Optimal. Leaf size=136 \[ \frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a+b x^{2 (m+1)}}}\right )}{16 \sqrt {b} (m+1)}+\frac {5 a^2 x^{m+1} \sqrt {a+b x^{2 (m+1)}}}{16 (m+1)}+\frac {x^{m+1} \left (a+b x^{2 (m+1)}\right )^{5/2}}{6 (m+1)}+\frac {5 a x^{m+1} \left (a+b x^{2 (m+1)}\right )^{3/2}}{24 (m+1)} \]
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Rubi [A] time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {345, 195, 217, 206} \[ \frac {5 a^2 x^{m+1} \sqrt {a+b x^{2 (m+1)}}}{16 (m+1)}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a+b x^{2 (m+1)}}}\right )}{16 \sqrt {b} (m+1)}+\frac {x^{m+1} \left (a+b x^{2 (m+1)}\right )^{5/2}}{6 (m+1)}+\frac {5 a x^{m+1} \left (a+b x^{2 (m+1)}\right )^{3/2}}{24 (m+1)} \]
Antiderivative was successfully verified.
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Rule 195
Rule 206
Rule 217
Rule 345
Rubi steps
\begin {align*} \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b x^2\right )^{5/2} \, dx,x,x^{1+m}\right )}{1+m}\\ &=\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {(5 a) \operatorname {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,x^{1+m}\right )}{6 (1+m)}\\ &=\frac {5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^{1+m}\right )}{8 (1+m)}\\ &=\frac {5 a^2 x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{16 (1+m)}+\frac {5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {\left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^{1+m}\right )}{16 (1+m)}\\ &=\frac {5 a^2 x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{16 (1+m)}+\frac {5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {\left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{1+m}}{\sqrt {a+b x^{2+2 m}}}\right )}{16 (1+m)}\\ &=\frac {5 a^2 x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{16 (1+m)}+\frac {5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a+b x^{2 (1+m)}}}\right )}{16 \sqrt {b} (1+m)}\\ \end {align*}
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Mathematica [C] time = 0.09, size = 88, normalized size = 0.65 \[ \frac {a^2 x^{m+1} \sqrt {a+b x^{2 m+2}} \, _2F_1\left (-\frac {5}{2},\frac {m+1}{2 m+2};\frac {m+1}{2 m+2}+1;-\frac {b x^{2 m+2}}{a}\right )}{(m+1) \sqrt {\frac {b x^{2 m+2}}{a}+1}} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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 \[ \int {\left (b x^{2 \, m + 2} + a\right )}^{\frac {5}{2}} x^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.23, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2 m +2}+a \right )^{\frac {5}{2}} x^{m}\, dx \]
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 \[ \int {\left (b x^{2 \, m + 2} + a\right )}^{\frac {5}{2}} x^{m}\,{d x} \]
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 \[ \int x^m\,{\left (a+b\,x^{2\,m+2}\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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