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

Optimal. Leaf size=77 \[ \frac {a^2 \log (a+b x)}{b (b c-a d)^2}+\frac {c^2}{d^2 (c+d x) (b c-a d)}+\frac {c (b c-2 a d) \log (c+d x)}{d^2 (b c-a d)^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {88} \begin {gather*} \frac {a^2 \log (a+b x)}{b (b c-a d)^2}+\frac {c^2}{d^2 (c+d x) (b c-a d)}+\frac {c (b c-2 a d) \log (c+d x)}{d^2 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x)*(c + d*x)^2),x]

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.00 \begin {gather*} \frac {a^2 d^2 (c+d x) \log (a+b x)+b c ((c+d x) (b c-2 a d) \log (c+d x)+c (b c-a d))}{b d^2 (c+d x) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.26, size = 148, normalized size = 1.92 \begin {gather*} \frac {b^{2} c^{3} - a b c^{2} d + {\left (a^{2} d^{3} x + a^{2} c d^{2}\right )} \log \left (b x + a\right ) + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + {\left (b^{2} c^{2} d - 2 \, a b c d^{2}\right )} x\right )} \log \left (d x + c\right )}{b^{3} c^{3} d^{2} - 2 \, a b^{2} c^{2} d^{3} + a^{2} b c d^{4} + {\left (b^{3} c^{2} d^{3} - 2 \, a b^{2} c d^{4} + a^{2} b d^{5}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.03, size = 114, normalized size = 1.48 \begin {gather*} \frac {a^{2} d \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}} + \frac {c^{2} d}{{\left (b c d^{3} - a d^{4}\right )} {\left (d x + c\right )}} - \frac {\log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{b d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 97, normalized size = 1.26 \begin {gather*} \frac {a^{2} \ln \left (b x +a \right )}{\left (a d -b c \right )^{2} b}-\frac {2 a c \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} d}+\frac {b \,c^{2} \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} d^{2}}-\frac {c^{2}}{\left (a d -b c \right ) \left (d x +c \right ) d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.03, size = 120, normalized size = 1.56 \begin {gather*} \frac {a^{2} \log \left (b x + a\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} + \frac {c^{2}}{b c^{2} d^{2} - a c d^{3} + {\left (b c d^{3} - a d^{4}\right )} x} + \frac {{\left (b c^{2} - 2 \, a c d\right )} \log \left (d x + c\right )}{b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(b*x + a)/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2) + c^2/(b*c^2*d^2 - a*c*d^3 + (b*c*d^3 - a*d^4)*x) + (b*c^
2 - 2*a*c*d)*log(d*x + c)/(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)

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mupad [B]  time = 0.22, size = 80, normalized size = 1.04 \begin {gather*} \frac {a^2\,\ln \left (a+b\,x\right )}{b\,{\left (a\,d-b\,c\right )}^2}-\frac {c^2}{d^2\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )}-\frac {c\,\ln \left (c+d\,x\right )\,\left (2\,a\,d-b\,c\right )}{d^2\,{\left (a\,d-b\,c\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x)*(c + d*x)^2),x)

[Out]

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

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sympy [B]  time = 2.23, size = 333, normalized size = 4.32 \begin {gather*} \frac {a^{2} \log {\left (x + \frac {\frac {a^{5} d^{4}}{b \left (a d - b c\right )^{2}} - \frac {3 a^{4} c d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{3} b c^{2} d^{2}}{\left (a d - b c\right )^{2}} - \frac {a^{2} b^{2} c^{3} d}{\left (a d - b c\right )^{2}} + 3 a^{2} c d - a b c^{2}}{a^{2} d^{2} + 2 a b c d - b^{2} c^{2}} \right )}}{b \left (a d - b c\right )^{2}} - \frac {c^{2}}{a c d^{3} - b c^{2} d^{2} + x \left (a d^{4} - b c d^{3}\right )} - \frac {c \left (2 a d - b c\right ) \log {\left (x + \frac {- \frac {a^{3} c d^{2} \left (2 a d - b c\right )}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b c^{2} d \left (2 a d - b c\right )}{\left (a d - b c\right )^{2}} + 3 a^{2} c d - \frac {3 a b^{2} c^{3} \left (2 a d - b c\right )}{\left (a d - b c\right )^{2}} - a b c^{2} + \frac {b^{3} c^{4} \left (2 a d - b c\right )}{d \left (a d - b c\right )^{2}}}{a^{2} d^{2} + 2 a b c d - b^{2} c^{2}} \right )}}{d^{2} \left (a d - b c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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