3.1761 \(\int \frac{(A+B x) (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=134 \[ \frac{(a+b x) (A b-a B) (b d-a e) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) (A+B x)^2}{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x (a+b x) (b d-a e)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(B*(b*d - a*e)*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e*(a + b*x)*(A + B*x)^2)/(2*b*B*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + ((A*b - a*B)*(b*d - a*e)*(a + b*x)*Log[a + b*x])/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.078713, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {770, 77} \[ \frac{(a+b x) (A b-a B) (b d-a e) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) (A+B x)^2}{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x (a+b x) (b d-a e)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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]

(B*(b*d - a*e)*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e*(a + b*x)*(A + B*x)^2)/(2*b*B*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + ((A*b - a*B)*(b*d - a*e)*(a + b*x)*Log[a + b*x])/(b^3*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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) (d+e x)}{a b+b^2 x} \, 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)}{b^3}+\frac{(A b-a B) (b d-a e)}{b^3 (a+b x)}+\frac{e (A+B x)}{b^2}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{B (b d-a e) x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) (A+B x)^2}{2 b B \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) (b d-a e) (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0469599, size = 72, normalized size = 0.54 \[ \frac{(a+b x) (b x (b (2 A e+2 B d+B e x)-2 a B e)+2 (A b-a B) (b d-a e) \log (a+b x))}{2 b^3 \sqrt{(a+b x)^2}} \]

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)*(b*x*(-2*a*B*e + b*(2*B*d + 2*A*e + B*e*x)) + 2*(A*b - a*B)*(b*d - a*e)*Log[a + b*x]))/(2*b^3*Sqrt[
(a + b*x)^2])

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Maple [A]  time = 0.007, size = 104, normalized size = 0.8 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -B{x}^{2}{b}^{2}e+2\,A\ln \left ( bx+a \right ) abe-2\,A\ln \left ( bx+a \right ){b}^{2}d-2\,Ax{b}^{2}e-2\,B\ln \left ( bx+a \right ){a}^{2}e+2\,B\ln \left ( bx+a \right ) abd+2\,Bxabe-2\,Bx{b}^{2}d \right ) }{2\,{b}^{3}}{\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]

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

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Maxima [A]  time = 0.965489, size = 159, normalized size = 1.19 \begin{align*} \frac{B a^{2} b^{2} e \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{B a b e x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{B e x^{2}}{2 \, \sqrt{b^{2}}} + A \sqrt{\frac{1}{b^{2}}} d \log \left (x + \frac{a}{b}\right ) - \frac{{\left (B d + A e\right )} a \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{b} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}{\left (B d + A e\right )}}{b^{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="maxima")

[Out]

B*a^2*b^2*e*log(x + a/b)/(b^2)^(5/2) - B*a*b*e*x/(b^2)^(3/2) + 1/2*B*e*x^2/sqrt(b^2) + A*sqrt(b^(-2))*d*log(x
+ a/b) - (B*d + A*e)*a*sqrt(b^(-2))*log(x + a/b)/b + sqrt(b^2*x^2 + 2*a*b*x + a^2)*(B*d + A*e)/b^2

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Fricas [A]  time = 1.36369, size = 157, normalized size = 1.17 \begin{align*} \frac{B b^{2} e x^{2} + 2 \,{\left (B b^{2} d -{\left (B a b - A b^{2}\right )} e\right )} x - 2 \,{\left ({\left (B a b - A b^{2}\right )} d -{\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{2 \, b^{3}} \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]

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

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Sympy [A]  time = 0.474171, size = 53, normalized size = 0.4 \begin{align*} \frac{B e x^{2}}{2 b} - \frac{x \left (- A b e + B a e - B b d\right )}{b^{2}} + \frac{\left (- A b + B a\right ) \left (a e - b d\right ) \log{\left (a + b x \right )}}{b^{3}} \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]

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

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Giac [A]  time = 1.09746, size = 165, normalized size = 1.23 \begin{align*} \frac{B b x^{2} e \mathrm{sgn}\left (b x + a\right ) + 2 \, B b d x \mathrm{sgn}\left (b x + a\right ) - 2 \, B a x e \mathrm{sgn}\left (b x + a\right ) + 2 \, A b x e \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} - \frac{{\left (B a b d \mathrm{sgn}\left (b x + a\right ) - A b^{2} d \mathrm{sgn}\left (b x + a\right ) - B a^{2} e \mathrm{sgn}\left (b x + a\right ) + A a b e \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \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*b*x^2*e*sgn(b*x + a) + 2*B*b*d*x*sgn(b*x + a) - 2*B*a*x*e*sgn(b*x + a) + 2*A*b*x*e*sgn(b*x + a))/b^2 -
(B*a*b*d*sgn(b*x + a) - A*b^2*d*sgn(b*x + a) - B*a^2*e*sgn(b*x + a) + A*a*b*e*sgn(b*x + a))*log(abs(b*x + a))/
b^3