Optimal. Leaf size=130 \[ \frac {2 (1+p) x \left (a+b x^{-\frac {1}{2 (1+p)}}\right ) \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a (1+2 p)}-\frac {x \left (a+b x^{-\frac {1}{2 (1+p)}}\right )^2 \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a^2 (1+2 p)} \]
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Rubi [A]
time = 0.04, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1357, 198, 197}
\begin {gather*} \frac {2 (p+1) x \left (a+b x^{-\frac {1}{2 (p+1)}}\right ) \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p}{a (2 p+1)}-\frac {x \left (a+b x^{-\frac {1}{2 (p+1)}}\right )^2 \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p}{a^2 (2 p+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 197
Rule 198
Rule 1357
Rubi steps
\begin {align*} \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx &=\left (\left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \left (2 a b+2 b^2 x^{-\frac {1}{2 (1+p)}}\right )^{-2 p}\right ) \int \left (2 a b+2 b^2 x^{-\frac {1}{2 (1+p)}}\right )^{2 p} \, dx\\ &=\frac {2 (1+p) x \left (a+b x^{-\frac {1}{2 (1+p)}}\right ) \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a (1+2 p)}-\frac {\left (\left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \left (2 a b+2 b^2 x^{-\frac {1}{2 (1+p)}}\right )^{-2 p}\right ) \int \left (2 a b+2 b^2 x^{-\frac {1}{2 (1+p)}}\right )^{1+2 p} \, dx}{2 a b (1+2 p)}\\ &=\frac {2 (1+p) x \left (a+b x^{-\frac {1}{2 (1+p)}}\right ) \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a (1+2 p)}-\frac {x \left (a+b x^{-\frac {1}{2 (1+p)}}\right )^2 \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a^2 (1+2 p)}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 80, normalized size = 0.62 \begin {gather*} \frac {x^{\frac {p}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right ) \left (x^{-\frac {1}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right )^2\right )^p \left (-b+a (1+2 p) x^{\frac {1}{2+2 p}}\right )}{a^2 (1+2 p)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (a^{2}+b^{2} x^{-\frac {1}{1+p}}+2 a b \,x^{-\frac {1}{2 \left (1+p \right )}}\right )^{p}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 103, normalized size = 0.79 \begin {gather*} \frac {{\left (2 \, a b p x x^{\frac {1}{2 \, {\left (p + 1\right )}}} - b^{2} x + {\left (2 \, a^{2} p + a^{2}\right )} x x^{\left (\frac {1}{p + 1}\right )}\right )} \left (\frac {2 \, a b x^{\frac {1}{2 \, {\left (p + 1\right )}}} + a^{2} x^{\left (\frac {1}{p + 1}\right )} + b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )^{p}}{{\left (2 \, a^{2} p + a^{2}\right )} x^{\left (\frac {1}{p + 1}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {b^2}{x^{\frac {1}{p+1}}}+a^2+\frac {2\,a\,b}{x^{\frac {1}{2\,\left (p+1\right )}}}\right )}^p \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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