Optimal. Leaf size=172 \[ \frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}-\frac {3 b^3 \sqrt {a+b x} (A b-2 a B)}{128 a^3 x}+\frac {b^2 \sqrt {a+b x} (A b-2 a B)}{64 a^2 x^2}+\frac {(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}+\frac {b \sqrt {a+b x} (A b-2 a B)}{16 a x^3}-\frac {A (a+b x)^{5/2}}{5 a x^5} \]
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Rubi [A] time = 0.08, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 51, 63, 208} \[ \frac {b^2 \sqrt {a+b x} (A b-2 a B)}{64 a^2 x^2}-\frac {3 b^3 \sqrt {a+b x} (A b-2 a B)}{128 a^3 x}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {b \sqrt {a+b x} (A b-2 a B)}{16 a x^3}+\frac {(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5} \]
Antiderivative was successfully verified.
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Rule 47
Rule 51
Rule 63
Rule 78
Rule 208
Rubi steps
\begin {align*} \int \frac {(a+b x)^{3/2} (A+B x)}{x^6} \, dx &=-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {\left (-\frac {5 A b}{2}+5 a B\right ) \int \frac {(a+b x)^{3/2}}{x^5} \, dx}{5 a}\\ &=\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {(3 b (A b-2 a B)) \int \frac {\sqrt {a+b x}}{x^4} \, dx}{16 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (b^2 (A b-2 a B)\right ) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{32 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {\left (3 b^3 (A b-2 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{128 a^2}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (3 b^4 (A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{256 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}-\frac {\left (3 b^3 (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}\right )}{128 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x}}{16 a x^3}+\frac {b^2 (A b-2 a B) \sqrt {a+b x}}{64 a^2 x^2}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x}}{128 a^3 x}+\frac {(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac {A (a+b x)^{5/2}}{5 a x^5}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.03, size = 57, normalized size = 0.33 \[ -\frac {(a+b x)^{5/2} \left (a^5 A+b^4 x^5 (2 a B-A b) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b x}{a}+1\right )\right )}{5 a^6 x^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 306, normalized size = 1.78 \[ \left [-\frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {a} x^{5} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (128 \, A a^{5} - 15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{1280 \, a^{4} x^{5}}, \frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (128 \, A a^{5} - 15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{640 \, a^{4} x^{5}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.34, size = 192, normalized size = 1.12 \[ \frac {\frac {15 \, {\left (2 \, B a b^{5} - A b^{6}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {30 \, {\left (b x + a\right )}^{\frac {9}{2}} B a b^{5} - 140 \, {\left (b x + a\right )}^{\frac {7}{2}} B a^{2} b^{5} + 140 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 30 \, \sqrt {b x + a} B a^{5} b^{5} - 15 \, {\left (b x + a\right )}^{\frac {9}{2}} A b^{6} + 70 \, {\left (b x + a\right )}^{\frac {7}{2}} A a b^{6} - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 70 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 15 \, \sqrt {b x + a} A a^{4} b^{6}}{a^{3} b^{5} x^{5}}}{640 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 129, normalized size = 0.75 \[ 2 \left (\frac {3 \left (A b -2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {7}{2}}}+\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}} A b}{10 a}+\frac {3 \left (A b -2 B a \right ) \sqrt {b x +a}\, a}{256}+\frac {7 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{2}}-\frac {3 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{3}}+\left (-\frac {7 A b}{128}+\frac {7 B a}{64}\right ) \left (b x +a \right )^{\frac {3}{2}}}{b^{5} x^{5}}\right ) b^{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.02, size = 224, normalized size = 1.30 \[ -\frac {1}{1280} \, b^{5} {\left (\frac {2 \, {\left (128 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b - 15 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 70 \, {\left (2 \, B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 70 \, {\left (2 \, B a^{4} - A a^{3} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (2 \, B a^{5} - A a^{4} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{5} a^{3} b - 5 \, {\left (b x + a\right )}^{4} a^{4} b + 10 \, {\left (b x + a\right )}^{3} a^{5} b - 10 \, {\left (b x + a\right )}^{2} a^{6} b + 5 \, {\left (b x + a\right )} a^{7} b - a^{8} b} - \frac {15 \, {\left (2 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.44, size = 203, normalized size = 1.18 \[ \frac {\left (\frac {7\,A\,b^5}{64}-\frac {7\,B\,a\,b^4}{32}\right )\,{\left (a+b\,x\right )}^{3/2}+\left (\frac {3\,B\,a^2\,b^4}{64}-\frac {3\,A\,a\,b^5}{128}\right )\,\sqrt {a+b\,x}-\frac {7\,\left (A\,b^5-2\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^2}+\frac {3\,\left (A\,b^5-2\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{9/2}}{128\,a^3}+\frac {A\,b^5\,{\left (a+b\,x\right )}^{5/2}}{5\,a}}{5\,a\,{\left (a+b\,x\right )}^4-5\,a^4\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^5-10\,a^2\,{\left (a+b\,x\right )}^3+10\,a^3\,{\left (a+b\,x\right )}^2+a^5}+\frac {3\,b^4\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-2\,B\,a\right )}{128\,a^{7/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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