3.18.96 \(\int \frac {a+b x}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=38 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x) (b d-a e)} \]

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {767} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x) (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 767

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Sim
p[(f*g*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)*(e*f - d*g)), x] /; FreeQ[{a, b, c, d, e, f, g,
m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && EqQ[2*c*f - b*g, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.74 \begin {gather*} -\frac {a+b x}{e \sqrt {(a+b x)^2} (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.49, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 13, normalized size = 0.34 \begin {gather*} -\frac {1}{e^{2} x + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(e^2*x + d*e)

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giac [A]  time = 0.17, size = 18, normalized size = 0.47 \begin {gather*} -\frac {e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right )}{x e + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(-1)*sgn(b*x + a)/(x*e + d)

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maple [A]  time = 0.05, size = 27, normalized size = 0.71 \begin {gather*} -\frac {b x +a}{\left (e x +d \right ) \sqrt {\left (b x +a \right )^{2}}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(e*x+d)^2/((b*x+a)^2)^(1/2),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 details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.10, size = 28, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}}{e\,\left (a+b\,x\right )\,\left (d+e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.17, size = 10, normalized size = 0.26 \begin {gather*} - \frac {1}{d e + e^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*e + e**2*x)

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