Optimal. Leaf size=135 \[ -\frac {1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {\left (a+b x^2\right )^{5/2} (2 a B+5 A b)}{10 a}+\frac {1}{6} \left (a+b x^2\right )^{3/2} (2 a B+5 A b)+\frac {1}{2} a \sqrt {a+b x^2} (2 a B+5 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2} \]
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Rubi [A] time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 78, 50, 63, 208} \[ -\frac {1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {\left (a+b x^2\right )^{5/2} (2 a B+5 A b)}{10 a}+\frac {1}{6} \left (a+b x^2\right )^{3/2} (2 a B+5 A b)+\frac {1}{2} a \sqrt {a+b x^2} (2 a B+5 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 78
Rule 208
Rule 446
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {\left (\frac {5 A b}{2}+a B\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,x^2\right )}{2 a}\\ &=\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} (5 A b+2 a B) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} (a (5 A b+2 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} \left (a^2 (5 A b+2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {\left (a^2 (5 A b+2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 b}\\ &=\frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}-\frac {1}{2} a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}
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Mathematica [A] time = 0.07, size = 105, normalized size = 0.78 \[ \frac {\sqrt {a+b x^2} \left (a^2 \left (46 B x^2-15 A\right )+a \left (70 A b x^2+22 b B x^4\right )+2 b^2 x^4 \left (5 A+3 B x^2\right )\right )}{30 x^2}-\frac {1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 221, normalized size = 1.64 \[ \left [\frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{6} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{60 \, x^{2}}, \frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, B b^{2} x^{6} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{2}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.37, size = 139, normalized size = 1.03 \[ \frac {6 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b + 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b + 30 \, \sqrt {b x^{2} + a} B a^{2} b + 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} + 60 \, \sqrt {b x^{2} + a} A a b^{2} - \frac {15 \, \sqrt {b x^{2} + a} A a^{2} b}{x^{2}} + \frac {15 \, {\left (2 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{30 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 161, normalized size = 1.19 \[ -\frac {5 A \,a^{\frac {3}{2}} b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}-B \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {5 \sqrt {b \,x^{2}+a}\, A a b}{2}+\sqrt {b \,x^{2}+a}\, B \,a^{2}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A b}{6}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B a}{3}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A b}{2 a}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{5}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{2 a \,x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.06, size = 138, normalized size = 1.02 \[ -B a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {5}{2} \, A a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + \sqrt {b x^{2} + a} B a^{2} + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} A a b - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{2 \, a x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.85, size = 132, normalized size = 0.98 \[ \frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5}+B\,a^2\,\sqrt {b\,x^2+a}+B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}+\frac {A\,b\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+2\,A\,a\,b\,\sqrt {b\,x^2+a}-\frac {A\,a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {A\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 68.24, size = 296, normalized size = 2.19 \[ - \frac {5 A a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {2 A a^{2} \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 A a b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A b^{2} \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) - B a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 B a b \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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