\(\int \frac {(b x+c x^2)^{3/2}}{x^{3/2}} \, dx\) [87]

Optimal result

Integrand size = 19, antiderivative size = 25 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \]

[Out]

2/5*(c*x^2+b*x)^(5/2)/c/x^(5/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {662} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^(3/2),x]

[Out]

(2*(b*x + c*x^2)^(5/2))/(5*c*x^(5/2))

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 (x (b+c x))^{5/2}}{5 c x^{5/2}} \]

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2))/(5*c*x^(5/2))

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c \,x^{\frac {3}{2}}}\)	\(25\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2}}{5 \sqrt {x}\, c}\)	\(25\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (c^{2} x^{2}+2 b c x +b^{2}\right )}{5 \sqrt {x \left (c x +b \right )}\, c}\)	\(39\)

[In]

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

[Out]

2/5*(c*x+b)/c*(c*x^2+b*x)^(3/2)/x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \, {\left (c^{2} x^{2} + 2 \, b c x + b^{2}\right )} \sqrt {c x^{2} + b x}}{5 \, c \sqrt {x}} \]

[In]

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

[Out]

2/5*(c^2*x^2 + 2*b*c*x + b^2)*sqrt(c*x^2 + b*x)/(c*sqrt(x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).

Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2 \, {\left (5 \, b c x^{2} + 5 \, b^{2} x + {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} x\right )} \sqrt {c x + b}}{15 \, c x} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (19) = 38\).

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.40 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx=\frac {2}{15} \, c {\left (\frac {2 \, b^{\frac {5}{2}}}{c^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b}{c^{2}}\right )} + \frac {2}{3} \, b {\left (\frac {{\left (c x + b\right )}^{\frac {3}{2}}}{c} - \frac {b^{\frac {3}{2}}}{c}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b*x + c*x^2)^(3/2)/x^(3/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(3/2), x)