Optimal. Leaf size=99 \[ \frac {x^{m+1} (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{a b^2 (m+1)}+\frac {d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac {c d x^{m+1}}{b (m+1)}+\frac {d^2 x^{m+2}}{b (m+2)} \]
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Rubi [A] time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {88, 64, 43} \[ \frac {x^{m+1} (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{a b^2 (m+1)}+\frac {d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac {c d x^{m+1}}{b (m+1)}+\frac {d^2 x^{m+2}}{b (m+2)} \]
Antiderivative was successfully verified.
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Rule 43
Rule 64
Rule 88
Rubi steps
\begin {align*} \int \frac {x^m (c+d x)^2}{a+b x} \, dx &=\int \left (\frac {d (b c-a d) x^m}{b^2}+\frac {(b c-a d)^2 x^m}{b^2 (a+b x)}+\frac {d x^m (c+d x)}{b}\right ) \, dx\\ &=\frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d \int x^m (c+d x) \, dx}{b}+\frac {(b c-a d)^2 \int \frac {x^m}{a+b x} \, dx}{b^2}\\ &=\frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a b^2 (1+m)}+\frac {d \int \left (c x^m+d x^{1+m}\right ) \, dx}{b}\\ &=\frac {c d x^{1+m}}{b (1+m)}+\frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d^2 x^{2+m}}{b (2+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a b^2 (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 77, normalized size = 0.78 \[ \frac {x^{m+1} \left ((m+2) (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )+a d (-a d (m+2)+2 b c (m+2)+b d (m+1) x)\right )}{a b^2 (m+1) (m+2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} x^{m}}{b x + a}, x\right ) \]
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 \[ \int \frac {{\left (d x + c\right )}^{2} x^{m}}{b x + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{2} x^{m}}{b x +a}\, dx \]
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 \[ \int \frac {{\left (d x + c\right )}^{2} x^{m}}{b x + a}\,{d x} \]
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 \[ \int \frac {x^m\,{\left (c+d\,x\right )}^2}{a+b\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 4.20, size = 219, normalized size = 2.21 \[ \frac {c^{2} m x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {c^{2} x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {2 c d m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {4 c d x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {d^{2} m x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac {3 d^{2} x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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