3.1770 \(\int \frac{(A+B x) (d+e x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{-2 a B e+A b e+b B d}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0988835, antiderivative size = 127, 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} \[ -\frac{(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{-2 a B e+A b e+b B d}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

-((b*B*d + A*b*e - 2*a*B*e)/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - ((A*b - a*B)*(b*d - a*e))/(2*b^3*(a + b*x)*
Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*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)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac{(A b-a B) (b d-a e)}{b^5 (a+b x)^3}+\frac{b B d+A b e-2 a B e}{b^5 (a+b x)^2}+\frac{B e}{b^5 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{b B d+A b e-2 a B e}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B 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 = 0.0483492, size = 86, normalized size = 0.68 \[ \frac{B \left (3 a^2 e-a b d+4 a b e x-2 b^2 d x\right )-A b (a e+b d+2 b e x)+2 B e (a+b x)^2 \log (a+b x)}{2 b^3 (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(-(A*b*(b*d + a*e + 2*b*e*x)) + B*(-(a*b*d) + 3*a^2*e - 2*b^2*d*x + 4*a*b*e*x) + 2*B*e*(a + b*x)^2*Log[a + b*x
])/(2*b^3*(a + b*x)*Sqrt[(a + b*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 109, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -2\,B\ln \left ( bx+a \right ){x}^{2}{b}^{2}e-4\,B\ln \left ( bx+a \right ) xabe+2\,Ax{b}^{2}e-2\,B\ln \left ( bx+a \right ){a}^{2}e-4\,Bxabe+2\,Bx{b}^{2}d+aAeb+Ad{b}^{2}-3\,Be{a}^{2}+Bdab \right ) \left ( bx+a \right ) }{2\,{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/2*(-2*B*ln(b*x+a)*x^2*b^2*e-4*B*ln(b*x+a)*x*a*b*e+2*A*x*b^2*e-2*B*ln(b*x+a)*a^2*e-4*B*x*a*b*e+2*B*x*b^2*d+a
*A*e*b+A*d*b^2-3*B*e*a^2+B*d*a*b)*(b*x+a)/b^3/((b*x+a)^2)^(3/2)

________________________________________________________________________________________

Maxima [A]  time = 0.995631, size = 185, normalized size = 1.46 \begin{align*} \frac{B e \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{3 \, B a^{2} b^{2} e}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, B a b e x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{B d + A e}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{A d}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{{\left (B d + A e\right )} a}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

B*e*log(x + a/b)/(b^2)^(3/2) + 3/2*B*a^2*b^2*e/((b^2)^(7/2)*(x + a/b)^2) + 2*B*a*b*e*x/((b^2)^(5/2)*(x + a/b)^
2) - (B*d + A*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 1/2*A*d/((b^2)^(3/2)*(x + a/b)^2) + 1/2*(B*d + A*e)*a/(
(b^2)^(3/2)*b*(x + a/b)^2)

________________________________________________________________________________________

Fricas [A]  time = 1.29137, size = 234, normalized size = 1.84 \begin{align*} -\frac{{\left (B a b + A b^{2}\right )} d -{\left (3 \, B a^{2} - A a b\right )} e + 2 \,{\left (B b^{2} d -{\left (2 \, B a b - A b^{2}\right )} e\right )} x - 2 \,{\left (B b^{2} e x^{2} + 2 \, B a b e x + B a^{2} e\right )} \log \left (b x + a\right )}{2 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/2*((B*a*b + A*b^2)*d - (3*B*a^2 - A*a*b)*e + 2*(B*b^2*d - (2*B*a*b - A*b^2)*e)*x - 2*(B*b^2*e*x^2 + 2*B*a*b
*e*x + B*a^2*e)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((A + B*x)*(d + e*x)/((a + b*x)**2)**(3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x