3.1549 \(\int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=85 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x) (d+e x)}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0405717, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x) (d+e x)}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 646

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]

Rule 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0194631, size = 50, normalized size = 0.59 \[ \frac{\sqrt{(a+b x)^2} (-a e+b (d+e x) \log (d+e x)+b d)}{e^2 (a+b x) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.227, size = 51, normalized size = 0.6 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ( \ln \left ( bxe+bd \right ) xbe+\ln \left ( bxe+bd \right ) bd-ae+bd \right ) }{{e}^{2} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57975, size = 78, normalized size = 0.92 \begin{align*} \frac{b d - a e +{\left (b e x + b d\right )} \log \left (e x + d\right )}{e^{3} x + d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.379246, size = 27, normalized size = 0.32 \begin{align*} \frac{b \log{\left (d + e x \right )}}{e^{2}} - \frac{a e - b d}{d e^{2} + e^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22229, size = 69, normalized size = 0.81 \begin{align*} b e^{\left (-2\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (b d \mathrm{sgn}\left (b x + a\right ) - a e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{x e + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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