3.1764 \(\int \frac{A+B x}{(d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (B d-A e)}{(d+e x) (b d-a e)^2}+\frac{(a+b x) (A b-a B) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{(a+b x) (A b-a B) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[Out]

((B*d - A*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/((b*d - a*e)^2*(d + e*x)) + ((A*b - a*B)*(a + b*x)*Log[a + b*x])/(
(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((A*b - a*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 = 0.0808597, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {769, 646, 36, 31} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (B d-A e)}{(d+e x) (b d-a e)^2}+\frac{(a+b x) (A b-a B) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{(a+b x) (A b-a B) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*d - A*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/((b*d - a*e)^2*(d + e*x)) + ((A*b - a*B)*(a + b*x)*Log[a + b*x])/(
(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((A*b - a*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 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

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{A+B x}{(d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{(B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac{(A b-a B) \int \frac{1}{(d+e x) \sqrt{a^2+2 a b x+b^2 x^2}} \, dx}{b d-a e}\\ &=\frac{(B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac{\left ((A b-a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)} \, dx}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac{\left (b (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{a b+b^2 x} \, dx}{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left ((A b-a B) e \left (a b+b^2 x\right )\right ) \int \frac{1}{d+e x} \, dx}{b (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac{(A b-a 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{(A b-a 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 = 0.128731, size = 97, normalized size = 0.66 \[ \frac{(a+b x) \left (\frac{B d-A e}{e (d+e x) (a e-b d)}+\frac{(A b-a B) \log (a+b x)}{(b d-a e)^2}+\frac{(a B-A b) \log (d+e x)}{(b d-a e)^2}\right )}{\sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 162, normalized size = 1.1 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( A\ln \left ( ex+d \right ) xb{e}^{2}-A\ln \left ( bx+a \right ) xb{e}^{2}-B\ln \left ( ex+d \right ) xa{e}^{2}+B\ln \left ( bx+a \right ) xa{e}^{2}+A\ln \left ( ex+d \right ) bde-A\ln \left ( bx+a \right ) bde-B\ln \left ( ex+d \right ) ade+B\ln \left ( bx+a \right ) ade+aA{e}^{2}-Abde-aBde+Bb{d}^{2} \right ) }{ \left ( ae-bd \right ) ^{2}e \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((B*x+A)/(e*x+d)^2/((b*x+a)^2)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.26387, size = 308, normalized size = 2.08 \begin{align*} -\frac{B b d^{2} + A a e^{2} -{\left (B a + A b\right )} d e +{\left ({\left (B a - A b\right )} e^{2} x +{\left (B a - A b\right )} d e\right )} \log \left (b x + a\right ) -{\left ({\left (B a - A b\right )} e^{2} x +{\left (B a - A b\right )} d e\right )} \log \left (e x + d\right )}{b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3} +{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e + ((B*a - A*b)*e^2*x + (B*a - A*b)*d*e)*log(b*x + a) - ((B*a - A*b)*e^2*
x + (B*a - A*b)*d*e)*log(e*x + d))/(b^2*d^3*e - 2*a*b*d^2*e^2 + a^2*d*e^3 + (b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e
^4)*x)

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Sympy [B]  time = 1.54498, size = 355, normalized size = 2.4 \begin{align*} \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b e - A b^{2} d + B a^{2} e + B a b d - \frac{a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac{3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac{3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac{b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} - \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b e - A b^{2} d + B a^{2} e + B a b d + \frac{a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac{3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac{3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac{b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} + \frac{- A e + B d}{a d e^{2} - b d^{2} e + x \left (a e^{3} - b d e^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11963, size = 257, normalized size = 1.74 \begin{align*} -\frac{{\left (B a b \mathrm{sgn}\left (b x + a\right ) - A b^{2} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} + \frac{{\left (B a e \mathrm{sgn}\left (b x + a\right ) - A b e \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} - \frac{{\left (B b d^{2} \mathrm{sgn}\left (b x + a\right ) - B a d e \mathrm{sgn}\left (b x + a\right ) - A b d e \mathrm{sgn}\left (b x + a\right ) + A a e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}}{{\left (b d - a e\right )}^{2}{\left (x e + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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