3.1936 \(\int \frac{(a+b x) (d+e x)^4}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=102 \[ \frac{e^3 x (4 b d-3 a e)}{b^4}+\frac{6 e^2 (b d-a e)^2 \log (a+b x)}{b^5}-\frac{4 e (b d-a e)^3}{b^5 (a+b x)}-\frac{(b d-a e)^4}{2 b^5 (a+b x)^2}+\frac{e^4 x^2}{2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0939681, antiderivative size = 102, 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 = {27, 43} \[ \frac{e^3 x (4 b d-3 a e)}{b^4}+\frac{6 e^2 (b d-a e)^2 \log (a+b x)}{b^5}-\frac{4 e (b d-a e)^3}{b^5 (a+b x)}-\frac{(b d-a e)^4}{2 b^5 (a+b x)^2}+\frac{e^4 x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^4}{(a+b x)^3} \, dx\\ &=\int \left (\frac{e^3 (4 b d-3 a e)}{b^4}+\frac{e^4 x}{b^3}+\frac{(b d-a e)^4}{b^4 (a+b x)^3}+\frac{4 e (b d-a e)^3}{b^4 (a+b x)^2}+\frac{6 e^2 (b d-a e)^2}{b^4 (a+b x)}\right ) \, dx\\ &=\frac{e^3 (4 b d-3 a e) x}{b^4}+\frac{e^4 x^2}{2 b^3}-\frac{(b d-a e)^4}{2 b^5 (a+b x)^2}-\frac{4 e (b d-a e)^3}{b^5 (a+b x)}+\frac{6 e^2 (b d-a e)^2 \log (a+b x)}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0599371, size = 163, normalized size = 1.6 \[ \frac{a^2 b^2 e^2 \left (18 d^2-16 d e x-11 e^2 x^2\right )+2 a^3 b e^3 (e x-10 d)+7 a^4 e^4-4 a b^3 e \left (-6 d^2 e x+d^3-4 d e^2 x^2+e^3 x^3\right )+12 e^2 (a+b x)^2 (b d-a e)^2 \log (a+b x)+b^4 \left (-8 d^3 e x-d^4+8 d e^3 x^3+e^4 x^4\right )}{2 b^5 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(7*a^4*e^4 + 2*a^3*b*e^3*(-10*d + e*x) + a^2*b^2*e^2*(18*d^2 - 16*d*e*x - 11*e^2*x^2) - 4*a*b^3*e*(d^3 - 6*d^2
*e*x - 4*d*e^2*x^2 + e^3*x^3) + b^4*(-d^4 - 8*d^3*e*x + 8*d*e^3*x^3 + e^4*x^4) + 12*e^2*(b*d - a*e)^2*(a + b*x
)^2*Log[a + b*x])/(2*b^5*(a + b*x)^2)

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Maple [B]  time = 0.009, size = 245, normalized size = 2.4 \begin{align*}{\frac{{e}^{4}{x}^{2}}{2\,{b}^{3}}}-3\,{\frac{a{e}^{4}x}{{b}^{4}}}+4\,{\frac{d{e}^{3}x}{{b}^{3}}}+4\,{\frac{{a}^{3}{e}^{4}}{{b}^{5} \left ( bx+a \right ) }}-12\,{\frac{{e}^{3}{a}^{2}d}{{b}^{4} \left ( bx+a \right ) }}+12\,{\frac{a{d}^{2}{e}^{2}}{{b}^{3} \left ( bx+a \right ) }}-4\,{\frac{e{d}^{3}}{{b}^{2} \left ( bx+a \right ) }}-{\frac{{a}^{4}{e}^{4}}{2\,{b}^{5} \left ( bx+a \right ) ^{2}}}+2\,{\frac{{a}^{3}d{e}^{3}}{{b}^{4} \left ( bx+a \right ) ^{2}}}-3\,{\frac{{a}^{2}{d}^{2}{e}^{2}}{{b}^{3} \left ( bx+a \right ) ^{2}}}+2\,{\frac{a{d}^{3}e}{{b}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{{d}^{4}}{2\,b \left ( bx+a \right ) ^{2}}}+6\,{\frac{{e}^{4}\ln \left ( bx+a \right ){a}^{2}}{{b}^{5}}}-12\,{\frac{{e}^{3}\ln \left ( bx+a \right ) ad}{{b}^{4}}}+6\,{\frac{{e}^{2}\ln \left ( bx+a \right ){d}^{2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

