3.1592 $$\int \frac{1}{(d+e x) \sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$

Optimal. Leaf size=86 $\frac{(a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}$

[Out]

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

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Rubi [A]  time = 0.0307776, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.107, Rules used = {646, 36, 31} $\frac{(a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b \left (a b+b^2 x\right )\right ) \int \frac{1}{a b+b^2 x} \, dx}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (e \left (a b+b^2 x\right )\right ) \int \frac{1}{d+e x} \, dx}{b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(a+b x) \log (a+b x)}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(a+b x) \log (d+e x)}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0187443, size = 42, normalized size = 0.49 $\frac{(a+b x) (\log (a+b x)-\log (d+e x))}{\sqrt{(a+b x)^2} (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.157, size = 41, normalized size = 0.5 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( \ln \left ( ex+d \right ) -\ln \left ( bx+a \right ) \right ) }{ae-bd}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59554, size = 58, normalized size = 0.67 \begin{align*} \frac{\log \left (b x + a\right ) - \log \left (e x + d\right )}{b d - a e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.401514, size = 128, normalized size = 1.49 \begin{align*} \frac{\log{\left (x + \frac{- \frac{a^{2} e^{2}}{a e - b d} + \frac{2 a b d e}{a e - b d} + a e - \frac{b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} - \frac{\log{\left (x + \frac{\frac{a^{2} e^{2}}{a e - b d} - \frac{2 a b d e}{a e - b d} + a e + \frac{b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24142, size = 101, normalized size = 1.17 \begin{align*} \frac{\log \left (\frac{{\left | 2 \, b x e + b d + a e -{\left | b d - a e \right |} \right |}}{{\left | 2 \, b x e + b d + a e +{\left | b d - a e \right |} \right |}}\right ) \mathrm{sgn}\left (b x + a\right )}{{\left | b d - a e \right |}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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