3.2037 \(\int \frac{(a+b x) (d+e x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac{d+e x}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{e}{6 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-(d + e*x)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - e/(6*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0276638, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {768, 607} \[ -\frac{d+e x}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{e}{6 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-(d + e*x)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - e/(6*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 768

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

Rule 607

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*(a + b*x + c*x^2)^(p + 1))/((2*p + 1)*(b + 2
*c*x)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

\begin{align*} \int \frac{(a+b x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac{d+e x}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{e \int \frac{1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac{d+e x}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{e}{6 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0220302, size = 32, normalized size = 0.47 \[ \frac{-a e-2 b d-3 b e x}{6 b^2 \left ((a+b x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-2*b*d - a*e - 3*b*e*x)/(6*b^2*((a + b*x)^2)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 35, normalized size = 0.5 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{2} \left ( 3\,bex+ae+2\,bd \right ) }{6\,{b}^{2}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/6*(b*x+a)^2/b^2*(3*b*e*x+a*e+2*b*d)/((b*x+a)^2)^(5/2)

________________________________________________________________________________________

Maxima [B]  time = 0.979275, size = 186, normalized size = 2.74 \begin{align*} -\frac{b d + a e}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{a^{2} b^{3} e}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, a b^{2} e}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{b e}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{a d}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{{\left (b d + a e\right )} a}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(b*d + a*e)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 1/4*a^2*b^3*e/((b^2)^(9/2)*(x + a/b)^4) + 2/3*a*b^2*e
/((b^2)^(7/2)*(x + a/b)^3) - 1/2*b*e/((b^2)^(5/2)*(x + a/b)^2) - 1/4*a*d/((b^2)^(5/2)*(x + a/b)^4) + 1/4*(b*d
+ a*e)*a/((b^2)^(5/2)*b*(x + a/b)^4)

________________________________________________________________________________________

Fricas [A]  time = 1.51839, size = 105, normalized size = 1.54 \begin{align*} -\frac{3 \, b e x + 2 \, b d + a e}{6 \,{\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\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)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(3*b*e*x + 2*b*d + a*e)/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)

________________________________________________________________________________________

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{5}{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)**(5/2),x)

[Out]

Integral((a + b*x)*(d + e*x)/((a + b*x)**2)**(5/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (e x + d\right )}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d 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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(e*x + d)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)