3.1779 \(\int \frac{(A+B x) (d+e x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{-2 a B e+A b e+b B d}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B e}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.100936, antiderivative size = 135, 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} \[ -\frac{-2 a B e+A b e+b B d}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B e}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

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]

-((A*b - a*B)*(b*d - a*e))/(4*b^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*B*d + A*b*e - 2*a*B*e)/(3*b^
3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (B*e)/(2*b^3*(a + b*x)*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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac{(A b-a B) (b d-a e)}{b^7 (a+b x)^5}+\frac{b B d+A b e-2 a B e}{b^7 (a+b x)^4}+\frac{B e}{b^7 (a+b x)^3}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b B d+A b e-2 a B e}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B e}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.051965, size = 75, normalized size = 0.56 \[ \frac{-B \left (a^2 e+a b (d+4 e x)+2 b^2 x (2 d+3 e x)\right )-A b (a e+3 b d+4 b e x)}{12 b^3 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(3*b*d + a*e + 4*b*e*x)) - B*(a^2*e + 2*b^2*x*(2*d + 3*e*x) + a*b*(d + 4*e*x)))/(12*b^3*(a + b*x)^3*Sqr
t[(a + b*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 77, normalized size = 0.6 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 6\,B{x}^{2}{b}^{2}e+4\,Ax{b}^{2}e+4\,Bxabe+4\,Bx{b}^{2}d+aAeb+3\,Ad{b}^{2}+Be{a}^{2}+Bdab \right ) }{12\,{b}^{3}} \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]

-1/12*(b*x+a)/b^3*(6*B*b^2*e*x^2+4*A*b^2*e*x+4*B*a*b*e*x+4*B*b^2*d*x+A*a*b*e+3*A*b^2*d+B*a^2*e+B*a*b*d)/((b*x+
a)^2)^(5/2)

________________________________________________________________________________________

Maxima [A]  time = 0.960257, size = 186, normalized size = 1.38 \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{B a^{2} b^{2} e}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, B a b 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*B*a^2*b^2*e/((b^2)^(9/2)*(x + a/b)^4) + 2/3*B*a*b
*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 = 1.29469, size = 223, normalized size = 1.65 \begin{align*} -\frac{6 \, B b^{2} e x^{2} +{\left (B a b + 3 \, A b^{2}\right )} d +{\left (B a^{2} + A a b\right )} e + 4 \,{\left (B b^{2} d +{\left (B a b + A b^{2}\right )} e\right )} x}{12 \,{\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(6*B*b^2*e*x^2 + (B*a*b + 3*A*b^2)*d + (B*a^2 + A*a*b)*e + 4*(B*b^2*d + (B*a*b + A*b^2)*e)*x)/(b^7*x^4 +
4*a*b^6*x^3 + 6*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*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{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*} \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)^(5/2),x, algorithm="giac")

[Out]

sage0*x