3.2020 \(\int \frac{a+b x}{(d+e x) \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=35 \[ \frac{(a+b x) \log (d+e x)}{e \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0276059, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 31} \[ \frac{(a+b x) \log (d+e x)}{e \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

Mathematica [A]  time = 0.0085886, size = 26, normalized size = 0.74 \[ \frac{(a+b x) \log (d+e x)}{e \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.001, size = 25, normalized size = 0.7 \begin{align*}{\frac{ \left ( bx+a \right ) \ln \left ( ex+d \right ) }{e}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47393, size = 22, normalized size = 0.63 \begin{align*} \frac{\log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e*x + d)/e

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Sympy [A]  time = 0.095012, size = 7, normalized size = 0.2 \begin{align*} \frac{\log{\left (d + e x \right )}}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(d + e*x)/e

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Giac [A]  time = 1.15446, size = 23, normalized size = 0.66 \begin{align*} e^{\left (-1\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-1)*log(abs(x*e + d))*sgn(b*x + a)