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

Optimal. Leaf size=25 \[ \frac {(b d-a e) \log (a+b x)}{b^2}+\frac {e x}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 43} \begin {gather*} \frac {(b d-a e) \log (a+b x)}{b^2}+\frac {e x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {(b d-a e) \log (a+b x)}{b^2}+\frac {e x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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)}{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))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

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

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fricas [A]  time = 0.38, size = 24, normalized size = 0.96 \begin {gather*} \frac {b e x + {\left (b d - a e\right )} \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 28, normalized size = 1.12 \begin {gather*} \frac {x e}{b} + \frac {{\left (b d - a e\right )} \log \left ({\left | b x + a \right |}\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 32, normalized size = 1.28 \begin {gather*} -\frac {a e \ln \left (b x +a \right )}{b^{2}}+\frac {d \ln \left (b x +a \right )}{b}+\frac {e x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} \frac {e x}{b} + \frac {{\left (b d - a e\right )} \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x/b + (b*d - a*e)*log(b*x + a)/b^2

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mupad [B]  time = 0.04, size = 26, normalized size = 1.04 \begin {gather*} \frac {e\,x}{b}-\frac {\ln \left (a+b\,x\right )\,\left (a\,e-b\,d\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.19, size = 20, normalized size = 0.80 \begin {gather*} \frac {e x}{b} - \frac {\left (a e - b d\right ) \log {\left (a + b x \right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x/b - (a*e - b*d)*log(a + b*x)/b**2

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