\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx\) [1551]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^4,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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)^4} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b (b d-a e)}{e (d+e x)^4}+\frac {b^2}{e (d+e x)^3}\right ) \, dx}{a b+b^2 x} \\ & = \frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {\sqrt {(a+b x)^2} (2 a e+b (d+3 e x))}{6 e^2 (a+b x) (d+e x)^3} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^4,x]

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(2*a*e + b*(d + 3*e*x)))/(e^2*(a + b*x)*(d + e*x)^3)

Maple [C] (warning: unable to verify)

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

Time = 2.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (3 b e x +2 a e +b d \right )}{6 e^{2} \left (e x +d \right )^{3}}\)	\(32\)
gosper	\(-\frac {\left (3 b e x +2 a e +b d \right ) \sqrt {\left (b x +a \right )^{2}}}{6 e^{2} \left (e x +d \right )^{3} \left (b x +a \right )}\)	\(42\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b x}{2 e}-\frac {2 a e +b d}{6 e^{2}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{3}}\)	\(46\)

[In]

int(((b*x+a)^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/6*csgn(b*x+a)*(3*b*e*x+2*a*e+b*d)/e^2/(e*x+d)^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {3 \, b e x + b d + 2 \, a e}{6 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/6*(3*b*e*x + b*d + 2*a*e)/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\text {Timed out} \]

[In]

integrate(((b*x+a)**2)**(1/2)/(e*x+d)**4,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^4,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 [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {b^{3} \mathrm {sgn}\left (b x + a\right )}{6 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} - \frac {3 \, b e x \mathrm {sgn}\left (b x + a\right ) + b d \mathrm {sgn}\left (b x + a\right ) + 2 \, a e \mathrm {sgn}\left (b x + a\right )}{6 \, {\left (e x + d\right )}^{3} e^{2}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.67 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,a\,e+b\,d+3\,b\,e\,x\right )}{6\,e^2\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \]

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x)^4,x)

[Out]

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