Integrand size = 20, antiderivative size = 91 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {(A b-a B) e (b d-a e) x}{b^3}+\frac {(A b-a B) (d+e x)^2}{2 b^2}+\frac {B (d+e x)^3}{3 b e}+\frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4} \]
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Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4}+\frac {e x (A b-a B) (b d-a e)}{b^3}+\frac {(d+e x)^2 (A b-a B)}{2 b^2}+\frac {B (d+e x)^3}{3 b e} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(A b-a B) e (b d-a e)}{b^3}+\frac {(A b-a B) (b d-a e)^2}{b^3 (a+b x)}+\frac {(A b-a B) e (d+e x)}{b^2}+\frac {B (d+e x)^2}{b}\right ) \, dx \\ & = \frac {(A b-a B) e (b d-a e) x}{b^3}+\frac {(A b-a B) (d+e x)^2}{2 b^2}+\frac {B (d+e x)^3}{3 b e}+\frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {b x \left (6 a^2 B e^2-3 a b e (4 B d+2 A e+B e x)+b^2 \left (3 A e (4 d+e x)+2 B \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+6 (A b-a B) (b d-a e)^2 \log (a+b x)}{6 b^4} \]
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Time = 0.71 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.65
method | result | size |
norman | \(-\frac {\left (A a b \,e^{2}-2 A \,b^{2} d e -B \,a^{2} e^{2}+2 B a b d e -b^{2} B \,d^{2}\right ) x}{b^{3}}+\frac {e \left (A b e -B a e +2 B b d \right ) x^{2}}{2 b^{2}}+\frac {B \,e^{2} x^{3}}{3 b}+\frac {\left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(150\) |
default | \(-\frac {-\frac {1}{3} b^{2} B \,x^{3} e^{2}-\frac {1}{2} A \,b^{2} e^{2} x^{2}+\frac {1}{2} B a b \,e^{2} x^{2}-B \,b^{2} d e \,x^{2}+A a b \,e^{2} x -2 A \,b^{2} d e x -B \,a^{2} e^{2} x +2 B a b d e x -b^{2} B \,d^{2} x}{b^{3}}+\frac {\left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(164\) |
risch | \(\frac {B \,e^{2} x^{3}}{3 b}+\frac {A \,e^{2} x^{2}}{2 b}-\frac {B a \,e^{2} x^{2}}{2 b^{2}}+\frac {B d e \,x^{2}}{b}-\frac {A a \,e^{2} x}{b^{2}}+\frac {2 A d e x}{b}+\frac {B \,a^{2} e^{2} x}{b^{3}}-\frac {2 B a d e x}{b^{2}}+\frac {B \,d^{2} x}{b}+\frac {\ln \left (b x +a \right ) A \,a^{2} e^{2}}{b^{3}}-\frac {2 \ln \left (b x +a \right ) A a d e}{b^{2}}+\frac {\ln \left (b x +a \right ) A \,d^{2}}{b}-\frac {\ln \left (b x +a \right ) B \,a^{3} e^{2}}{b^{4}}+\frac {2 \ln \left (b x +a \right ) B \,a^{2} d e}{b^{3}}-\frac {\ln \left (b x +a \right ) B a \,d^{2}}{b^{2}}\) | \(197\) |
parallelrisch | \(\frac {2 B \,e^{2} x^{3} b^{3}+3 A \,x^{2} b^{3} e^{2}-3 B \,x^{2} a \,b^{2} e^{2}+6 B \,x^{2} b^{3} d e +6 A \ln \left (b x +a \right ) a^{2} b \,e^{2}-12 A \ln \left (b x +a \right ) a \,b^{2} d e +6 A \ln \left (b x +a \right ) b^{3} d^{2}-6 A x a \,b^{2} e^{2}+12 A x \,b^{3} d e -6 B \ln \left (b x +a \right ) a^{3} e^{2}+12 B \ln \left (b x +a \right ) a^{2} b d e -6 B \ln \left (b x +a \right ) a \,b^{2} d^{2}+6 B x \,a^{2} b \,e^{2}-12 B x a \,b^{2} d e +6 B x \,b^{3} d^{2}}{6 b^{4}}\) | \(198\) |
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Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{3} e^{2} x^{3} + 3 \, {\left (2 \, B b^{3} d e - {\left (B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{3} d^{2} - 2 \, {\left (B a b^{2} - A b^{3}\right )} d e + {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x - 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {B e^{2} x^{3}}{3 b} + x^{2} \left (\frac {A e^{2}}{2 b} - \frac {B a e^{2}}{2 b^{2}} + \frac {B d e}{b}\right ) + x \left (- \frac {A a e^{2}}{b^{2}} + \frac {2 A d e}{b} + \frac {B a^{2} e^{2}}{b^{3}} - \frac {2 B a d e}{b^{2}} + \frac {B d^{2}}{b}\right ) - \frac {\left (- A b + B a\right ) \left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{4}} \]
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Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} + 3 \, {\left (2 \, B b^{2} d e - {\left (B a b - A b^{2}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} - 2 \, {\left (B a b - A b^{2}\right )} d e + {\left (B a^{2} - A a b\right )} e^{2}\right )} x}{6 \, b^{3}} - \frac {{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} + 6 \, B b^{2} d e x^{2} - 3 \, B a b e^{2} x^{2} + 3 \, A b^{2} e^{2} x^{2} + 6 \, B b^{2} d^{2} x - 12 \, B a b d e x + 12 \, A b^{2} d e x + 6 \, B a^{2} e^{2} x - 6 \, A a b e^{2} x}{6 \, b^{3}} - \frac {{\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \]
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Time = 0.08 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=x\,\left (\frac {B\,d^2+2\,A\,e\,d}{b}-\frac {a\,\left (\frac {A\,e^2+2\,B\,d\,e}{b}-\frac {B\,a\,e^2}{b^2}\right )}{b}\right )+x^2\,\left (\frac {A\,e^2+2\,B\,d\,e}{2\,b}-\frac {B\,a\,e^2}{2\,b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2\right )}{b^4}+\frac {B\,e^2\,x^3}{3\,b} \]
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