \, A b c^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (7 \, B b^{3} c - 10 \, A b^{2} c^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (7 \, B b^{4} - 10 \, A b^{3} c\right )}}{c^{4}}\right )} - \frac {{\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*x + (B*b*c^3 + 10*A*c^4)/c^4)*x - (7*B*b^2*c^2 - 10*A*b*c^3)/c^4)*x + 5
*(7*B*b^3*c - 10*A*b^2*c^2)/c^4)*x - 15*(7*B*b^4 - 10*A*b^3*c)/c^4) - 1/256*(7*B*b^5 - 10*A*b^4*c)*log(abs(-2*
(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

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maple [A]  time = 0.05, size = 245, normalized size = 1.48 \begin {gather*} -\frac {5 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {7 B \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{2} x}{32 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{3} x}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,x^{2}}{5 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{3}}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A x}{4 c}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{4}}{128 c^{4}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b x}{40 c^{2}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b}{24 c^{2}}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2}}{48 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*x^2*(c*x^2+b*x)^(3/2)/c-7/40*B*b/c^2*x*(c*x^2+b*x)^(3/2)+7/48*B*b^2/c^3*(c*x^2+b*x)^(3/2)-7/64*B*b^3/c^3
*(c*x^2+b*x)^(1/2)*x-7/128*B*b^4/c^4*(c*x^2+b*x)^(1/2)+7/256*B*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^
(1/2))+1/4*A*x*(c*x^2+b*x)^(3/2)/c-5/24*A*b/c^2*(c*x^2+b*x)^(3/2)+5/32*A*b^2/c^2*(c*x^2+b*x)^(1/2)*x+5/64*A*b^
3/c^3*(c*x^2+b*x)^(1/2)-5/128*A*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.86, size = 242, normalized size = 1.47 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x^{2}}{5 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x}{40 \, c^{2}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x}{4 \, c} + \frac {7 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{48 \, c^{3}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{24 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c*x^2 + b*x)^(3/2)*B*x^2/c - 7/64*sqrt(c*x^2 + b*x)*B*b^3*x/c^3 - 7/40*(c*x^2 + b*x)^(3/2)*B*b*x/c^2 + 5/
32*sqrt(c*x^2 + b*x)*A*b^2*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*A*x/c + 7/256*B*b^5*log(2*c*x + b + 2*sqrt(c*x^2 +
b*x)*sqrt(c))/c^(9/2) - 5/128*A*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 7/128*sqrt(c*x^2 +
b*x)*B*b^4/c^4 + 7/48*(c*x^2 + b*x)^(3/2)*B*b^2/c^3 + 5/64*sqrt(c*x^2 + b*x)*A*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3
/2)*A*b/c^2

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mupad [B]  time = 1.48, size = 215, normalized size = 1.30 \begin {gather*} \frac {A\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,A\,b\,\left (\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}\right )}{8\,c}-\frac {7\,B\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\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}\right )}{8\,c}\right )}{10\,c}+\frac {B\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*x*(b*x + c*x^2)^(3/2))/(4*c) - (5*A*b*((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)))/(8*c) - (7*B*b*((x*(b*x + c*x^2)^(3/2))/(4*c)
- (5*b*((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)))/(8*c)))/(10*c) + (B*x^2*(b*x + c*x^2)^(3/2))/(5*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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