3.104 \(\int \frac{x^{9/2}}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=108 \[ \frac{16 b^2 x^{3/2}}{5 c^3 \sqrt{b x+c x^2}}+\frac{32 b^3 \sqrt{x}}{5 c^4 \sqrt{b x+c x^2}}-\frac{4 b x^{5/2}}{5 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}} \]

[Out]

(32*b^3*Sqrt[x])/(5*c^4*Sqrt[b*x + c*x^2]) + (16*b^2*x^(3/2))/(5*c^3*Sqrt[b*x + c*x^2]) - (4*b*x^(5/2))/(5*c^2
*Sqrt[b*x + c*x^2]) + (2*x^(7/2))/(5*c*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.0413416, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {656, 648} \[ \frac{16 b^2 x^{3/2}}{5 c^3 \sqrt{b x+c x^2}}+\frac{32 b^3 \sqrt{x}}{5 c^4 \sqrt{b x+c x^2}}-\frac{4 b x^{5/2}}{5 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(32*b^3*Sqrt[x])/(5*c^4*Sqrt[b*x + c*x^2]) + (16*b^2*x^(3/2))/(5*c^3*Sqrt[b*x + c*x^2]) - (4*b*x^(5/2))/(5*c^2
*Sqrt[b*x + c*x^2]) + (2*x^(7/2))/(5*c*Sqrt[b*x + c*x^2])

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]

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]

Rubi steps

\begin{align*} \int \frac{x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx &=\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}}-\frac{(6 b) \int \frac{x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c}\\ &=-\frac{4 b x^{5/2}}{5 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}}+\frac{\left (8 b^2\right ) \int \frac{x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c^2}\\ &=\frac{16 b^2 x^{3/2}}{5 c^3 \sqrt{b x+c x^2}}-\frac{4 b x^{5/2}}{5 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}}-\frac{\left (16 b^3\right ) \int \frac{x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c^3}\\ &=\frac{32 b^3 \sqrt{x}}{5 c^4 \sqrt{b x+c x^2}}+\frac{16 b^2 x^{3/2}}{5 c^3 \sqrt{b x+c x^2}}-\frac{4 b x^{5/2}}{5 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{7/2}}{5 c \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0237896, size = 52, normalized size = 0.48 \[ \frac{2 \sqrt{x} \left (8 b^2 c x+16 b^3-2 b c^2 x^2+c^3 x^3\right )}{5 c^4 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(9/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(16*b^3 + 8*b^2*c*x - 2*b*c^2*x^2 + c^3*x^3))/(5*c^4*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.048, size = 54, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ({x}^{3}{c}^{3}-2\,b{x}^{2}{c}^{2}+8\,{b}^{2}xc+16\,{b}^{3} \right ) }{5\,{c}^{4}}{x}^{{\frac{3}{2}}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(c*x^2+b*x)^(3/2),x)

[Out]

2/5*(c*x+b)*(c^3*x^3-2*b*c^2*x^2+8*b^2*c*x+16*b^3)*x^(3/2)/c^4/(c*x^2+b*x)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left ({\left (3 \, c^{4} x^{3} - b c^{3} x^{2} + 4 \, b^{2} c^{2} x + 8 \, b^{3} c\right )} x^{3} - 2 \,{\left (b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 7 \, b^{3} c x - 4 \, b^{4}\right )} x^{2} + 10 \,{\left (b^{2} c^{2} x^{3} + 2 \, b^{3} c x^{2} + b^{4} x\right )} x\right )}}{15 \,{\left (c^{5} x^{3} + b c^{4} x^{2}\right )} \sqrt{c x + b}} - \int \frac{2 \,{\left (b^{3} c x + b^{4}\right )} x}{{\left (c^{5} x^{3} + 2 \, b c^{4} x^{2} + b^{2} c^{3} x\right )} \sqrt{c x + b}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/15*((3*c^4*x^3 - b*c^3*x^2 + 4*b^2*c^2*x + 8*b^3*c)*x^3 - 2*(b*c^3*x^3 - 2*b^2*c^2*x^2 - 7*b^3*c*x - 4*b^4)*
x^2 + 10*(b^2*c^2*x^3 + 2*b^3*c*x^2 + b^4*x)*x)/((c^5*x^3 + b*c^4*x^2)*sqrt(c*x + b)) - integrate(2*(b^3*c*x +
b^4)*x/((c^5*x^3 + 2*b*c^4*x^2 + b^2*c^3*x)*sqrt(c*x + b)), x)

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Fricas [A]  time = 2.00206, size = 130, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (c^{3} x^{3} - 2 \, b c^{2} x^{2} + 8 \, b^{2} c x + 16 \, b^{3}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{5 \,{\left (c^{5} x^{2} + b c^{4} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/5*(c^3*x^3 - 2*b*c^2*x^2 + 8*b^2*c*x + 16*b^3)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^5*x^2 + b*c^4*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.22841, size = 76, normalized size = 0.7 \begin{align*} -\frac{32 \, b^{\frac{5}{2}}}{5 \, c^{4}} + \frac{2 \,{\left ({\left (c x + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x + b\right )}^{\frac{3}{2}} b + 15 \, \sqrt{c x + b} b^{2} + \frac{5 \, b^{3}}{\sqrt{c x + b}}\right )}}{5 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-32/5*b^(5/2)/c^4 + 2/5*((c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b + 15*sqrt(c*x + b)*b^2 + 5*b^3/sqrt(c*x + b))/c
^4