Optimal. Leaf size=62 \[ \frac {x}{2 a b p}-\frac {x}{2 a \left (b+a e^{2 p x}\right ) p}-\frac {\log \left (b+a e^{2 p x}\right )}{4 a b p^2} \]
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Rubi [A]
time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2321, 2222,
2320, 36, 29, 31} \begin {gather*} -\frac {\log \left (a e^{2 p x}+b\right )}{4 a b p^2}+\frac {x}{2 a b p}-\frac {x}{2 a p \left (a e^{2 p x}+b\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 2222
Rule 2320
Rule 2321
Rubi steps
\begin {align*} \int \frac {x}{\left (b e^{-p x}+a e^{p x}\right )^2} \, dx &=\int \frac {e^{2 p x} x}{\left (b+a e^{2 p x}\right )^2} \, dx\\ &=-\frac {x}{2 a \left (b+a e^{2 p x}\right ) p}+\frac {\int \frac {1}{b+a e^{2 p x}} \, dx}{2 a p}\\ &=-\frac {x}{2 a \left (b+a e^{2 p x}\right ) p}+\frac {\text {Subst}\left (\int \frac {1}{x (b+a x)} \, dx,x,e^{2 p x}\right )}{4 a p^2}\\ &=-\frac {x}{2 a \left (b+a e^{2 p x}\right ) p}-\frac {\text {Subst}\left (\int \frac {1}{b+a x} \, dx,x,e^{2 p x}\right )}{4 b p^2}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 p x}\right )}{4 a b p^2}\\ &=\frac {x}{2 a b p}-\frac {x}{2 a \left (b+a e^{2 p x}\right ) p}-\frac {\log \left (b+a e^{2 p x}\right )}{4 a b p^2}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 49, normalized size = 0.79 \begin {gather*} \frac {\frac {2 e^{2 p x} p x}{b+a e^{2 p x}}-\frac {\log \left (b+a e^{2 p x}\right )}{a}}{4 b p^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 50, normalized size = 0.81
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (b +a \,{\mathrm e}^{2 p x}\right )}{4 b a}+\frac {p x \,{\mathrm e}^{2 p x}}{2 b \left (b +a \,{\mathrm e}^{2 p x}\right )}}{p^{2}}\) | \(50\) |
default | \(\frac {-\frac {\ln \left (b +a \,{\mathrm e}^{2 p x}\right )}{4 b a}+\frac {p x \,{\mathrm e}^{2 p x}}{2 b \left (b +a \,{\mathrm e}^{2 p x}\right )}}{p^{2}}\) | \(50\) |
norman | \(\frac {x \,{\mathrm e}^{2 p x}}{2 b p \left (b +a \,{\mathrm e}^{2 p x}\right )}-\frac {\ln \left (b +a \,{\mathrm e}^{2 p x}\right )}{4 a b \,p^{2}}\) | \(51\) |
risch | \(\frac {x}{2 a b p}-\frac {x}{2 a \left (b +a \,{\mathrm e}^{2 p x}\right ) p}-\frac {\ln \left ({\mathrm e}^{2 p x}+\frac {b}{a}\right )}{4 a b \,p^{2}}\) | \(57\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.36, size = 51, normalized size = 0.82 \begin {gather*} \frac {x e^{\left (2 \, p x\right )}}{2 \, {\left (a b p e^{\left (2 \, p x\right )} + b^{2} p\right )}} - \frac {\log \left (\frac {a e^{\left (2 \, p x\right )} + b}{a}\right )}{4 \, a b p^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.82, size = 58, normalized size = 0.94 \begin {gather*} \frac {2 \, a p x e^{\left (2 \, p x\right )} - {\left (a e^{\left (2 \, p x\right )} + b\right )} \log \left (a e^{\left (2 \, p x\right )} + b\right )}{4 \, {\left (a^{2} b p^{2} e^{\left (2 \, p x\right )} + a b^{2} p^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 49, normalized size = 0.79 \begin {gather*} - \frac {x}{2 a^{2} p e^{2 p x} + 2 a b p} + \frac {x}{2 a b p} - \frac {\log {\left (e^{2 p x} + \frac {b}{a} \right )}}{4 a b p^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.81, size = 74, normalized size = 1.19 \begin {gather*} \frac {2 \, a p x e^{\left (2 \, p x\right )} - a e^{\left (2 \, p x\right )} \log \left (-a e^{\left (2 \, p x\right )} - b\right ) - b \log \left (-a e^{\left (2 \, p x\right )} - b\right )}{4 \, {\left (a^{2} b p^{2} e^{\left (2 \, p x\right )} + a b^{2} p^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.41, size = 47, normalized size = 0.76 \begin {gather*} \frac {x\,{\mathrm {e}}^{2\,p\,x}}{2\,b\,p\,\left (b+a\,{\mathrm {e}}^{2\,p\,x}\right )}-\frac {\ln \left (b+a\,{\mathrm {e}}^{2\,p\,x}\right )}{4\,a\,b\,p^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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