### 3.1593 $$\int \frac{1}{(d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$

Optimal. Leaf size=131 $\frac{a+b x}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)}+\frac{b (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{b (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}$

[Out]

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

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Rubi [A]  time = 0.0620819, antiderivative size = 131, 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, 44} $\frac{a+b x}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)}+\frac{b (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{b (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a + b*x)/((b*d - a*e)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b*(a + b*x)*Log[a + b*x])/((b*d - a*e)^2*Sq
rt[a^2 + 2*a*b*x + b^2*x^2]) - (b*(a + b*x)*Log[d + e*x])/((b*d - a*e)^2*Sqrt[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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0439544, size = 69, normalized size = 0.53 $\frac{(a+b x) (b (d+e x) \log (a+b x)-a e-b (d+e x) \log (d+e x)+b d)}{\sqrt{(a+b x)^2} (d+e x) (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.162, size = 82, normalized size = 0.6 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( \ln \left ( ex+d \right ) xbe-\ln \left ( bx+a \right ) xbe+\ln \left ( ex+d \right ) bd-\ln \left ( bx+a \right ) bd+ae-bd \right ) }{ \left ( ae-bd \right ) ^{2} \left ( ex+d \right ) }{\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)^2/((b*x+a)^2)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59596, size = 198, normalized size = 1.51 \begin{align*} \frac{b d - a e +{\left (b e x + b d\right )} \log \left (b x + a\right ) -{\left (b e x + b d\right )} \log \left (e x + d\right )}{b^{2} d^{3} - 2 \, a b d^{2} e + a^{2} d e^{2} +{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.95312, size = 233, normalized size = 1.78 \begin{align*} - \frac{b \log{\left (x + \frac{- \frac{a^{3} b e^{3}}{\left (a e - b d\right )^{2}} + \frac{3 a^{2} b^{2} d e^{2}}{\left (a e - b d\right )^{2}} - \frac{3 a b^{3} d^{2} e}{\left (a e - b d\right )^{2}} + a b e + \frac{b^{4} d^{3}}{\left (a e - b d\right )^{2}} + b^{2} d}{2 b^{2} e} \right )}}{\left (a e - b d\right )^{2}} + \frac{b \log{\left (x + \frac{\frac{a^{3} b e^{3}}{\left (a e - b d\right )^{2}} - \frac{3 a^{2} b^{2} d e^{2}}{\left (a e - b d\right )^{2}} + \frac{3 a b^{3} d^{2} e}{\left (a e - b d\right )^{2}} + a b e - \frac{b^{4} d^{3}}{\left (a e - b d\right )^{2}} + b^{2} d}{2 b^{2} e} \right )}}{\left (a e - b d\right )^{2}} - \frac{1}{a d e - b d^{2} + x \left (a e^{2} - b d e\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19021, size = 139, normalized size = 1.06 \begin{align*}{\left (\frac{b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} - \frac{b e \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} + \frac{1}{{\left (b d - a e\right )}{\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(1/(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

(b^2*log(abs(b*x + a))/(b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2) - b*e*log(abs(x*e + d))/(b^2*d^2*e - 2*a*b*d*e^2 +
a^2*e^3) + 1/((b*d - a*e)*(x*e + d)))*sgn(b*x + a)