3.1763 \(\int \frac{A+B x}{(d+e x) \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=107 \[ \frac{(a+b x) (A b-a B) \log (a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac{(a+b x) (B d-A e) \log (d+e x)}{e \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0770278, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 72} \[ \frac{(a+b x) (A b-a B) \log (a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac{(a+b x) (B d-A e) \log (d+e x)}{e \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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 - a*B)*(a + b*x)*Log[a + b*x])/(b*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((B*d - A*e)*(a + b*x)*Lo
g[d + e*x])/(e*(b*d - a*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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

Mathematica [A]  time = 0.043808, size = 66, normalized size = 0.62 \[ \frac{(a+b x) (e (A b-a B) \log (a+b x)+b (B d-A e) \log (d+e x))}{b e \sqrt{(a+b x)^2} (b d-a e)} \]

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

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Maple [A]  time = 0.01, size = 75, normalized size = 0.7 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( A\ln \left ( ex+d \right ) be-A\ln \left ( bx+a \right ) be-B\ln \left ( ex+d \right ) bd+B\ln \left ( bx+a \right ) ae \right ) }{e \left ( ae-bd \right ) b}{\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)*(A*ln(e*x+d)*b*e-A*ln(b*x+a)*b*e-B*ln(e*x+d)*b*d+B*ln(b*x+a)*a*e)/((b*x+a)^2)^(1/2)/e/(a*e-b*d)/b

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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 = 1.30536, size = 111, normalized size = 1.04 \begin{align*} -\frac{{\left (B a - A b\right )} e \log \left (b x + a\right ) -{\left (B b d - A b e\right )} \log \left (e x + d\right )}{b^{2} d e - a b e^{2}} \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]

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

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Sympy [B]  time = 1.47977, size = 226, normalized size = 2.11 \begin{align*} - \frac{\left (- A e + B d\right ) \log{\left (x + \frac{- A a e - A b d + 2 B a d - \frac{a^{2} e \left (- A e + B d\right )}{a e - b d} + \frac{2 a b d \left (- A e + B d\right )}{a e - b d} - \frac{b^{2} d^{2} \left (- A e + B d\right )}{e \left (a e - b d\right )}}{- 2 A b e + B a e + B b d} \right )}}{e \left (a e - b d\right )} + \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a e - A b d + 2 B a d + \frac{a^{2} e^{2} \left (- A b + B a\right )}{b \left (a e - b d\right )} - \frac{2 a d e \left (- A b + B a\right )}{a e - b d} + \frac{b d^{2} \left (- A b + B a\right )}{a e - b d}}{- 2 A b e + B a e + B b d} \right )}}{b \left (a e - b d\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)

[Out]

-(-A*e + B*d)*log(x + (-A*a*e - A*b*d + 2*B*a*d - a**2*e*(-A*e + B*d)/(a*e - b*d) + 2*a*b*d*(-A*e + B*d)/(a*e
- b*d) - b**2*d**2*(-A*e + B*d)/(e*(a*e - b*d)))/(-2*A*b*e + B*a*e + B*b*d))/(e*(a*e - b*d)) + (-A*b + B*a)*lo
g(x + (-A*a*e - A*b*d + 2*B*a*d + a**2*e**2*(-A*b + B*a)/(b*(a*e - b*d)) - 2*a*d*e*(-A*b + B*a)/(a*e - b*d) +
b*d**2*(-A*b + B*a)/(a*e - b*d))/(-2*A*b*e + B*a*e + B*b*d))/(b*(a*e - b*d))

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Giac [A]  time = 1.14964, size = 197, normalized size = 1.84 \begin{align*} \frac{B e^{\left (-1\right )} \log \left ({\left | b x^{2} e + b d x + a x e + a d \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{2 \, b} - \frac{{\left (B b d \mathrm{sgn}\left (b x + a\right ) + B a e \mathrm{sgn}\left (b x + a\right ) - 2 \, A b e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \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 )}{2 \, b{\left | b d - a e \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]

1/2*B*e^(-1)*log(abs(b*x^2*e + b*d*x + a*x*e + a*d))*sgn(b*x + a)/b - 1/2*(B*b*d*sgn(b*x + a) + B*a*e*sgn(b*x
+ a) - 2*A*b*e*sgn(b*x + a))*e^(-1)*log(abs(2*b*x*e + b*d + a*e - abs(b*d - a*e))/abs(2*b*x*e + b*d + a*e + ab
s(b*d - a*e)))/(b*abs(b*d - a*e))