Optimal. Leaf size=132 \[ -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\frac {1}{16} a^2 \sqrt {a+c x^2} (16 A+5 B x)+\frac {1}{24} a \left (a+c x^2\right )^{3/2} (8 A+5 B x)+\frac {1}{30} \left (a+c x^2\right )^{5/2} (6 A+5 B x) \]
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Rubi [A] time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {815, 844, 217, 206, 266, 63, 208} \begin {gather*} \frac {1}{16} a^2 \sqrt {a+c x^2} (16 A+5 B x)-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\frac {1}{24} a \left (a+c x^2\right )^{3/2} (8 A+5 B x)+\frac {1}{30} \left (a+c x^2\right )^{5/2} (6 A+5 B x) \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 206
Rule 208
Rule 217
Rule 266
Rule 815
Rule 844
Rubi steps
\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x} \, dx &=\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {\int \frac {(6 a A c+5 a B c x) \left (a+c x^2\right )^{3/2}}{x} \, dx}{6 c}\\ &=\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {\int \frac {\left (24 a^2 A c^2+15 a^2 B c^2 x\right ) \sqrt {a+c x^2}}{x} \, dx}{24 c^2}\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+c x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {\int \frac {48 a^3 A c^3+15 a^3 B c^3 x}{x \sqrt {a+c x^2}} \, dx}{48 c^3}\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+c x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\left (a^3 A\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\frac {1}{16} \left (5 a^3 B\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+c x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {1}{2} \left (a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\frac {1}{16} \left (5 a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+c x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}+\frac {\left (a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c}\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+c x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+c x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+c x^2\right )^{5/2}+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 \sqrt {c}}-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}
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Mathematica [A] time = 0.33, size = 139, normalized size = 1.05 \begin {gather*} -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {5 a^{7/2} B \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 \sqrt {c} \sqrt {a+c x^2}}+\frac {1}{240} \sqrt {a+c x^2} \left (a^2 (368 A+165 B x)+2 a c x^2 (88 A+65 B x)+8 c^2 x^4 (6 A+5 B x)\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.47, size = 138, normalized size = 1.05 \begin {gather*} 2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {5 a^3 B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 \sqrt {c}}+\frac {1}{240} \sqrt {a+c x^2} \left (368 a^2 A+165 a^2 B x+176 a A c x^2+130 a B c x^3+48 A c^2 x^4+40 B c^2 x^5\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 539, normalized size = 4.08 \begin {gather*} \left [\frac {75 \, B a^{3} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 240 \, A a^{\frac {5}{2}} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{480 \, c}, -\frac {75 \, B a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 120 \, A a^{\frac {5}{2}} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{240 \, c}, \frac {480 \, A \sqrt {-a} a^{2} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 75 \, 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} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{480 \, c}, -\frac {75 \, B a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 240 \, A \sqrt {-a} a^{2} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 130 \, B a c^{2} x^{3} + 176 \, A a c^{2} x^{2} + 165 \, B a^{2} c x + 368 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{240 \, c}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 125, normalized size = 0.95 \begin {gather*} \frac {2 \, A a^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, B a^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, \sqrt {c}} + \frac {1}{240} \, {\left (368 \, A a^{2} + {\left (165 \, B a^{2} + 2 \, {\left (88 \, A a c + {\left (65 \, B a c + 4 \, {\left (5 \, B c^{2} x + 6 \, A c^{2}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} + a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 138, normalized size = 1.05 \begin {gather*} -A \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {5 B \,a^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 \sqrt {c}}+\frac {5 \sqrt {c \,x^{2}+a}\, B \,a^{2} x}{16}+\sqrt {c \,x^{2}+a}\, A \,a^{2}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A a}{3}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B x}{6}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 119, normalized size = 0.90 \begin {gather*} \frac {1}{6} \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B x + \frac {5}{24} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B a x + \frac {5}{16} \, \sqrt {c x^{2} + a} B a^{2} x + \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {c}} - A a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A + \frac {1}{3} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A a + \sqrt {c x^{2} + a} A a^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.43, size = 101, normalized size = 0.77 \begin {gather*} \frac {A\,{\left (c\,x^2+a\right )}^{5/2}}{5}+A\,a^2\,\sqrt {c\,x^2+a}+\frac {A\,a\,{\left (c\,x^2+a\right )}^{3/2}}{3}+\frac {B\,x\,{\left (c\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {c\,x^2}{a}\right )}{{\left (\frac {c\,x^2}{a}+1\right )}^{5/2}}+A\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 29.89, size = 323, normalized size = 2.45 \begin {gather*} - A a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )} + \frac {A a^{3}}{\sqrt {c} x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A a^{2} \sqrt {c} x}{\sqrt {\frac {a}{c x^{2}} + 1}} + 2 A a c \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^{2} \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 \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {3 B a^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {35 B a^{\frac {3}{2}} c x^{3}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 B \sqrt {a} c^{2} x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 \sqrt {c}} + \frac {B c^{3} 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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