Optimal. Leaf size=130 \[ \frac {e^4 \log (a+b x)}{(b d-a e)^5}-\frac {e^4 \log (d+e x)}{(b d-a e)^5}+\frac {e^3}{(a+b x) (b d-a e)^4}-\frac {e^2}{2 (a+b x)^2 (b d-a e)^3}+\frac {e}{3 (a+b x)^3 (b d-a e)^2}-\frac {1}{4 (a+b x)^4 (b d-a e)} \]
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Rubi [A] time = 0.09, antiderivative size = 130, 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, 44} \begin {gather*} \frac {e^3}{(a+b x) (b d-a e)^4}-\frac {e^2}{2 (a+b x)^2 (b d-a e)^3}+\frac {e^4 \log (a+b x)}{(b d-a e)^5}-\frac {e^4 \log (d+e x)}{(b d-a e)^5}+\frac {e}{3 (a+b x)^3 (b d-a e)^2}-\frac {1}{4 (a+b x)^4 (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 44
Rubi steps
\begin {align*} \int \frac {a+b x}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {1}{(a+b x)^5 (d+e x)} \, dx\\ &=\int \left (\frac {b}{(b d-a e) (a+b x)^5}-\frac {b e}{(b d-a e)^2 (a+b x)^4}+\frac {b e^2}{(b d-a e)^3 (a+b x)^3}-\frac {b e^3}{(b d-a e)^4 (a+b x)^2}+\frac {b e^4}{(b d-a e)^5 (a+b x)}-\frac {e^5}{(b d-a e)^5 (d+e x)}\right ) \, dx\\ &=-\frac {1}{4 (b d-a e) (a+b x)^4}+\frac {e}{3 (b d-a e)^2 (a+b x)^3}-\frac {e^2}{2 (b d-a e)^3 (a+b x)^2}+\frac {e^3}{(b d-a e)^4 (a+b x)}+\frac {e^4 \log (a+b x)}{(b d-a e)^5}-\frac {e^4 \log (d+e x)}{(b d-a e)^5}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 130, normalized size = 1.00 \begin {gather*} \frac {e^4 \log (a+b x)}{(b d-a e)^5}-\frac {e^4 \log (d+e x)}{(b d-a e)^5}+\frac {e^3}{(a+b x) (b d-a e)^4}-\frac {e^2}{2 (a+b x)^2 (b d-a e)^3}+\frac {e}{3 (a+b x)^3 (b d-a e)^2}+\frac {1}{4 (a+b x)^4 (a e-b d)} \end {gather*}
Antiderivative was successfully verified.
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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) \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [B] time = 0.41, size = 657, normalized size = 5.05 \begin {gather*} -\frac {3 \, b^{4} d^{4} - 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} - 48 \, a^{3} b d e^{3} + 25 \, a^{4} e^{4} - 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 8 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 18 \, a^{2} b^{2} d e^{3} - 13 \, a^{3} b e^{4}\right )} x - 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (a^{4} b^{5} d^{5} - 5 \, a^{5} b^{4} d^{4} e + 10 \, a^{6} b^{3} d^{3} e^{2} - 10 \, a^{7} b^{2} d^{2} e^{3} + 5 \, a^{8} b d e^{4} - a^{9} e^{5} + {\left (b^{9} d^{5} - 5 \, a b^{8} d^{4} e + 10 \, a^{2} b^{7} d^{3} e^{2} - 10 \, a^{3} b^{6} d^{2} e^{3} + 5 \, a^{4} b^{5} d e^{4} - a^{5} b^{4} e^{5}\right )} x^{4} + 4 \, {\left (a b^{8} d^{5} - 5 \, a^{2} b^{7} d^{4} e + 10 \, a^{3} b^{6} d^{3} e^{2} - 10 \, a^{4} b^{5} d^{2} e^{3} + 5 \, a^{5} b^{4} d e^{4} - a^{6} b^{3} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{7} d^{5} - 5 \, a^{3} b^{6} d^{4} e + 10 \, a^{4} b^{5} d^{3} e^{2} - 10 \, a^{5} b^{4} d^{2} e^{3} + 5 \, a^{6} b^{3} d e^{4} - a^{7} b^{2} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{6} d^{5} - 