\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx\) [1550]

Optimal result

Integrand size = 28, antiderivative size = 46 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (b d-a e) (d+e x)^2} \]

[Out]

1/2*(b*x+a)*((b*x+a)^2)^(1/2)/(-a*e+b*d)/(e*x+d)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 37} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (d+e x)^2 (b d-a e)} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^3,x]

[Out]

((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(b*d - a*e)*(d + e*x)^2)

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 {a b+b^2 x}{(d+e x)^3} \, dx}{a b+b^2 x} \\ & = \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (b d-a e) (d+e x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\frac {(a+b x) \sqrt {(a+b x)^2}}{2 (b d-a e) (d+e x)^2} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^3,x]

[Out]

((a + b*x)*Sqrt[(a + b*x)^2])/(2*(b*d - a*e)*(d + e*x)^2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 2.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (2 b e x +a e +b d \right )}{2 e^{2} \left (e x +d \right )^{2}}\)	\(31\)
gosper	\(-\frac {\left (2 b e x +a e +b d \right ) \sqrt {\left (b x +a \right )^{2}}}{2 \left (e x +d \right )^{2} e^{2} \left (b x +a \right )}\)	\(41\)
risch	\(\frac {\left (-\frac {b x}{e}-\frac {a e +b d}{2 e^{2}}\right ) \sqrt {\left (b x +a \right )^{2}}}{\left (e x +d \right )^{2} \left (b x +a \right )}\)	\(45\)

[In]

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

[Out]

-1/2*csgn(b*x+a)*(2*b*e*x+a*e+b*d)/e^2/(e*x+d)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=-\frac {2 \, b e x + b d + a e}{2 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\text {Timed out} \]

[In]

integrate(((b*x+a)**2)**(1/2)/(e*x+d)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (33) = 66\).

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=\frac {b^{2} \mathrm {sgn}\left (b x + a\right )}{2 \, {\left (b d e^{2} - a e^{3}\right )}} - \frac {2 \, b e x \mathrm {sgn}\left (b x + a\right ) + b d \mathrm {sgn}\left (b x + a\right ) + a e \mathrm {sgn}\left (b x + a\right )}{2 \, {\left (e x + d\right )}^{2} e^{2}} \]

[In]

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

[Out]

1/2*b^2*sgn(b*x + a)/(b*d*e^2 - a*e^3) - 1/2*(2*b*e*x*sgn(b*x + a) + b*d*sgn(b*x + a) + a*e*sgn(b*x + a))/((e*
x + d)^2*e^2)

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (a\,e+b\,d+2\,b\,e\,x\right )}{2\,e^2\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2} \]

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x)^3,x)

[Out]

-(((a + b*x)^2)^(1/2)*(a*e + b*d + 2*b*e*x))/(2*e^2*(a + b*x)*(d + e*x)^2)