Optimal. Leaf size=231 \[ \frac {\text {Li}_2\left (-\frac {(a+b) e^{2 c+2 d x}}{a-b-2 \sqrt {-a} \sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b} d^2}-\frac {\text {Li}_2\left (-\frac {(a+b) e^{2 c+2 d x}}{a-b+2 \sqrt {-a} \sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b} d^2}+\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{-2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 \sqrt {-a} \sqrt {b} d} \]
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Rubi [A] time = 0.54, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5632, 3320, 2264, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{-2 \sqrt {-a} \sqrt {b}+a-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2}-\frac {\text {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{2 \sqrt {-a} \sqrt {b}+a-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2}+\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{-2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 \sqrt {-a} \sqrt {b} d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 3320
Rule 5632
Rubi steps
\begin {align*} \int \frac {x \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=2 \int \frac {x}{a-b+(a+b) \cosh (2 c+2 d x)} \, dx\\ &=4 \int \frac {e^{2 c+2 d x} x}{a+b+2 (a-b) e^{2 c+2 d x}+(a+b) e^{2 (2 c+2 d x)}} \, dx\\ &=\frac {(2 (a+b)) \int \frac {e^{2 c+2 d x} x}{2 (a-b)-4 \sqrt {-a} \sqrt {b}+2 (a+b) e^{2 c+2 d x}} \, dx}{\sqrt {-a} \sqrt {b}}-\frac {(2 (a+b)) \int \frac {e^{2 c+2 d x} x}{2 (a-b)+4 \sqrt {-a} \sqrt {b}+2 (a+b) e^{2 c+2 d x}} \, dx}{\sqrt {-a} \sqrt {b}}\\ &=\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {\int \log \left (1+\frac {2 (a+b) e^{2 c+2 d x}}{2 (a-b)-4 \sqrt {-a} \sqrt {b}}\right ) \, dx}{2 \sqrt {-a} \sqrt {b} d}+\frac {\int \log \left (1+\frac {2 (a+b) e^{2 c+2 d x}}{2 (a-b)+4 \sqrt {-a} \sqrt {b}}\right ) \, dx}{2 \sqrt {-a} \sqrt {b} d}\\ &=\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 (a+b) x}{2 (a-b)-4 \sqrt {-a} \sqrt {b}}\right )}{x} \, dx,x,e^{2 c+2 d x}\right )}{4 \sqrt {-a} \sqrt {b} d^2}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 (a+b) x}{2 (a-b)+4 \sqrt {-a} \sqrt {b}}\right )}{x} \, dx,x,e^{2 c+2 d x}\right )}{4 \sqrt {-a} \sqrt {b} d^2}\\ &=\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}+\frac {\text {Li}_2\left (-\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2}-\frac {\text {Li}_2\left (-\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2}\\ \end {align*}
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Mathematica [A] time = 2.79, size = 250, normalized size = 1.08 \[ \frac {\sqrt {a} \text {Li}_2\left (-\frac {(a+b) e^{2 (c+d x)}}{a-b-2 \sqrt {-a} \sqrt {b}}\right )-\sqrt {a} \text {Li}_2\left (-\frac {(a+b) e^{2 (c+d x)}}{a-b+2 \sqrt {-a} \sqrt {b}}\right )+2 \sqrt {a} (c+d x) \log \left (\frac {(a+b) e^{2 (c+d x)}}{-2 \sqrt {-a} \sqrt {b}+a-b}+1\right )-2 \sqrt {a} (c+d x) \log \left (\frac {(a+b) e^{2 (c+d x)}}{2 \sqrt {-a} \sqrt {b}+a-b}+1\right )-4 \sqrt {-a} c \tan ^{-1}\left (\frac {(a+b) e^{2 (c+d x)}+a-b}{2 \sqrt {a} \sqrt {b}}\right )}{4 \sqrt {-a^2} \sqrt {b} d^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 1516, normalized size = 6.56 \[ \text {result too large to display} \]
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 \[ \int \frac {x \operatorname {sech}\left (d x + c\right )^{2}}{b \tanh \left (d x + c\right )^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.48, size = 953, normalized size = 4.13 \[ -\frac {c^{2}}{2 d^{2} \sqrt {-a b}}-\frac {c^{2}}{d^{2} \left (-2 \sqrt {-a b}-a +b \right )}-\frac {c x}{d \sqrt {-a b}}-\frac {2 c x}{d \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right ) c}{2 d^{2} \sqrt {-a b}}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) c}{d^{2} \left (-2 \sqrt {-a b}-a +b \right )}-\frac {c \arctan \left (\frac {2 \left (a +b \right ) {\mathrm e}^{2 d x +2 c}+2 a -2 b}{4 \sqrt {a b}}\right )}{d^{2} \sqrt {a b}}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right ) x}{2 d \sqrt {-a b}}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) x}{d \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b \,x^{2}}{2 \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {a \,x^{2}}{2 \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right )}{4 d^{2} \sqrt {-a b}}+\frac {\polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right )}{2 d^{2} \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b c x}{d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {a \,c^{2}}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b \,c^{2}}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a}{4 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b}{4 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b c}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a x}{2 d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b x}{2 d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {x^{2}}{2 \sqrt {-a b}}-\frac {x^{2}}{-2 \sqrt {-a b}-a +b}-\frac {a c x}{d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a c}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )} \]
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 \[ \int \frac {x \operatorname {sech}\left (d x + c\right )^{2}}{b \tanh \left (d x + c\right )^{2} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {sech}^{2}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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