\(\int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx\) [95]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {16 b^2 \sqrt {b x+c x^2}}{15 c^3 \sqrt {x}}-\frac {8 b \sqrt {x} \sqrt {b x+c x^2}}{15 c^2}+\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c} \]

[Out]

2/5*x^(3/2)*(c*x^2+b*x)^(1/2)/c+16/15*b^2*(c*x^2+b*x)^(1/2)/c^3/x^(1/2)-8/15*b*x^(1/2)*(c*x^2+b*x)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {16 b^2 \sqrt {b x+c x^2}}{15 c^3 \sqrt {x}}-\frac {8 b \sqrt {x} \sqrt {b x+c x^2}}{15 c^2}+\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c} \]

[In]

Int[x^(5/2)/Sqrt[b*x + c*x^2],x]

[Out]

(16*b^2*Sqrt[b*x + c*x^2])/(15*c^3*Sqrt[x]) - (8*b*Sqrt[x]*Sqrt[b*x + c*x^2])/(15*c^2) + (2*x^(3/2)*Sqrt[b*x +
c*x^2])/(5*c)

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^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {(4 b) \int \frac {x^{3/2}}{\sqrt {b x+c x^2}} \, dx}{5 c} \\ & = -\frac {8 b \sqrt {x} \sqrt {b x+c x^2}}{15 c^2}+\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {x}}{\sqrt {b x+c x^2}} \, dx}{15 c^2} \\ & = \frac {16 b^2 \sqrt {b x+c x^2}}{15 c^3 \sqrt {x}}-\frac {8 b \sqrt {x} \sqrt {b x+c x^2}}{15 c^2}+\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {x (b+c x)} \left (8 b^2-4 b c x+3 c^2 x^2\right )}{15 c^3 \sqrt {x}} \]

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(8*b^2 - 4*b*c*x + 3*c^2*x^2))/(15*c^3*Sqrt[x])

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (3 c^{2} x^{2}-4 b c x +8 b^{2}\right )}{15 \sqrt {x}\, c^{3}}\)	\(37\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (3 c^{2} x^{2}-4 b c x +8 b^{2}\right )}{15 \sqrt {x \left (c x +b \right )}\, c^{3}}\)	\(42\)
gosper	\(\frac {2 \left (c x +b \right ) \left (3 c^{2} x^{2}-4 b c x +8 b^{2}\right ) \sqrt {x}}{15 c^{3} \sqrt {c \,x^{2}+b x}}\)	\(44\)

[In]

int(x^(5/2)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15/x^(1/2)*(x*(c*x+b))^(1/2)*(3*c^2*x^2-4*b*c*x+8*b^2)/c^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.48 \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (3 \, c^{2} x^{2} - 4 \, b c x + 8 \, b^{2}\right )} \sqrt {c x^{2} + b x}}{15 \, c^{3} \sqrt {x}} \]

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*c^2*x^2 - 4*b*c*x + 8*b^2)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

Sympy [F]

\[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{\frac {5}{2}}}{\sqrt {x \left (b + c x\right )}}\, dx \]

[In]

integrate(x**(5/2)/(c*x**2+b*x)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (3 \, c^{3} x^{3} - b c^{2} x^{2} + 4 \, b^{2} c x + 8 \, b^{3}\right )}}{15 \, \sqrt {c x + b} c^{3}} \]

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*c^3*x^3 - b*c^2*x^2 + 4*b^2*c*x + 8*b^3)/(sqrt(c*x + b)*c^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61 \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, \sqrt {c x + b} b^{2}}{c^{3}} - \frac {16 \, b^{\frac {5}{2}}}{15 \, c^{3}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 10 \, {\left (c x + b\right )}^{\frac {3}{2}} b\right )}}{15 \, c^{3}} \]

[In]

integrate(x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2*sqrt(c*x + b)*b^2/c^3 - 16/15*b^(5/2)/c^3 + 2/15*(3*(c*x + b)^(5/2) - 10*(c*x + b)^(3/2)*b)/c^3

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{5/2}}{\sqrt {c\,x^2+b\,x}} \,d x \]

[In]

int(x^(5/2)/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(5/2)/(b*x + c*x^2)^(1/2), x)