3.2036 \(\int \frac{(a+b x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{(d+e x)^3}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0236631, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.03, Rules used = {767} \[ -\frac{(d+e x)^3}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)} \]

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]

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

Rule 767

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

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{(d+e x)^3}{3 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0332619, size = 60, normalized size = 1.46 \[ \frac{-a^2 e^2-a b e (d+3 e x)+b^2 \left (-\left (d^2+3 d e x+3 e^2 x^2\right )\right )}{3 b^3 \left ((a+b x)^2\right )^{3/2}} \]

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

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Maple [A]  time = 0.007, size = 69, normalized size = 1.7 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{2} \left ( 3\,{x}^{2}{b}^{2}{e}^{2}+3\,xab{e}^{2}+3\,x{b}^{2}de+{a}^{2}{e}^{2}+abde+{b}^{2}{d}^{2} \right ) }{3\,{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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.01561, size = 440, normalized size = 10.73 \begin{align*} -\frac{e^{2} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b} - \frac{2 \, a^{2} e^{2}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{3}} - \frac{a^{3} b^{2} e^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, a^{2} b e^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{a e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\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{a^{3} e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b^{2}{\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]

-e^2*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b) - 2/3*a^2*e^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^3) - 1/4*a^3*b^2
*e^2/((b^2)^(9/2)*(x + a/b)^4) + 2/3*a^2*b*e^2/((b^2)^(7/2)*(x + a/b)^3) - 1/2*a*e^2/((b^2)^(5/2)*(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*a^3*e^2/((b^2)^(5/2)*b^2*(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 = 1.5933, size = 170, normalized size = 4.15 \begin{align*} -\frac{3 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + a b d e + a^{2} e^{2} + 3 \,{\left (b^{2} d e + a b e^{2}\right )} x}{3 \,{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\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]

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

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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*} \int \frac{{\left (b x + a\right )}{\left (e x + d\right )}^{2}}{{\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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

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