Optimal. Leaf size=103 \[ -\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}+\frac {\left (a+c x^2\right )^{5/2} (6 A+5 B x)}{30 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c} \]
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Rubi [A] time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {780, 195, 217, 206} \begin {gather*} -\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}+\frac {\left (a+c x^2\right )^{5/2} (6 A+5 B x)}{30 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 195
Rule 206
Rule 217
Rule 780
Rubi steps
\begin {align*} \int x (A+B x) \left (a+c x^2\right )^{3/2} \, dx &=\frac {(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac {(a B) \int \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac {\left (a^2 B\right ) \int \sqrt {a+c x^2} \, dx}{8 c}\\ &=-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac {\left (a^3 B\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c}\\ &=-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac {\left (a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{16 c}\\ &=-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}\\ \end {align*}
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Mathematica [A] time = 0.18, size = 107, normalized size = 1.04 \begin {gather*} \frac {\sqrt {a+c x^2} \left (\sqrt {c} \left (3 a^2 (16 A+5 B x)+2 a c x^2 (48 A+35 B x)+8 c^2 x^4 (6 A+5 B x)\right )-\frac {15 a^{5/2} B \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {\frac {c x^2}{a}+1}}\right )}{240 c^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.36, size = 101, normalized size = 0.98 \begin {gather*} \frac {a^3 B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 c^{3/2}}+\frac {\sqrt {a+c x^2} \left (48 a^2 A+15 a^2 B x+96 a A c x^2+70 a B c x^3+48 A c^2 x^4+40 B c^2 x^5\right )}{240 c} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 205, normalized size = 1.99 \begin {gather*} \left [\frac {15 \, B a^{3} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{480 \, c^{2}}, \frac {15 \, B a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{240 \, c^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 89, normalized size = 0.86 \begin {gather*} \frac {B a^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} + \frac {1}{240} \, \sqrt {c x^{2} + a} {\left (\frac {48 \, A a^{2}}{c} + {\left (\frac {15 \, B a^{2}}{c} + 2 \, {\left (48 \, A a + {\left (35 \, B a + 4 \, {\left (5 \, B c x + 6 \, A c\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 94, normalized size = 0.91 \begin {gather*} -\frac {B \,a^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+a}\, B \,a^{2} x}{16 c}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B x}{6 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{5 c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 86, normalized size = 0.83 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, c} - \frac {\sqrt {c x^{2} + a} B a^{2} x}{16 \, c} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, c} \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 x\,{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 12.11, size = 223, normalized size = 2.17 \begin {gather*} A a \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + A c \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {B a^{\frac {5}{2}} x}{16 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 B a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {11 B \sqrt {a} c x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {3}{2}}} + \frac {B c^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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