3.1.53 \(\int \frac {x^3}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {2 x^2}{c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {668, 640, 620, 206} \begin {gather*} \frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}-\frac {2 x^2}{c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*x^2)/(c*Sqrt[b*x + c*x^2]) + (3*Sqrt[b*x + c*x^2])/c^2 - (3*b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(5
/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 668

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*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{c}\\ &=-\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {(3 b) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{2 c^2}\\ &=-\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{c^2}\\ &=-\frac {2 x^2}{c \sqrt {b x+c x^2}}+\frac {3 \sqrt {b x+c x^2}}{c^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.72 \begin {gather*} \frac {2 x^3 \sqrt {\frac {c x}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {c x}{b}\right )}{5 b \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(b*x + c*x^2)^(3/2),x]

[Out]

(2*x^3*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 5/2, 7/2, -((c*x)/b)])/(5*b*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.37, size = 75, normalized size = 1.09 \begin {gather*} \frac {\sqrt {b x+c x^2} (3 b+c x)}{c^2 (b+c x)}+\frac {3 b \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b + c*x)*Sqrt[b*x + c*x^2])/(c^2*(b + c*x)) + (3*b*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[b*x + c*x^2]])/(2*
c^(5/2))

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fricas [A]  time = 0.41, size = 152, normalized size = 2.20 \begin {gather*} \left [\frac {3 \, {\left (b c x + b^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (c^{2} x + 3 \, b c\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{4} x + b c^{3}\right )}}, \frac {3 \, {\left (b c x + b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (c^{2} x + 3 \, b c\right )} \sqrt {c x^{2} + b x}}{c^{4} x + b c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b*c*x + b^2)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(c^2*x + 3*b*c)*sqrt(c*x^2 + b*
x))/(c^4*x + b*c^3), (3*(b*c*x + b^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (c^2*x + 3*b*c)*sqrt
(c*x^2 + b*x))/(c^4*x + b*c^3)]

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giac [A]  time = 0.25, size = 89, normalized size = 1.29 \begin {gather*} \frac {3 \, b \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} + \frac {2 \, b^{2}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} c + b \sqrt {c}\right )} c^{2}} + \frac {\sqrt {c x^{2} + b x}}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*b*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) + 2*b^2/(((sqrt(c)*x - sqrt(c*x^2 + b*x
))*c + b*sqrt(c))*c^2) + sqrt(c*x^2 + b*x)/c^2

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maple [A]  time = 0.05, size = 68, normalized size = 0.99 \begin {gather*} \frac {x^{2}}{\sqrt {c \,x^{2}+b x}\, c}+\frac {3 b x}{\sqrt {c \,x^{2}+b x}\, c^{2}}-\frac {3 b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(c*x^2+b*x)^(3/2),x)

[Out]

x^2/c/(c*x^2+b*x)^(1/2)+3*b/c^2/(c*x^2+b*x)^(1/2)*x-3/2*b/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 = 1.31, size = 66, normalized size = 0.96 \begin {gather*} \frac {x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {3 \, b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

x^2/(sqrt(c*x^2 + b*x)*c) + 3*b*x/(sqrt(c*x^2 + b*x)*c^2) - 3/2*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))
/c^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^3/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/(x*(b + c*x))**(3/2), x)

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