3.1778 \(\int \frac{(A+B x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{(d+e x)^4 (A b-a B)}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{(d+e x)^3 (B d-A e)}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.0624928, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {769, 646, 37} \[ -\frac{(d+e x)^4 (A b-a B)}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{(d+e x)^3 (B d-A e)}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*d - A*e)*(d + e*x)^3)/(3*(b*d - a*e)^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - ((A*b - a*B)*(d + e*x)^4)/(4*(b
*d - a*e)^2*(a + b*x)^3*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0815082, size = 142, normalized size = 1.34 \[ \frac{-A b \left (a^2 e^2+2 a b e (d+2 e x)+b^2 \left (3 d^2+8 d e x+6 e^2 x^2\right )\right )-B \left (2 a^2 b e (d+6 e x)+3 a^3 e^2+a b^2 \left (d^2+8 d e x+18 e^2 x^2\right )+4 b^3 x \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{12 b^4 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(a^2*e^2 + 2*a*b*e*(d + 2*e*x) + b^2*(3*d^2 + 8*d*e*x + 6*e^2*x^2))) - B*(3*a^3*e^2 + 2*a^2*b*e*(d + 6*
e*x) + 4*b^3*x*(d^2 + 3*d*e*x + 3*e^2*x^2) + a*b^2*(d^2 + 8*d*e*x + 18*e^2*x^2)))/(12*b^4*(a + b*x)^3*Sqrt[(a
+ b*x)^2])

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Maple [A]  time = 0.008, size = 174, normalized size = 1.6 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 12\,B{x}^{3}{b}^{3}{e}^{2}+6\,A{x}^{2}{b}^{3}{e}^{2}+18\,B{x}^{2}a{b}^{2}{e}^{2}+12\,B{x}^{2}{b}^{3}de+4\,Axa{b}^{2}{e}^{2}+8\,Ax{b}^{3}de+12\,Bx{a}^{2}b{e}^{2}+8\,Bxa{b}^{2}de+4\,Bx{b}^{3}{d}^{2}+Ab{a}^{2}{e}^{2}+2\,Aa{b}^{2}de+3\,A{b}^{3}{d}^{2}+3\,B{a}^{3}{e}^{2}+2\,B{a}^{2}bde+Ba{b}^{2}{d}^{2} \right ) }{12\,{b}^{4}} \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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(b*x+a)*(12*B*b^3*e^2*x^3+6*A*b^3*e^2*x^2+18*B*a*b^2*e^2*x^2+12*B*b^3*d*e*x^2+4*A*a*b^2*e^2*x+8*A*b^3*d*
e*x+12*B*a^2*b*e^2*x+8*B*a*b^2*d*e*x+4*B*b^3*d^2*x+A*a^2*b*e^2+2*A*a*b^2*d*e+3*A*b^3*d^2+3*B*a^3*e^2+2*B*a^2*b
*d*e+B*a*b^2*d^2)/b^4/((b*x+a)^2)^(5/2)

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Maxima [B]  time = 0.996534, size = 448, normalized size = 4.23 \begin{align*} -\frac{B e^{2} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{2 \, B a^{2} e^{2}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{4}} - \frac{B a^{3} b e^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, B a^{2} e^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{B a e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{B d^{2} + 2 \, A d e}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{{\left (2 \, B d e + A e^{2}\right )} a^{2} b^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} - \frac{A d^{2}}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{B a^{3} e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \,{\left (2 \, B d e + A e^{2}\right )} a b}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{2 \, B d e + A e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{{\left (B d^{2} + 2 \, A d 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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.32759, size = 392, normalized size = 3.7 \begin{align*} -\frac{12 \, B b^{3} e^{2} x^{3} +{\left (B a b^{2} + 3 \, A b^{3}\right )} d^{2} + 2 \,{\left (B a^{2} b + A a b^{2}\right )} d e +{\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + 6 \,{\left (2 \, B b^{3} d e +{\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 4 \,{\left (B b^{3} d^{2} + 2 \,{\left (B a b^{2} + A b^{3}\right )} d e +{\left (3 \, B a^{2} b + A a b^{2}\right )} e^{2}\right )} x}{12 \,{\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(12*B*b^3*e^2*x^3 + (B*a*b^2 + 3*A*b^3)*d^2 + 2*(B*a^2*b + A*a*b^2)*d*e + (3*B*a^3 + A*a^2*b)*e^2 + 6*(2
*B*b^3*d*e + (3*B*a*b^2 + A*b^3)*e^2)*x^2 + 4*(B*b^3*d^2 + 2*(B*a*b^2 + A*b^3)*d*e + (3*B*a^2*b + A*a*b^2)*e^2
)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)

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

[Out]

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

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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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x