Optimal. Leaf size=127 \[ \frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}+\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}-\frac {a \left (a+b x^2\right )^{3/2} (16 A+15 B x)}{120 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]
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Rubi [A] time = 0.08, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {833, 780, 195, 217, 206} \[ \frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}-\frac {a \left (a+b x^2\right )^{3/2} (16 A+15 B x)}{120 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]
Antiderivative was successfully verified.
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Rule 195
Rule 206
Rule 217
Rule 780
Rule 833
Rubi steps
\begin {align*} \int x^3 (A+B x) \sqrt {a+b x^2} \, dx &=\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {\int x^2 (-3 a B+6 A b x) \sqrt {a+b x^2} \, dx}{6 b}\\ &=\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {\int x (-12 a A b-15 a b B x) \sqrt {a+b x^2} \, dx}{30 b^2}\\ &=\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^2 B\right ) \int \sqrt {a+b x^2} \, dx}{8 b^2}\\ &=\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^3 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^2}\\ &=\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b^2}\\ &=\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 107, normalized size = 0.84 \[ \frac {\sqrt {a+b x^2} \left (\frac {15 a^{5/2} B \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} \left (-a^2 (32 A+15 B x)+2 a b x^2 (8 A+5 B x)+8 b^2 x^4 (6 A+5 B x)\right )\right )}{240 b^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.94, size = 206, normalized size = 1.62 \[ \left [\frac {15 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, -\frac {15 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.51, size = 93, normalized size = 0.73 \[ -\frac {B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} + \frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B x + 6 \, A\right )} x + \frac {5 \, B a}{b}\right )} x + \frac {8 \, A a}{b}\right )} x - \frac {15 \, B a^{2}}{b^{2}}\right )} x - \frac {32 \, A a^{2}}{b^{2}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 115, normalized size = 0.91 \[ \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,x^{3}}{6 b}+\frac {B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,x^{2}}{5 b}+\frac {\sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16 b^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B a x}{8 b^{2}}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A a}{15 b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.41, size = 107, normalized size = 0.84 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, b^{2}} \]
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 \[ \int x^3\,\sqrt {b\,x^2+a}\,\left (A+B\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 16.25, size = 192, normalized size = 1.51 \[ A \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - \frac {B a^{\frac {5}{2}} x}{16 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {3}{2}} x^{3}}{48 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B \sqrt {a} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {5}{2}}} + \frac {B b x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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