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

Optimal. Leaf size=46 \[ \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (b d-a e) (d+e x)^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 37} \begin {gather*} \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (d+e x)^2 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.01, size = 44, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {(a+b x)^2} (a e+b (d+2 e x))}{2 e^2 (a+b x) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 2.
time = 0.46, size = 31, normalized size = 0.67

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.81, size = 36, normalized size = 0.78 \begin {gather*} -\frac {b d + {\left (2 \, b x + a\right )} e}{2 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 39, normalized size = 0.85 \begin {gather*} \frac {- a e - b d - 2 b e x}{2 d^{2} e^{2} + 4 d e^{3} x + 2 e^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.99, size = 44, normalized size = 0.96 \begin {gather*} -\frac {{\left (2 \, b x e \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 )\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.57, size = 40, normalized size = 0.87 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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