3.15.98 \(\int \frac {(a+b x)^4}{(a c+(b c+a d) x+b d x^2)^2} \, dx\)

Optimal. Leaf size=51 \[ -\frac {(b c-a d)^2}{d^3 (c+d x)}-\frac {2 b (b c-a d) \log (c+d x)}{d^3}+\frac {b^2 x}{d^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 51, 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 = {626, 43} \begin {gather*} -\frac {(b c-a d)^2}{d^3 (c+d x)}-\frac {2 b (b c-a d) \log (c+d x)}{d^3}+\frac {b^2 x}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]

[Out]

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

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])

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {(a+b x)^4}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx &=\int \frac {(a+b x)^2}{(c+d x)^2} \, dx\\ &=\int \left (\frac {b^2}{d^2}+\frac {(-b c+a d)^2}{d^2 (c+d x)^2}-\frac {2 b (b c-a d)}{d^2 (c+d x)}\right ) \, dx\\ &=\frac {b^2 x}{d^2}-\frac {(b c-a d)^2}{d^3 (c+d x)}-\frac {2 b (b c-a d) \log (c+d x)}{d^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.92 \begin {gather*} \frac {-\frac {(b c-a d)^2}{c+d x}+2 b (a d-b c) \log (c+d x)+b^2 d x}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^4}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]

[Out]

IntegrateAlgebraic[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2)^2, x]

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fricas [A]  time = 0.40, size = 92, normalized size = 1.80 \begin {gather*} \frac {b^{2} d^{2} x^{2} + b^{2} c d x - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} - 2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \log \left (d x + c\right )}{d^{4} x + c d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 65, normalized size = 1.27 \begin {gather*} \frac {b^{2} x}{d^{2}} - \frac {2 \, {\left (b^{2} c - a b d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{3}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{{\left (d x + c\right )} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 86, normalized size = 1.69 \begin {gather*} -\frac {a^{2}}{\left (d x +c \right ) d}+\frac {2 a b c}{\left (d x +c \right ) d^{2}}+\frac {2 a b \ln \left (d x +c \right )}{d^{2}}-\frac {b^{2} c^{2}}{\left (d x +c \right ) d^{3}}-\frac {2 b^{2} c \ln \left (d x +c \right )}{d^{3}}+\frac {b^{2} x}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x)

[Out]

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

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maxima [A]  time = 1.06, size = 67, normalized size = 1.31 \begin {gather*} \frac {b^{2} x}{d^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{4} x + c d^{3}} - \frac {2 \, {\left (b^{2} c - a b d\right )} \log \left (d x + c\right )}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.61, size = 71, normalized size = 1.39 \begin {gather*} \frac {b^2\,x}{d^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{d\,\left (x\,d^3+c\,d^2\right )}-\frac {\ln \left (c+d\,x\right )\,\left (2\,b^2\,c-2\,a\,b\,d\right )}{d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^4/(a*c + x*(a*d + b*c) + b*d*x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.40, size = 60, normalized size = 1.18 \begin {gather*} \frac {b^{2} x}{d^{2}} + \frac {2 b \left (a d - b c\right ) \log {\left (c + d x \right )}}{d^{3}} + \frac {- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}}{c d^{3} + d^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**4/(a*c+(a*d+b*c)*x+b*d*x**2)**2,x)

[Out]

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

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