\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx\) [1552]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^2 (a+b x) (d+e x)^4}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3} \]

[Out]

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

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} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{4 e^2 (a+b x) (d+e x)^4}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^5,x]

[Out]

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

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)^5} \, 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)^5}+\frac {b^2}{e (d+e x)^4}\right ) \, dx}{a b+b^2 x} \\ & = \frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^2 (a+b x) (d+e x)^4}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {\sqrt {(a+b x)^2} (3 a e+b (d+4 e x))}{12 e^2 (a+b x) (d+e x)^4} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(d + e*x)^5,x]

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(3*a*e + b*(d + 4*e*x)))/(e^2*(a + b*x)*(d + e*x)^4)

Maple [C] (warning: unable to verify)

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

Time = 2.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (4 b e x +3 a e +b d \right )}{12 e^{2} \left (e x +d \right )^{4}}\)	\(32\)
gosper	\(-\frac {\left (4 b e x +3 a e +b d \right ) \sqrt {\left (b x +a \right )^{2}}}{12 e^{2} \left (e x +d \right )^{4} \left (b x +a \right )}\)	\(42\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b x}{3 e}-\frac {3 a e +b d}{12 e^{2}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{4}}\)	\(46\)

[In]

int(((b*x+a)^2)^(1/2)/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

-1/12*csgn(b*x+a)*(4*b*e*x+3*a*e+b*d)/e^2/(e*x+d)^4

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {4 \, b e x + b d + 3 \, a e}{12 \, {\left (e^{6} x^{4} + 4 \, d e^{5} x^{3} + 6 \, d^{2} e^{4} x^{2} + 4 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/12*(4*b*e*x + b*d + 3*a*e)/(e^6*x^4 + 4*d*e^5*x^3 + 6*d^2*e^4*x^2 + 4*d^3*e^3*x + d^4*e^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\text {Timed out} \]

[In]

integrate(((b*x+a)**2)**(1/2)/(e*x+d)**5,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^5,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 = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\frac {b^{4} \mathrm {sgn}\left (b x + a\right )}{12 \, {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )}} - \frac {4 \, b e x \mathrm {sgn}\left (b x + a\right ) + b d \mathrm {sgn}\left (b x + a\right ) + 3 \, a e \mathrm {sgn}\left (b x + a\right )}{12 \, {\left (e x + d\right )}^{4} e^{2}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

1/12*b^4*sgn(b*x + a)/(b^3*d^3*e^2 - 3*a*b^2*d^2*e^3 + 3*a^2*b*d*e^4 - a^3*e^5) - 1/12*(4*b*e*x*sgn(b*x + a) +
b*d*sgn(b*x + a) + 3*a*e*sgn(b*x + a))/((e*x + d)^4*e^2)

Mupad [B] (verification not implemented)

Time = 9.54 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (3\,a\,e+b\,d+4\,b\,e\,x\right )}{12\,e^2\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \]

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x)^5,x)

[Out]

-(((a + b*x)^2)^(1/2)*(3*a*e + b*d + 4*b*e*x))/(12*e^2*(a + b*x)*(d + e*x)^4)