Optimal. Leaf size=73 \[ \frac {x^{m+1} (A b-a B)}{a b (a+b x)}-\frac {x^{m+1} (A b m-a B (m+1)) \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{a^2 b (m+1)} \]
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Rubi [A] time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {78, 64} \[ \frac {x^{m+1} (A b-a B)}{a b (a+b x)}-\frac {x^{m+1} (A b m-a B (m+1)) \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{a^2 b (m+1)} \]
Antiderivative was successfully verified.
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Rule 64
Rule 78
Rubi steps
\begin {align*} \int \frac {x^m (A+B x)}{(a+b x)^2} \, dx &=\frac {(A b-a B) x^{1+m}}{a b (a+b x)}-\frac {(A b m-a B (1+m)) \int \frac {x^m}{a+b x} \, dx}{a b}\\ &=\frac {(A b-a B) x^{1+m}}{a b (a+b x)}-\frac {(A b m-a B (1+m)) x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a^2 b (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 63, normalized size = 0.86 \[ \frac {x^{m+1} \left (\frac {(a B (m+1)-A b m) \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{m+1}+\frac {a (A b-a B)}{a+b x}\right )}{a^2 b} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} x^{m}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} x^{m}}{{\left (b x + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) x^{m}}{\left (b x +a \right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} x^{m}}{{\left (b x + a\right )}^{2}}\,{d x} \]
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 \[ \int \frac {x^m\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 5.76, size = 639, normalized size = 8.75 \[ A \left (- \frac {a m^{2} x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {a m x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a m x x^{m} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a x x^{m} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m^{2} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )}\right ) + B \left (- \frac {a m^{2} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {3 a m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} + \frac {a m x^{2} x^{m} \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {2 a x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} + \frac {2 a x^{2} x^{m} \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {b m^{2} x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {3 b m x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {2 b x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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