Optimal. Leaf size=210 \[ -\frac {(b d-a e) x}{2 b}-\frac {(b c d+e-a c e) x}{4 b c}-\frac {(d+e x)^2}{8 e}-\frac {(b c d+e-a c e)^2 \log (1-a c-b c x)}{4 b^2 c^2 e}-\frac {(b d-a e) (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}+\frac {(d+e x)^2 \log (1-a c-b c x)}{4 e}-\frac {(b d-a e)^2 \text {PolyLog}(2,c (a+b x))}{2 b^2 e}+\frac {(d+e x)^2 \text {PolyLog}(2,c (a+b x))}{2 e} \]
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Rubi [A]
time = 0.14, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6733, 2465,
2436, 2332, 2440, 2438, 2442, 45} \begin {gather*} -\frac {(-a c e+b c d+e)^2 \log (-a c-b c x+1)}{4 b^2 c^2 e}-\frac {(b d-a e)^2 \text {Li}_2(c (a+b x))}{2 b^2 e}-\frac {(-a c-b c x+1) (b d-a e) \log (-a c-b c x+1)}{2 b^2 c}+\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}-\frac {x (-a c e+b c d+e)}{4 b c}+\frac {(d+e x)^2 \log (-a c-b c x+1)}{4 e}-\frac {x (b d-a e)}{2 b}-\frac {(d+e x)^2}{8 e} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 2332
Rule 2436
Rule 2438
Rule 2440
Rule 2442
Rule 2465
Rule 6733
Rubi steps
\begin {align*} \int (d+e x) \text {Li}_2(c (a+b x)) \, dx &=\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}+\frac {b \int \frac {(d+e x)^2 \log (1-a c-b c x)}{a+b x} \, dx}{2 e}\\ &=\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}+\frac {b \int \left (\frac {e (b d-a e) \log (1-a c-b c x)}{b^2}+\frac {(b d-a e)^2 \log (1-a c-b c x)}{b^2 (a+b x)}+\frac {e (d+e x) \log (1-a c-b c x)}{b}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}+\frac {1}{2} \int (d+e x) \log (1-a c-b c x) \, dx+\frac {(b d-a e) \int \log (1-a c-b c x) \, dx}{2 b}+\frac {(b d-a e)^2 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 b e}\\ &=\frac {(d+e x)^2 \log (1-a c-b c x)}{4 e}+\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}+\frac {(b c) \int \frac {(d+e x)^2}{1-a c-b c x} \, dx}{4 e}-\frac {(b d-a e) \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{2 b^2 c}+\frac {(b d-a e)^2 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 b^2 e}\\ &=-\frac {(b d-a e) x}{2 b}-\frac {(b d-a e) (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}+\frac {(d+e x)^2 \log (1-a c-b c x)}{4 e}-\frac {(b d-a e)^2 \text {Li}_2(c (a+b x))}{2 b^2 e}+\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}+\frac {(b c) \int \left (-\frac {e (b c d+e-a c e)}{b^2 c^2}+\frac {(b c d+e-a c e)^2}{b^2 c^2 (1-a c-b c x)}-\frac {e (d+e x)}{b c}\right ) \, dx}{4 e}\\ &=-\frac {(b d-a e) x}{2 b}-\frac {(b c d+e-a c e) x}{4 b c}-\frac {(d+e x)^2}{8 e}-\frac {(b c d+e-a c e)^2 \log (1-a c-b c x)}{4 b^2 c^2 e}-\frac {(b d-a e) (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}+\frac {(d+e x)^2 \log (1-a c-b c x)}{4 e}-\frac {(b d-a e)^2 \text {Li}_2(c (a+b x))}{2 b^2 e}+\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 161, normalized size = 0.77 \begin {gather*} \frac {e \left (-b c x (2-6 a c+b c x)+\left (-2-6 a^2 c^2+2 b^2 c^2 x^2-4 a c (-2+b c x)\right ) \log (1-a c-b c x)-4 a^2 c^2 \text {PolyLog}(2,c (a+b x))\right )}{8 b^2 c^2}+\frac {d (-c (a+b x)+(-1+c (a+b x)) \log (1-c (a+b x))+c (a+b x) \text {PolyLog}(2,c (a+b x)))}{b c}+\frac {1}{2} e x^2 \text {PolyLog}(2,a c+b c x) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.66, size = 249, normalized size = 1.19
method | result | size |
