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

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

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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 = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {24, 43} \begin {gather*} \frac {(b c-a d) \log (a+b x)}{b^2}+\frac {d x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)/b + ((b*c - a*d)*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 c+(b c+a d) x+b d x^2}{(a+b x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 24, normalized size = 0.96 \begin {gather*} \frac {b d x + {\left (b c - a d\right )} \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 117, normalized size = 4.68 \begin {gather*} b d {\left (\frac {2 \, a \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b x + a}{b^{3}} - \frac {a^{2}}{{\left (b x + a\right )} b^{3}}\right )} - \frac {{\left (b c + a d\right )} {\left (\frac {\log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x + a\right )} b}\right )}}{b} - \frac {a c}{{\left (b x + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 32, normalized size = 1.28 \begin {gather*} -\frac {a d \ln \left (b x +a \right )}{b^{2}}+\frac {c \ln \left (b x +a \right )}{b}+\frac {d x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 25, normalized size = 1.00 \begin {gather*} \frac {d x}{b} + \frac {{\left (b c - a d\right )} \log \left (b x + a\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x/b + (b*c - a*d)*log(b*x + a)/b^2

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mupad [B]  time = 0.05, size = 26, normalized size = 1.04 \begin {gather*} \frac {d\,x}{b}-\frac {\ln \left (a+b\,x\right )\,\left (a\,d-b\,c\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.18, size = 20, normalized size = 0.80 \begin {gather*} \frac {d x}{b} - \frac {\left (a d - b c\right ) \log {\left (a + b x \right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x/b - (a*d - b*c)*log(a + b*x)/b**2

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