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

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

[Out]

16/5*b^2*x^(3/2)/c^3/(c*x^2+b*x)^(1/2)-4/5*b*x^(5/2)/c^2/(c*x^2+b*x)^(1/2)+2/5*x^(7/2)/c/(c*x^2+b*x)^(1/2)+32/
5*b^3*x^(1/2)/c^4/(c*x^2+b*x)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.48 \[ \frac {2 \sqrt {x} \left (16 b^3+8 b^2 c x-2 b c^2 x^2+c^3 x^3\right )}{5 c^4 \sqrt {x (b+c x)}} \]

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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fricas [A]  time = 1.21, size = 61, normalized size = 0.56 \[ \frac {2 \, {\left (c^{3} x^{3} - 2 \, b c^{2} x^{2} + 8 \, b^{2} c x + 16 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{5 \, {\left (c^{5} x^{2} + b c^{4} x\right )}} \]

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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giac [A]  time = 0.18, size = 69, normalized size = 0.64 \[ -\frac {32 \, b^{\frac {5}{2}}}{5 \, c^{4}} + \frac {2 \, b^{3}}{\sqrt {c x + b} c^{4}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {5}{2}} c^{16} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{16} + 15 \, \sqrt {c x + b} b^{2} c^{16}\right )}}{5 \, c^{20}} \]

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*b^3/(sqrt(c*x + b)*c^4) + 2/5*((c*x + b)^(5/2)*c^16 - 5*(c*x + b)^(3/2)*b*c^16 + 15*sqrt
(c*x + b)*b^2*c^16)/c^20

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maple [A]  time = 0.04, size = 54, normalized size = 0.50 \[ \frac {2 \left (c x +b \right ) \left (x^{3} c^{3}-2 b \,x^{2} c^{2}+8 b^{2} x c +16 b^{3}\right ) x^{\frac {3}{2}}}{5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{4}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left ({\left (3 \, c^{4} x^{3} - b c^{3} x^{2} + 4 \, b^{2} c^{2} x + 8 \, b^{3} c\right )} x^{3} - 2 \, {\left (b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 7 \, b^{3} c x - 4 \, b^{4}\right )} x^{2} + 10 \, {\left (b^{2} c^{2} x^{3} + 2 \, b^{3} c x^{2} + b^{4} x\right )} x\right )}}{15 \, {\left (c^{5} x^{3} + b c^{4} x^{2}\right )} \sqrt {c x + b}} - \int \frac {2 \, {\left (b^{3} c x + b^{4}\right )} x}{{\left (c^{5} x^{3} + 2 \, b c^{4} x^{2} + b^{2} c^{3} x\right )} \sqrt {c x + b}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{9/2}}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(9/2)/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {9}{2}}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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