\(\int x^{3/2} \sqrt {b x+c x^2} \, dx\) [74]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \]

[Out]

16/105*b^2*(c*x^2+b*x)^(3/2)/c^3/x^(3/2)-8/35*b*(c*x^2+b*x)^(3/2)/c^2/x^(1/2)+2/7*(c*x^2+b*x)^(3/2)*x^(1/2)/c

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} \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \]

[In]

Int[x^(3/2)*Sqrt[b*x + c*x^2],x]

[Out]

(16*b^2*(b*x + c*x^2)^(3/2))/(105*c^3*x^(3/2)) - (8*b*(b*x + c*x^2)^(3/2))/(35*c^2*Sqrt[x]) + (2*Sqrt[x]*(b*x
+ c*x^2)^(3/2))/(7*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 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {(4 b) \int \sqrt {x} \sqrt {b x+c x^2} \, dx}{7 c} \\ & = -\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{35 c^2} \\ & = \frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\frac {2 (x (b+c x))^{3/2} \left (8 b^2-12 b c x+15 c^2 x^2\right )}{105 c^3 x^{3/2}} \]

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(8*b^2 - 12*b*c*x + 15*c^2*x^2))/(105*c^3*x^(3/2))

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {2 \left (c x +b \right ) \left (15 c^{2} x^{2}-12 b c x +8 b^{2}\right ) \sqrt {x \left (c x +b \right )}}{105 c^{3} \sqrt {x}}\)	\(42\)
gosper	\(\frac {2 \left (c x +b \right ) \left (15 c^{2} x^{2}-12 b c x +8 b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 c^{3} \sqrt {x}}\)	\(44\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (15 c^{3} x^{3}+3 b \,c^{2} x^{2}-4 b^{2} c x +8 b^{3}\right )}{105 \sqrt {x \left (c x +b \right )}\, c^{3}}\)	\(53\)

[In]

int(x^(3/2)*(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/105*(c*x+b)*(15*c^2*x^2-12*b*c*x+8*b^2)*(x*(c*x+b))^(1/2)/c^3/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61 \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{105 \, c^{3} \sqrt {x}} \]

[In]

integrate(x^(3/2)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

Sympy [F]

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

[In]

integrate(x**(3/2)*(c*x**2+b*x)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b}}{105 \, c^{3}} \]

[In]

integrate(x^(3/2)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/105*(15*c^3*x^3 + 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 = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=-\frac {16 \, b^{\frac {7}{2}}}{105 \, c^{3}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )}}{105 \, c^{3}} \]

[In]

integrate(x^(3/2)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-16/105*b^(7/2)/c^3 + 2/105*(15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)/c^3

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} \sqrt {b x+c x^2} \, dx=\int x^{3/2}\,\sqrt {c\,x^2+b\,x} \,d x \]

[In]

int(x^(3/2)*(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(3/2)*(b*x + c*x^2)^(1/2), x)