Optimal. Leaf size=188 \[ \frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}}-\frac {3 a^3 x \sqrt {a+b x^2} (2 A b-a B)}{256 b^3}+\frac {a^2 x^3 \sqrt {a+b x^2} (2 A b-a B)}{128 b^2}+\frac {x^5 \left (a+b x^2\right )^{3/2} (2 A b-a B)}{16 b}+\frac {a x^5 \sqrt {a+b x^2} (2 A b-a B)}{32 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b} \]
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Rubi [A] time = 0.10, antiderivative size = 188, 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 = {459, 279, 321, 217, 206} \begin {gather*} \frac {a^2 x^3 \sqrt {a+b x^2} (2 A b-a B)}{128 b^2}-\frac {3 a^3 x \sqrt {a+b x^2} (2 A b-a B)}{256 b^3}+\frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}}+\frac {x^5 \left (a+b x^2\right )^{3/2} (2 A b-a B)}{16 b}+\frac {a x^5 \sqrt {a+b x^2} (2 A b-a B)}{32 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 217
Rule 279
Rule 321
Rule 459
Rubi steps
\begin {align*} \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx &=\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}-\frac {(-10 A b+5 a B) \int x^4 \left (a+b x^2\right )^{3/2} \, dx}{10 b}\\ &=\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {(3 a (2 A b-a B)) \int x^4 \sqrt {a+b x^2} \, dx}{16 b}\\ &=\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {\left (a^2 (2 A b-a B)\right ) \int \frac {x^4}{\sqrt {a+b x^2}} \, dx}{32 b}\\ &=\frac {a^2 (2 A b-a B) x^3 \sqrt {a+b x^2}}{128 b^2}+\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}-\frac {\left (3 a^3 (2 A b-a B)\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{128 b^2}\\ &=-\frac {3 a^3 (2 A b-a B) x \sqrt {a+b x^2}}{256 b^3}+\frac {a^2 (2 A b-a B) x^3 \sqrt {a+b x^2}}{128 b^2}+\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{256 b^3}\\ &=-\frac {3 a^3 (2 A b-a B) x \sqrt {a+b x^2}}{256 b^3}+\frac {a^2 (2 A b-a B) x^3 \sqrt {a+b x^2}}{128 b^2}+\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{256 b^3}\\ &=-\frac {3 a^3 (2 A b-a B) x \sqrt {a+b x^2}}{256 b^3}+\frac {a^2 (2 A b-a B) x^3 \sqrt {a+b x^2}}{128 b^2}+\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 150, normalized size = 0.80 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (15 a^4 B-10 a^3 b \left (3 A+B x^2\right )+4 a^2 b^2 x^2 \left (5 A+2 B x^2\right )+16 a b^3 x^4 \left (15 A+11 B x^2\right )+32 b^4 x^6 \left (5 A+4 B x^2\right )\right )-\frac {15 a^{7/2} (a B-2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{1280 b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.21, size = 150, normalized size = 0.80 \begin {gather*} \frac {3 \left (a^5 B-2 a^4 A b\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{256 b^{7/2}}+\frac {\sqrt {a+b x^2} \left (15 a^4 B x-30 a^3 A b x-10 a^3 b B x^3+20 a^2 A b^2 x^3+8 a^2 b^2 B x^5+240 a A b^3 x^5+176 a b^3 B x^7+160 A b^4 x^7+128 b^4 B x^9\right )}{1280 b^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.03, size = 299, normalized size = 1.59 \begin {gather*} \left [-\frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (128 \, B b^{5} x^{9} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2560 \, b^{4}}, \frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (128 \, B b^{5} x^{9} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{1280 \, b^{4}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 159, normalized size = 0.85 \begin {gather*} \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B b x^{2} + \frac {11 \, B a b^{8} + 10 \, A b^{9}}{b^{8}}\right )} x^{2} + \frac {B a^{2} b^{7} + 30 \, A a b^{8}}{b^{8}}\right )} x^{2} - \frac {5 \, {\left (B a^{3} b^{6} - 2 \, A a^{2} b^{7}\right )}}{b^{8}}\right )} x^{2} + \frac {15 \, {\left (B a^{4} b^{5} - 2 \, A a^{3} b^{6}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x + \frac {3 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 219, normalized size = 1.16 \begin {gather*} \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,x^{5}}{10 b}+\frac {3 A \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}-\frac {3 B \,a^{5} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {7}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, A \,a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,x^{3}}{8 b}-\frac {3 \sqrt {b \,x^{2}+a}\, B \,a^{4} x}{256 b^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B a \,x^{3}}{16 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,a^{2} x}{64 b^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{3} x}{128 b^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A a x}{16 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,a^{2} x}{32 b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.01, size = 204, normalized size = 1.09 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{5}}{10 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x^{3}}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{3}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{32 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b^{2}} - \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \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^4\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 51.18, size = 345, normalized size = 1.84 \begin {gather*} - \frac {3 A a^{\frac {7}{2}} x}{128 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{\frac {5}{2}} x^{3}}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 A a^{\frac {3}{2}} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 A \sqrt {a} b x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {5}{2}}} + \frac {A b^{2} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{\frac {9}{2}} x}{256 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B a^{\frac {7}{2}} x^{3}}{256 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {5}{2}} x^{5}}{640 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 B a^{\frac {3}{2}} x^{7}}{160 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {19 B \sqrt {a} b x^{9}}{80 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 B a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{256 b^{\frac {7}{2}}} + \frac {B b^{2} x^{11}}{10 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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