3.1938 \(\int \frac {(a+b x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=59 \[ -\frac {2 e (b d-a e)}{b^3 (a+b x)}-\frac {(b d-a e)^2}{2 b^3 (a+b x)^2}+\frac {e^2 \log (a+b x)}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 e (b d-a e)}{b^3 (a+b x)}-\frac {(b d-a e)^2}{2 b^3 (a+b x)^2}+\frac {e^2 \log (a+b x)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.83 \[ \frac {2 e^2 \log (a+b x)-\frac {(b d-a e) (3 a e+b (d+4 e x))}{(a+b x)^2}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 99, normalized size = 1.68 \[ -\frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 67, normalized size = 1.14 \[ \frac {e^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {4 \, {\left (b d e - a e^{2}\right )} x + \frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

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)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 92, normalized size = 1.56 \[ -\frac {a^{2} e^{2}}{2 \left (b x +a \right )^{2} b^{3}}+\frac {a d e}{\left (b x +a \right )^{2} b^{2}}-\frac {d^{2}}{2 \left (b x +a \right )^{2} b}+\frac {2 a \,e^{2}}{\left (b x +a \right ) b^{3}}-\frac {2 d e}{\left (b x +a \right ) b^{2}}+\frac {e^{2} \ln \left (b x +a \right )}{b^{3}} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 0.58, size = 79, normalized size = 1.34 \[ -\frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {e^{2} \log \left (b x + a\right )}{b^{3}} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 77, normalized size = 1.31 \[ \frac {e^2\,\ln \left (a+b\,x\right )}{b^3}-\frac {\frac {-3\,a^2\,e^2+2\,a\,b\,d\,e+b^2\,d^2}{2\,b^3}-\frac {2\,e\,x\,\left (a\,e-b\,d\right )}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 80, normalized size = 1.36 \[ \frac {3 a^{2} e^{2} - 2 a b d e - b^{2} d^{2} + x \left (4 a b e^{2} - 4 b^{2} d e\right )}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {e^{2} \log {\left (a + b x \right )}}{b^{3}} \]

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)**2,x)

[Out]

(3*a**2*e**2 - 2*a*b*d*e - b**2*d**2 + x*(4*a*b*e**2 - 4*b**2*d*e))/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) +
e**2*log(a + b*x)/b**3

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