\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{x^5} \, dx\) [158]

Optimal result

Integrand size = 24, antiderivative size = 37 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4} \]

[Out]

-1/4*(b*x+a)^3*((b*x+a)^2)^(1/2)/a/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 37} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^5,x]

[Out]

-1/4*((a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a*x^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3}{x^5} \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = -\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {(a+b x)^3 \sqrt {(a+b x)^2}}{4 a x^4} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^5,x]

[Out]

-1/4*((a + b*x)^3*Sqrt[(a + b*x)^2])/(a*x^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).

Time = 2.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35

method	result	size
gosper	\(-\frac {\left (4 b^{3} x^{3}+6 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{4 x^{4} \left (b x +a \right )^{3}}\)	\(50\)
default	\(-\frac {\left (4 b^{3} x^{3}+6 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{4 x^{4} \left (b x +a \right )^{3}}\)	\(50\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-b^{3} x^{3}-\frac {3}{2} a \,b^{2} x^{2}-a^{2} b x -\frac {1}{4} a^{3}\right )}{\left (b x +a \right ) x^{4}}\)	\(51\)

[In]

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

[Out]

-1/4*(4*b^3*x^3+6*a*b^2*x^2+4*a^2*b*x+a^3)*((b*x+a)^2)^(3/2)/x^4/(b*x+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {4 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3}}{4 \, x^{4}} \]

[In]

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

[Out]

-1/4*(4*b^3*x^3 + 6*a*b^2*x^2 + 4*a^2*b*x + a^3)/x^4

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (24) = 48\).

Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.73 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}}{4 \, a^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{3}}{4 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{4 \, a^{4} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b}{4 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{4 \, a^{2} x^{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {b^{4} \mathrm {sgn}\left (b x + a\right )}{4 \, a} - \frac {4 \, b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + a^{3} \mathrm {sgn}\left (b x + a\right )}{4 \, x^{4}} \]

[In]

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

[Out]

-1/4*b^4*sgn(b*x + a)/a - 1/4*(4*b^3*x^3*sgn(b*x + a) + 6*a*b^2*x^2*sgn(b*x + a) + 4*a^2*b*x*sgn(b*x + a) + a^
3*sgn(b*x + a))/x^4

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 135, normalized size of antiderivative = 3.65 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x\,\left (a+b\,x\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^2\,\left (a+b\,x\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^3\,\left (a+b\,x\right )} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/x^5,x)

[Out]

- (a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^4*(a + b*x)) - (b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x*(a + b*x)
) - (3*a*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*x^2*(a + b*x)) - (a^2*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^3
*(a + b*x))