derivativedivides | \(\frac {-\frac {\polylog \left (2, x b c +a c \right ) a e \left (x b c +a c \right )}{b}+\polylog \left (2, x b c +a c \right ) d \left (x b c +a c \right )+\frac {\polylog \left (2, x b c +a c \right ) e \left (x b c +a c \right )^{2}}{2 c b}-\frac {-2 a c e \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )+2 d c b \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )-e \left (\frac {\left (-x b c -a c +1\right )^{2} \ln \left (-x b c -a c +1\right )}{2}-\frac {\left (-x b c -a c +1\right )^{2}}{4}\right )+e \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )}{2 b c}}{b c}\) | \(249\) |
default | \(\frac {-\frac {\polylog \left (2, x b c +a c \right ) a e \left (x b c +a c \right )}{b}+\polylog \left (2, x b c +a c \right ) d \left (x b c +a c \right )+\frac {\polylog \left (2, x b c +a c \right ) e \left (x b c +a c \right )^{2}}{2 c b}-\frac {-2 a c e \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )+2 d c b \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )-e \left (\frac {\left (-x b c -a c +1\right )^{2} \ln \left (-x b c -a c +1\right )}{2}-\frac {\left (-x b c -a c +1\right )^{2}}{4}\right )+e \left (\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )-1+x b c +a c \right )}{2 b c}}{b c}\) | \(249\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 219, normalized size = 1.04 \begin {gather*} -\frac {{\left (2 \, a b d - a^{2} e\right )} {\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )}}{2 \, b^{2}} - \frac {b^{2} c^{2} x^{2} e + 2 \, {\left (4 \, b^{2} c^{2} d - 3 \, a b c^{2} e + b c e\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} e + 2 \, b^{2} c^{2} d x\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} e - 3 \, a^{2} c^{2} e + 4 \, a c e + 4 \, {\left (a b c^{2} - b c\right )} d + 2 \, {\left (2 \, b^{2} c^{2} d - a b c^{2} e\right )} x - e\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 174, normalized size = 0.83 \begin {gather*} -\frac {8 \, b^{2} c^{2} d x - 4 \, {\left (2 \, b^{2} c^{2} d x + 2 \, a b c^{2} d + {\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )} e\right )} {\rm Li}_2\left (b c x + a c\right ) + {\left (b^{2} c^{2} x^{2} - 2 \, {\left (3 \, a b c^{2} - b c\right )} x\right )} e - 2 \, {\left (4 \, b^{2} c^{2} d x + 4 \, {\left (a b c^{2} - b c\right )} d + {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} e\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 3.33, size = 252, normalized size = 1.20 \begin {gather*} \begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \operatorname {Li}_{2}\left (a c\right ) & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {3 a^{2} e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2}} - \frac {a^{2} e \operatorname {Li}_{2}\left (a c + b c x\right )}{2 b^{2}} - \frac {a d \operatorname {Li}_{1}\left (a c + b c x\right )}{b} + \frac {a d \operatorname {Li}_{2}\left (a c + b c x\right )}{b} + \frac {a e x \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b} + \frac {3 a e x}{4 b} - \frac {a e \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{2} c} - d x \operatorname {Li}_{1}\left (a c + b c x\right ) + d x \operatorname {Li}_{2}\left (a c + b c x\right ) - d x - \frac {e x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4} + \frac {e x^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2} - \frac {e x^{2}}{8} + \frac {d \operatorname {Li}_{1}\left (a c + b c x\right )}{b c} - \frac {e x}{4 b c} + \frac {e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2} c^{2}} & \text {otherwise} \end {cases} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (d+e\,x\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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