Optimal. Leaf size=123 \[ \frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}}+\frac {3 (b B-5 A c)}{4 b^3 c \sqrt {x}}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2} \]
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Rubi [A] time = 0.06, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \begin {gather*} \frac {3 (b B-5 A c)}{4 b^3 c \sqrt {x}}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}+\frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 78
Rule 205
Rule 781
Rubi steps
\begin {align*} \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac {A+B x}{x^{3/2} (b+c x)^3} \, dx\\ &=-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}-\frac {\left (\frac {b B}{2}-\frac {5 A c}{2}\right ) \int \frac {1}{x^{3/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}-\frac {(3 (b B-5 A c)) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac {3 (b B-5 A c)}{4 b^3 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}+\frac {(3 (b B-5 A c)) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b^3}\\ &=\frac {3 (b B-5 A c)}{4 b^3 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}+\frac {(3 (b B-5 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=\frac {3 (b B-5 A c)}{4 b^3 c \sqrt {x}}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}-\frac {b B-5 A c}{4 b^2 c \sqrt {x} (b+c x)}+\frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}}\\ \end {align*}
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Mathematica [C] time = 0.02, size = 59, normalized size = 0.48 \begin {gather*} \frac {\frac {b^2 (A c-b B)}{(b+c x)^2}+(b B-5 A c) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {c x}{b}\right )}{2 b^3 c \sqrt {x}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.17, size = 96, normalized size = 0.78 \begin {gather*} \frac {3 (b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}}+\frac {-8 A b^2-25 A b c x-15 A c^2 x^2+5 b^2 B x+3 b B c x^2}{4 b^3 \sqrt {x} (b+c x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.43, size = 331, normalized size = 2.69 \begin {gather*} \left [\frac {3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b + 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) - 2 \, {\left (8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{8 \, {\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}, -\frac {3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{4 \, {\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 86, normalized size = 0.70 \begin {gather*} \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{3}} - \frac {2 \, A}{b^{3} \sqrt {x}} + \frac {3 \, B b c x^{\frac {3}{2}} - 7 \, A c^{2} x^{\frac {3}{2}} + 5 \, B b^{2} \sqrt {x} - 9 \, A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 125, normalized size = 1.02 \begin {gather*} -\frac {7 A \,c^{2} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{3}}+\frac {3 B c \,x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{2}}-\frac {9 A c \sqrt {x}}{4 \left (c x +b \right )^{2} b^{2}}+\frac {5 B \sqrt {x}}{4 \left (c x +b \right )^{2} b}-\frac {15 A c \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{3}}+\frac {3 B \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{2}}-\frac {2 A}{b^{3} \sqrt {x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.29, size = 98, normalized size = 0.80 \begin {gather*} -\frac {8 \, A b^{2} - 3 \, {\left (B b c - 5 \, A c^{2}\right )} x^{2} - 5 \, {\left (B b^{2} - 5 \, A b c\right )} x}{4 \, {\left (b^{3} c^{2} x^{\frac {5}{2}} + 2 \, b^{4} c x^{\frac {3}{2}} + b^{5} \sqrt {x}\right )}} + \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.12, size = 116, normalized size = 0.94 \begin {gather*} -\frac {\frac {2\,A}{b}+\frac {5\,x\,\left (5\,A\,c-B\,b\right )}{4\,b^2}+\frac {3\,c\,x^2\,\left (5\,A\,c-B\,b\right )}{4\,b^3}}{b^2\,\sqrt {x}+c^2\,x^{5/2}+2\,b\,c\,x^{3/2}}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {c}\,\sqrt {x}\,\left (5\,A\,c-B\,b\right )}{\sqrt {b}\,\left (15\,A\,c-3\,B\,b\right )}\right )\,\left (5\,A\,c-B\,b\right )}{4\,b^{7/2}\,\sqrt {c}} \end {gather*}
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 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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