\(\int \frac {(b x+c x^2)^{3/2}}{x^5} \, dx\) [19]

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 b x^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {664} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 b x^5} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^5,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(5*b*x^5)

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a +
b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e))), 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 + 2*p + 2, 0]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 (x (b+c x))^{5/2}}{5 b x^5} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^5,x]

[Out]

(-2*(x*(b + c*x))^(5/2))/(5*b*x^5)

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 b \,x^{5}}\)	\(20\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 x^{4} b}\)	\(25\)
pseudoelliptic	\(-\frac {2 \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}}{5 x^{3} b}\)	\(25\)
trager	\(-\frac {2 \left (c^{2} x^{2}+2 b c x +b^{2}\right ) \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}\)	\(36\)
risch	\(-\frac {2 \left (c x +b \right ) \left (c^{2} x^{2}+2 b c x +b^{2}\right )}{5 x^{2} \sqrt {x \left (c x +b \right )}\, b}\)	\(39\)

[In]

int((c*x^2+b*x)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \, {\left (c^{2} x^{2} + 2 \, b c x + b^{2}\right )} \sqrt {c x^{2} + b x}}{5 \, b x^{3}} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((c*x**2+b*x)**(3/2)/x**5,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**5, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (19) = 38\).

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{5 \, b x} + \frac {\sqrt {c x^{2} + b x} c}{5 \, x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} b}{5 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{4}} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^5,x, algorithm="maxima")

[Out]

-2/5*sqrt(c*x^2 + b*x)*c^2/(b*x) + 1/5*sqrt(c*x^2 + b*x)*c/x^2 + 3/5*sqrt(c*x^2 + b*x)*b/x^3 - (c*x^2 + b*x)^(
3/2)/x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (19) = 38\).

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 5.83 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=\frac {2 \, {\left (5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} c^{2} + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b c^{\frac {3}{2}} + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{2} c + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{3} \sqrt {c} + b^{4}\right )}}{5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^5,x, algorithm="giac")

[Out]

2/5*(5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*c^2 + 10*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c^(3/2) + 10*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^2*b^2*c + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*sqrt(c) + b^4)/(sqrt(c)*x - sqrt(c*x^2 +
b*x))^5

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,{\left (b+c\,x\right )}^2}{5\,b\,x^3} \]

[In]

int((b*x + c*x^2)^(3/2)/x^5,x)

[Out]

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