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

Optimal. Leaf size=49 \[ \frac {(b d-a e)^2 \log (a+b x)}{b^3}+\frac {e x (b d-a e)}{b^2}+\frac {(d+e x)^2}{2 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 49, 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} \begin {gather*} \frac {e x (b d-a e)}{b^2}+\frac {(b d-a e)^2 \log (a+b x)}{b^3}+\frac {(d+e x)^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.88 \begin {gather*} \frac {b e x (-2 a e+4 b d+b e x)+2 (b d-a e)^2 \log (a+b x)}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) (d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 63, normalized size = 1.29 \begin {gather*} \frac {b^{2} e^{2} x^{2} + 2 \, {\left (2 \, b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end {gather*}

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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 59, normalized size = 1.20 \begin {gather*} \frac {b x^{2} e^{2} + 4 \, b d x e - 2 \, a x e^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end {gather*}

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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 74, normalized size = 1.51 \begin {gather*} \frac {e^{2} x^{2}}{2 b}+\frac {a^{2} e^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {2 a d e \ln \left (b x +a \right )}{b^{2}}-\frac {a \,e^{2} x}{b^{2}}+\frac {d^{2} \ln \left (b x +a \right )}{b}+\frac {2 d e x}{b} \end {gather*}

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),x)

[Out]

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

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maxima [A]  time = 0.49, size = 61, normalized size = 1.24 \begin {gather*} \frac {b e^{2} x^{2} + 2 \, {\left (2 \, b d e - a e^{2}\right )} x}{2 \, b^{2}} + \frac {{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{b^{3}} \end {gather*}

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 62, normalized size = 1.27 \begin {gather*} \frac {\ln \left (a+b\,x\right )\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{b^3}-x\,\left (\frac {a\,e^2}{b^2}-\frac {2\,d\,e}{b}\right )+\frac {e^2\,x^2}{2\,b} \end {gather*}

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),x)

[Out]

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

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sympy [A]  time = 0.26, size = 44, normalized size = 0.90 \begin {gather*} x \left (- \frac {a e^{2}}{b^{2}} + \frac {2 d e}{b}\right ) + \frac {e^{2} x^{2}}{2 b} + \frac {\left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{3}} \end {gather*}

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),x)

[Out]

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

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