3.2.6 \(\int \frac {x^{5/2}}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {4 b \sqrt {x}}{c^2 \sqrt {b x+c x^2}}+\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {656, 648} \begin {gather*} \frac {4 b \sqrt {x}}{c^2 \sqrt {b x+c x^2}}+\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(4*b*Sqrt[x])/(c^2*Sqrt[b*x + c*x^2]) + (2*x^(3/2))/(c*Sqrt[b*x + c*x^2])

Rule 648

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 656

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[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx &=\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}-\frac {(2 b) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{c}\\ &=\frac {4 b \sqrt {x}}{c^2 \sqrt {b x+c x^2}}+\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt {x} (2 b+c x)}{c^2 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(2*b + c*x))/(c^2*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.47, size = 37, normalized size = 0.77 \begin {gather*} \frac {2 (2 b+c x) \sqrt {b x+c x^2}}{c^2 \sqrt {x} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*(2*b + c*x)*Sqrt[b*x + c*x^2])/(c^2*Sqrt[x]*(b + c*x))

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fricas [A]  time = 0.39, size = 39, normalized size = 0.81 \begin {gather*} \frac {2 \, \sqrt {c x^{2} + b x} {\left (c x + 2 \, b\right )} \sqrt {x}}{c^{3} x^{2} + b c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 38, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (\frac {\sqrt {c x + b}}{c} + \frac {b}{\sqrt {c x + b} c}\right )}}{c} - \frac {4 \, \sqrt {b}}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

2*(sqrt(c*x + b)/c + b/(sqrt(c*x + b)*c))/c - 4*sqrt(b)/c^2

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maple [A]  time = 0.04, size = 32, normalized size = 0.67 \begin {gather*} \frac {2 \left (c x +b \right ) \left (c x +2 b \right ) x^{\frac {3}{2}}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(c*x^2+b*x)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{5/2}}{{\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^(5/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(5/2)/(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^{\frac {5}{2}}}{\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**(5/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**(5/2)/(x*(b + c*x))**(3/2), x)

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