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

Optimal. Leaf size=127 \[ -\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}+\frac {b \sqrt {b x+c x^2} (5 b B-6 A c)}{8 c^3}-\frac {x \sqrt {b x+c x^2} (5 b B-6 A c)}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c} \]

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Rubi [A]  time = 0.12, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {794, 670, 640, 620, 206} \begin {gather*} -\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}+\frac {b \sqrt {b x+c x^2} (5 b B-6 A c)}{8 c^3}-\frac {x \sqrt {b x+c x^2} (5 b B-6 A c)}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*b*B - 6*A*c)*Sqrt[b*x + c*x^2])/(8*c^3) - ((5*b*B - 6*A*c)*x*Sqrt[b*x + c*x^2])/(12*c^2) + (B*x^2*Sqrt[b
*x + c*x^2])/(3*c) - (b^2*(5*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(7/2))

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 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 \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx &=\frac {B x^2 \sqrt {b x+c x^2}}{3 c}+\frac {\left (2 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{3 c}\\ &=-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}+\frac {(b (5 b B-6 A c)) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{8 c^2}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x+c x^2}}{8 c^3}-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {\left (b^2 (5 b B-6 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^3}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x+c x^2}}{8 c^3}-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {\left (b^2 (5 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^3}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x+c x^2}}{8 c^3}-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 116, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c} x (b+c x) \left (-2 b c (9 A+5 B x)+4 c^2 x (3 A+2 B x)+15 b^2 B\right )-3 b^{5/2} \sqrt {x} \sqrt {\frac {c x}{b}+1} (5 b B-6 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(b + c*x)*(15*b^2*B + 4*c^2*x*(3*A + 2*B*x) - 2*b*c*(9*A + 5*B*x)) - 3*b^(5/2)*(5*b*B - 6*A*c)*Sqrt
[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.55, size = 105, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-18 A b c+12 A c^2 x+15 b^2 B-10 b B c x+8 B c^2 x^2\right )}{24 c^3}+\frac {\left (5 b^3 B-6 A b^2 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{16 c^{7/2}} \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]*(15*b^2*B - 18*A*b*c - 10*b*B*c*x + 12*A*c^2*x + 8*B*c^2*x^2))/(24*c^3) + ((5*b^3*B - 6*A*b
^2*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(16*c^(7/2))

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fricas [A]  time = 0.45, size = 207, normalized size = 1.63 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\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]

[-1/48*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*x^2 + 15*B*b
^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(-c)*
arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*x^2 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)
*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.23, size = 109, normalized size = 0.86 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (\frac {4 \, B x}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{2} - 6 \, A b c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{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]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x/c - (5*B*b*c - 6*A*c^2)/c^3)*x + 3*(5*B*b^2 - 6*A*b*c)/c^3) + 1/16*(5*B*b^3 -
6*A*b^2*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

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maple [A]  time = 0.06, size = 163, normalized size = 1.28 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x}\, B \,x^{2}}{3 c}+\frac {3 A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}-\frac {5 B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}+\frac {\sqrt {c \,x^{2}+b x}\, A x}{2 c}-\frac {5 \sqrt {c \,x^{2}+b x}\, B b x}{12 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x}\, A b}{4 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2}}{8 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]

1/3*B*x^2*(c*x^2+b*x)^(1/2)/c-5/12*B*b/c^2*x*(c*x^2+b*x)^(1/2)+5/8*B*b^2/c^3*(c*x^2+b*x)^(1/2)-5/16*B*b^3/c^(7
/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*A*x/c*(c*x^2+b*x)^(1/2)-3/4*A*b/c^2*(c*x^2+b*x)^(1/2)+3/8*A*
b^2/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 = 0.92, size = 160, normalized size = 1.26 \begin {gather*} \frac {\sqrt {c x^{2} + b x} B x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} B b x}{12 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x}{2 \, c} - \frac {5 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{8 \, c^{3}} - \frac {3 \, \sqrt {c x^{2} + b x} A b}{4 \, 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]

1/3*sqrt(c*x^2 + b*x)*B*x^2/c - 5/12*sqrt(c*x^2 + b*x)*B*b*x/c^2 + 1/2*sqrt(c*x^2 + b*x)*A*x/c - 5/16*B*b^3*lo
g(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 3/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^
(5/2) + 5/8*sqrt(c*x^2 + b*x)*B*b^2/c^3 - 3/4*sqrt(c*x^2 + b*x)*A*b/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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