\(\int \frac {x^{9/2}}{(b x+c x^2)^{3/2}} \, dx\) [104]

Optimal result

Integrand size = 19, antiderivative size = 108 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {32 b^3 \sqrt {x}}{5 c^4 \sqrt {b x+c x^2}}+\frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {32 b^3 \sqrt {x}}{5 c^4 \sqrt {b x+c x^2}}+\frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}} \]

[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 662

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 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[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*} \text {integral}& = \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] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.48 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (16 b^3+8 b^2 c x-2 b c^2 x^2+c^3 x^3\right )}{5 c^4 \sqrt {x (b+c x)}} \]

[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)])

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (c^{3} x^{3}-2 b \,c^{2} x^{2}+8 b^{2} c x +16 b^{3}\right ) x^{\frac {3}{2}}}{5 c^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(54\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c^{3} x^{3}-2 b \,c^{2} x^{2}+8 b^{2} c x +16 b^{3}\right )}{5 \sqrt {x}\, \left (c x +b \right ) c^{4}}\)	\(54\)
risch	\(\frac {2 \left (c^{2} x^{2}-3 b c x +11 b^{2}\right ) \left (c x +b \right ) \sqrt {x}}{5 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {2 b^{3} \sqrt {x}}{c^{4} \sqrt {x \left (c x +b \right )}}\)	\(62\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c^{3} x^{3} - 2 \, b c^{2} x^{2} + 8 \, b^{2} c x + 16 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{5 \, {\left (c^{5} x^{2} + b c^{4} x\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{\frac {9}{2}}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {x^{\frac {9}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {32 \, b^{\frac {5}{2}}}{5 \, c^{4}} + \frac {2 \, b^{3}}{\sqrt {c x + b} c^{4}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {5}{2}} c^{16} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{16} + 15 \, \sqrt {c x + b} b^{2} c^{16}\right )}}{5 \, c^{20}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{9/2}}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

[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)