5 \, a^{4} b^{5} d^{4} e + 10 \, a^{5} b^{4} d^{3} e^{2} - 10 \, a^{6} b^{3} d^{2} e^{3} + 5 \, a^{7} b^{2} d e^{4} - a^{8} b e^{5}\right )} x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 322, normalized size = 2.48 \begin {gather*} \frac {b e^{4} \log \left ({\left | b x + a \right |}\right )}{b^{6} d^{5} - 5 \, a b^{5} d^{4} e + 10 \, a^{2} b^{4} d^{3} e^{2} - 10 \, a^{3} b^{3} d^{2} e^{3} + 5 \, a^{4} b^{2} d e^{4} - a^{5} b e^{5}} - \frac {e^{5} \log \left ({\left | x e + d \right |}\right )}{b^{5} d^{5} e - 5 \, a b^{4} d^{4} e^{2} + 10 \, a^{2} b^{3} d^{3} e^{3} - 10 \, a^{3} b^{2} d^{2} e^{4} + 5 \, a^{4} b d e^{5} - a^{5} e^{6}} - \frac {3 \, b^{4} d^{4} - 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} - 48 \, a^{3} b d e^{3} + 25 \, a^{4} e^{4} - 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 8 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 18 \, a^{2} b^{2} d e^{3} - 13 \, a^{3} b e^{4}\right )} x}{12 \, {\left (b d - a e\right )}^{5} {\left (b x + a\right )}^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 125, normalized size = 0.96 \begin {gather*} -\frac {e^{4} \ln \left (b x +a \right )}{\left (a e -b d \right )^{5}}+\frac {e^{4} \ln \left (e x +d \right )}{\left (a e -b d \right )^{5}}+\frac {e^{3}}{\left (a e -b d \right )^{4} \left (b x +a \right )}+\frac {e^{2}}{2 \left (a e -b d \right )^{3} \left (b x +a \right )^{2}}+\frac {e}{3 \left (a e -b d \right )^{2} \left (b x +a \right )^{3}}+\frac {1}{4 \left (a e -b d \right ) \left (b x +a \right )^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.78, size = 558, normalized size = 4.29 \begin {gather*} \frac {e^{4} \log \left (b x + a\right )}{b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}} - \frac {e^{4} \log \left (e x + d\right )}{b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}} + \frac {12 \, b^{3} e^{3} x^{3} - 3 \, b^{3} d^{3} + 13 \, a b^{2} d^{2} e - 23 \, a^{2} b d e^{2} + 25 \, a^{3} e^{3} - 6 \, {\left (b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 4 \, {\left (b^{3} d^{2} e - 5 \, a b^{2} d e^{2} + 13 \, a^{2} b e^{3}\right )} x}{12 \, {\left (a^{4} b^{4} d^{4} - 4 \, a^{5} b^{3} d^{3} e + 6 \, a^{6} b^{2} d^{2} e^{2} - 4 \, a^{7} b d e^{3} + a^{8} e^{4} + {\left (b^{8} d^{4} - 4 \, a b^{7} d^{3} e + 6 \, a^{2} b^{6} d^{2} e^{2} - 4 \, a^{3} b^{5} d e^{3} + a^{4} b^{4} e^{4}\right )} x^{4} + 4 \, {\left (a b^{7} d^{4} - 4 \, a^{2} b^{6} d^{3} e + 6 \, a^{3} b^{5} d^{2} e^{2} - 4 \, a^{4} b^{4} d e^{3} + a^{5} b^{3} e^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} d^{4} - 4 \, a^{3} b^{5} d^{3} e + 6 \, a^{4} b^{4} d^{2} e^{2} - 4 \, a^{5} b^{3} d e^{3} + a^{6} b^{2} e^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} d^{4} - 4 \, a^{4} b^{4} d^{3} e + 6 \, a^{5} b^{3} d^{2} e^{2} - 4 \, a^{6} b^{2} d e^{3} + a^{7} b e^{4}\right )} x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.27, size = 505, normalized size = 3.88 \begin {gather*} \frac {\frac {25\,a^3\,e^3-23\,a^2\,b\,d\,e^2+13\,a\,b^2\,d^2\,e-3\,b^3\,d^3}{12\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}+\frac {b^3\,e^3\,x^3}{a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4}-\frac {e^2\,x^2\,\left (b^3\,d-7\,a\,b^2\,e\right )}{2\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}+\frac {e\,x\,\left (13\,a^2\,b\,e^2-5\,a\,b^2\,d\,e+b^3\,d^2\right )}{3\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4}-\frac {2\,e^4\,\mathrm {atanh}\left (\frac {a^5\,e^5-3\,a^4\,b\,d\,e^4+2\,a^3\,b^2\,d^2\,e^3+2\,a^2\,b^3\,d^3\,e^2-3\,a\,b^4\,d^4\,e+b^5\,d^5}{{\left (a\,e-b\,d\right )}^5}+\frac {2\,b\,e\,x\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^5}\right )}{{\left (a\,e-b\,d\right )}^5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 2.14, size = 802, normalized size = 6.17 \begin {gather*} \frac {e^{4} \log {\left (x + \frac {- \frac {a^{6} e^{10}}{\left (a e - b d\right )^{5}} + \frac {6 a^{5} b d e^{9}}{\left (a e - b d\right )^{5}} - \frac {15 a^{4} b^{2} d^{2} e^{8}}{\left (a e - b d\right )^{5}} + \frac {20 a^{3} b^{3} d^{3} e^{7}}{\left (a e - b d\right )^{5}} - \frac {15 a^{2} b^{4} d^{4} e^{6}}{\left (a e - b d\right )^{5}} + \frac {6 a b^{5} d^{5} e^{5}}{\left (a e - b d\right )^{5}} + a e^{5} - \frac {b^{6} d^{6} e^{4}}{\left (a e - b d\right )^{5}} + b d e^{4}}{2 b e^{5}} \right )}}{\left (a e - b d\right )^{5}} - \frac {e^{4} \log {\left (x + \frac {\frac {a^{6} e^{10}}{\left (a e - b d\right )^{5}} - \frac {6 a^{5} b d e^{9}}{\left (a e - b d\right )^{5}} + \frac {15 a^{4} b^{2} d^{2} e^{8}}{\left (a e - b d\right )^{5}} - \frac {20 a^{3} b^{3} d^{3} e^{7}}{\left (a e - b d\right )^{5}} + \frac {15 a^{2} b^{4} d^{4} e^{6}}{\left (a e - b d\right )^{5}} - \frac {6 a b^{5} d^{5} e^{5}}{\left (a e - b d\right )^{5}} + a e^{5} + \frac {b^{6} d^{6} e^{4}}{\left (a e - b d\right )^{5}} + b d e^{4}}{2 b e^{5}} \right )}}{\left (a e - b d\right )^{5}} + \frac {25 a^{3} e^{3} - 23 a^{2} b d e^{2} + 13 a b^{2} d^{2} e - 3 b^{3} d^{3} + 12 b^{3} e^{3} x^{3} + x^{2} \left (42 a b^{2} e^{3} - 6 b^{3} d e^{2}\right ) + x \left (52 a^{2} b e^{3} - 20 a b^{2} d e^{2} + 4 b^{3} d^{2} e\right )}{12 a^{8} e^{4} - 48 a^{7} b d e^{3} + 72 a^{6} b^{2} d^{2} e^{2} - 48 a^{5} b^{3} d^{3} e + 12 a^{4} b^{4} d^{4} + x^{4} \left (12 a^{4} b^{4} e^{4} - 48 a^{3} b^{5} d e^{3} + 72 a^{2} b^{6} d^{2} e^{2} - 48 a b^{7} d^{3} e + 12 b^{8} d^{4}\right ) + x^{3} \left (48 a^{5} b^{3} e^{4} - 192 a^{4} b^{4} d e^{3} + 288 a^{3} b^{5} d^{2} e^{2} - 192 a^{2} b^{6} d^{3} e + 48 a b^{7} d^{4}\right ) + x^{2} \left (72 a^{6} b^{2} e^{4} - 288 a^{5} b^{3} d e^{3} + 432 a^{4} b^{4} d^{2} e^{2} - 288 a^{3} b^{5} d^{3} e + 72 a^{2} b^{6} d^{4}\right ) + x \left (48 a^{7} b e^{4} - 192 a^{6} b^{2} d e^{3} + 288 a^{5} b^{3} d^{2} e^{2} - 192 a^{4} b^{4} d^{3} e + 48 a^{3} b^{5} d^{4}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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