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

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Maxima [A]  time = 0.984445, size = 257, normalized size = 2.52 \begin{align*} -\frac{b^{4} d^{4} + 4 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 20 \, a^{3} b d e^{3} - 7 \, a^{4} e^{4} + 8 \,{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac{b e^{4} x^{2} + 2 \,{\left (4 \, b d e^{3} - 3 \, a e^{4}\right )} x}{2 \, b^{4}} + \frac{6 \,{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \log \left (b x + a\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

-1/2*(b^4*d^4 + 4*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 20*a^3*b*d*e^3 - 7*a^4*e^4 + 8*(b^4*d^3*e - 3*a*b^3*d^2*e
^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5) + 1/2*(b*e^4*x^2 + 2*(4*b*d*e^3 - 3*a*e^4
)*x)/b^4 + 6*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*log(b*x + a)/b^5

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Fricas [B]  time = 1.43212, size = 586, normalized size = 5.75 \begin{align*} \frac{b^{4} e^{4} x^{4} - b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 20 \, a^{3} b d e^{3} + 7 \, a^{4} e^{4} + 4 \,{\left (2 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} +{\left (16 \, a b^{3} d e^{3} - 11 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \,{\left (4 \, b^{4} d^{3} e - 12 \, a b^{3} d^{2} e^{2} + 8 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \,{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} +{\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \,{\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

1/2*(b^4*e^4*x^4 - b^4*d^4 - 4*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 20*a^3*b*d*e^3 + 7*a^4*e^4 + 4*(2*b^4*d*e^3
- a*b^3*e^4)*x^3 + (16*a*b^3*d*e^3 - 11*a^2*b^2*e^4)*x^2 - 2*(4*b^4*d^3*e - 12*a*b^3*d^2*e^2 + 8*a^2*b^2*d*e^3
- a^3*b*e^4)*x + 12*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*
x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [A]  time = 1.47961, size = 184, normalized size = 1.8 \begin{align*} \frac{7 a^{4} e^{4} - 20 a^{3} b d e^{3} + 18 a^{2} b^{2} d^{2} e^{2} - 4 a b^{3} d^{3} e - b^{4} d^{4} + x \left (8 a^{3} b e^{4} - 24 a^{2} b^{2} d e^{3} + 24 a b^{3} d^{2} e^{2} - 8 b^{4} d^{3} e\right )}{2 a^{2} b^{5} + 4 a b^{6} x + 2 b^{7} x^{2}} + \frac{e^{4} x^{2}}{2 b^{3}} - \frac{x \left (3 a e^{4} - 4 b d e^{3}\right )}{b^{4}} + \frac{6 e^{2} \left (a e - b d\right )^{2} \log{\left (a + b x \right )}}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

(7*a**4*e**4 - 20*a**3*b*d*e**3 + 18*a**2*b**2*d**2*e**2 - 4*a*b**3*d**3*e - b**4*d**4 + x*(8*a**3*b*e**4 - 24
*a**2*b**2*d*e**3 + 24*a*b**3*d**2*e**2 - 8*b**4*d**3*e))/(2*a**2*b**5 + 4*a*b**6*x + 2*b**7*x**2) + e**4*x**2
/(2*b**3) - x*(3*a*e**4 - 4*b*d*e**3)/b**4 + 6*e**2*(a*e - b*d)**2*log(a + b*x)/b**5

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Giac [A]  time = 1.13217, size = 232, normalized size = 2.27 \begin{align*} \frac{6 \,{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac{b^{3} x^{2} e^{4} + 8 \, b^{3} d x e^{3} - 6 \, a b^{2} x e^{4}}{2 \, b^{6}} - \frac{b^{4} d^{4} + 4 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 20 \, a^{3} b d e^{3} - 7 \, a^{4} e^{4} + 8 \,{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \,{\left (b x + a\right )}^{2} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

6*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*log(abs(b*x + a))/b^5 + 1/2*(b^3*x^2*e^4 + 8*b^3*d*x*e^3 - 6*a*b^2*x*e
^4)/b^6 - 1/2*(b^4*d^4 + 4*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 20*a^3*b*d*e^3 - 7*a^4*e^4 + 8*(b^4*d^3*e - 3*a*
b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)/((b*x + a)^2*b^5)