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

Optimal. Leaf size=162 \[ \frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}-\frac {5 b^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{64 c^4}+\frac {5 b x \sqrt {b x+c x^2} (7 b B-8 A c)}{96 c^3}-\frac {x^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c} \]

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Rubi [A]  time = 0.16, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {5 b^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{64 c^4}+\frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}+\frac {5 b x \sqrt {b x+c x^2} (7 b B-8 A c)}{96 c^3}-\frac {x^2 \sqrt {b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*(7*b*B - 8*A*c)*Sqrt[b*x + c*x^2])/(64*c^4) + (5*b*(7*b*B - 8*A*c)*x*Sqrt[b*x + c*x^2])/(96*c^3) - ((7
*b*B - 8*A*c)*x^2*Sqrt[b*x + c*x^2])/(24*c^2) + (B*x^3*Sqrt[b*x + c*x^2])/(4*c) + (5*b^3*(7*b*B - 8*A*c)*ArcTa
nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 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^3 (A+B x)}{\sqrt {b x+c x^2}} \, dx &=\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (3 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx}{4 c}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {(5 b (7 b B-8 A c)) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{48 c^2}\\ &=\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}-\frac {\left (5 b^2 (7 b B-8 A c)\right ) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{64 c^3}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (5 b^3 (7 b B-8 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^4}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (5 b^3 (7 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^4}\\ &=-\frac {5 b^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{64 c^4}+\frac {5 b (7 b B-8 A c) x \sqrt {b x+c x^2}}{96 c^3}-\frac {(7 b B-8 A c) x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {b x+c x^2}}{4 c}+\frac {5 b^3 (7 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 129, normalized size = 0.80 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{5/2} (7 b B-8 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^2 c (12 A+7 B x)-8 b c^2 x (10 A+7 B x)+16 c^3 x^2 (4 A+3 B x)-105 b^3 B\right )\right )}{192 c^{9/2}} \end {gather*}

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]*(-105*b^3*B + 16*c^3*x^2*(4*A + 3*B*x) - 8*b*c^2*x*(10*A + 7*B*x) + 10*b^2*c*(12*A
+ 7*B*x)) + (15*b^(5/2)*(7*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(19
2*c^(9/2))

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IntegrateAlgebraic [A]  time = 0.52, size = 129, normalized size = 0.80 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (120 A b^2 c-80 A b c^2 x+64 A c^3 x^2-105 b^3 B+70 b^2 B c x-56 b B c^2 x^2+48 B c^3 x^3\right )}{192 c^4}-\frac {5 \left (7 b^4 B-8 A b^3 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{128 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-105*b^3*B + 120*A*b^2*c + 70*b^2*B*c*x - 80*A*b*c^2*x - 56*b*B*c^2*x^2 + 64*A*c^3*x^2 + 4
8*B*c^3*x^3))/(192*c^4) - (5*(7*b^4*B - 8*A*b^3*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(128*c^(9/2))

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fricas [A]  time = 0.43, size = 256, normalized size = 1.58 \begin {gather*} \left [-\frac {15 \, {\left (7 \, 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} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{5}}, -\frac {15 \, {\left (7 \, 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} - 105 \, B b^{3} c + 120 \, A b^{2} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{5}}\right ] \end {gather*}

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]

[-1/384*(15*(7*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 - 105
*B*b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^
5, -1/192*(15*(7*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 - 105*B*
b^3*c + 120*A*b^2*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 10*(7*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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giac [A]  time = 0.27, size = 137, normalized size = 0.85 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (7 \, B b^{2} c - 8 \, A b c^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (7 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{4}}\right )} - \frac {5 \, {\left (7 \, 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 {9}{2}}} \end {gather*}

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]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x + 5*(7*B*b^2*c - 8*A*b*c^2)/c^4)*x - 15*
(7*B*b^3 - 8*A*b^2*c)/c^4) - 5/128*(7*B*b^4 - 8*A*b^3*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 = 209, normalized size = 1.29 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x}\, B \,x^{3}}{4 c}+\frac {\sqrt {c \,x^{2}+b x}\, A \,x^{2}}{3 c}-\frac {7 \sqrt {c \,x^{2}+b x}\, B b \,x^{2}}{24 c^{2}}-\frac {5 A \,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 {35 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {9}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x}\, A b x}{12 c^{2}}+\frac {35 \sqrt {c \,x^{2}+b x}\, B \,b^{2} x}{96 c^{3}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{2}}{8 c^{3}}-\frac {35 \sqrt {c \,x^{2}+b x}\, B \,b^{3}}{64 c^{4}} \end {gather*}

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]

1/4*B*x^3*(c*x^2+b*x)^(1/2)/c-7/24*B*b/c^2*x^2*(c*x^2+b*x)^(1/2)+35/96*B*b^2/c^3*x*(c*x^2+b*x)^(1/2)-35/64*B*b
^3/c^4*(c*x^2+b*x)^(1/2)+35/128*B*b^4/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*A*x^2/c*(c*x^2+b*x
)^(1/2)-5/12*A*b/c^2*x*(c*x^2+b*x)^(1/2)+5/8*A*b^2/c^3*(c*x^2+b*x)^(1/2)-5/16*A*b^3/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 = 0.87, size = 206, normalized size = 1.27 \begin {gather*} \frac {\sqrt {c x^{2} + b x} B x^{3}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b x^{2}}{24 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x^{2}}{3 \, c} + \frac {35 \, \sqrt {c x^{2} + b x} B b^{2} x}{96 \, c^{3}} - \frac {5 \, \sqrt {c x^{2} + b x} A b x}{12 \, c^{2}} + \frac {35 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{8 \, c^{3}} \end {gather*}

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]

1/4*sqrt(c*x^2 + b*x)*B*x^3/c - 7/24*sqrt(c*x^2 + b*x)*B*b*x^2/c^2 + 1/3*sqrt(c*x^2 + b*x)*A*x^2/c + 35/96*sqr
t(c*x^2 + b*x)*B*b^2*x/c^3 - 5/12*sqrt(c*x^2 + b*x)*A*b*x/c^2 + 35/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*
x)*sqrt(c))/c^(9/2) - 5/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 35/64*sqrt(c*x^2 + b*x
)*B*b^3/c^4 + 5/8*sqrt(c*x^2 + b*x)*A*b^2/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\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^3*(A + B*x))/(b*x + c*x^2)^(1/2),x)

[Out]

int((x^3*(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^{3} \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**3*(B*x+A)/(c*x**2+b*x)**(1/2),x)

[Out]

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

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