Optimal. Leaf size=152 \[ -\frac {(4 a+1) \tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}-\frac {(4 a+1) \tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}+\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (a+1) \left ((a-1) e^{4 x}+a+1\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.18, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2282, 390, 385, 212, 208, 205} \[ -\frac {(4 a+1) \tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}-\frac {(4 a+1) \tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}+\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (a+1) \left ((a-1) e^{4 x}+a+1\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 205
Rule 208
Rule 212
Rule 385
Rule 390
Rule 2282
Rubi steps
\begin {align*} \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx &=\operatorname {Subst}\left (\int \frac {\left (1+x^4\right )^2}{\left (1+a-(1-a) x^4\right )^2} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{(-1+a)^2}-\frac {4 \left (a-(1-a) x^4\right )}{(-1+a)^2 \left (1+a+(-1+a) x^4\right )^2}\right ) \, dx,x,e^x\right )\\ &=\frac {e^x}{(1-a)^2}-\frac {4 \operatorname {Subst}\left (\int \frac {a-(1-a) x^4}{\left (1+a+(-1+a) x^4\right )^2} \, dx,x,e^x\right )}{(1-a)^2}\\ &=\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (1+a) \left (1+a-(1-a) e^{4 x}\right )}-\frac {(1+4 a) \operatorname {Subst}\left (\int \frac {1}{1+a+(-1+a) x^4} \, dx,x,e^x\right )}{(1-a)^2 (1+a)}\\ &=\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (1+a) \left (1+a-(1-a) e^{4 x}\right )}-\frac {(1+4 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a}-\sqrt {1-a} x^2} \, dx,x,e^x\right )}{2 (1-a)^2 (1+a)^{3/2}}-\frac {(1+4 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a}+\sqrt {1-a} x^2} \, dx,x,e^x\right )}{2 (1-a)^2 (1+a)^{3/2}}\\ &=\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (1+a) \left (1+a-(1-a) e^{4 x}\right )}-\frac {(1+4 a) \tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}}-\frac {(1+4 a) \tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 0.17, size = 107, normalized size = 0.70 \[ \frac {(4 a+1) \text {RootSum}\left [\text {$\#$1}^4 a-\text {$\#$1}^4+a+1\& ,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\& \right ]+\frac {4 (a-1) e^x \left (a^2 \left (e^{4 x}+1\right )+2 a-e^{4 x}+2\right )}{a e^{4 x}+a-e^{4 x}+1}}{4 (a-1)^3 (a+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.61, size = 1163, normalized size = 7.65 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 0.15, size = 456, normalized size = 3.00 \[ -\frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} + 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} - 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (-\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {e^{x}}{a^{2} - 2 \, a + 1} + \frac {e^{x}}{{\left (a^{3} - a^{2} - a + 1\right )} {\left (a e^{\left (4 \, x\right )} + a - e^{\left (4 \, x\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [C] time = 0.36, size = 476, normalized size = 3.13 \[ -\frac {2}{\left (-1+a \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\tanh ^{3}\left (\frac {x}{2}\right )}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) \left (a +1\right )}-\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) a \left (a +1\right )}+\frac {3 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) \left (a +1\right )}-\frac {\tanh \left (\frac {x}{2}\right )}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) \left (a +1\right )}-\frac {2 \tanh \left (\frac {x}{2}\right )}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) a \left (a +1\right )}+\frac {1}{\left (-1+a \right )^{2} \left (a \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )-4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+6 a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-4 \tanh \left (\frac {x}{2}\right )+a \right ) \left (a +1\right )}-\frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 a \,\textit {\_Z}^{2}-4 \textit {\_Z} +a \right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} +1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{3} a -3 \textit {\_R}^{2}+3 \textit {\_R} a -1}}{4 \left (-1+a \right )^{2} \left (a +1\right )}-\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 a \,\textit {\_Z}^{2}-4 \textit {\_Z} +a \right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} +1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{3} a -3 \textit {\_R}^{2}+3 \textit {\_R} a -1}\right ) a}{\left (-1+a \right )^{2} \left (a +1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 23.35, size = 280, normalized size = 1.84 \[ \frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2}+\frac {\ln \left (\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}+\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}-\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2\,\left (a+1\right )\,\left (a+{\mathrm {e}}^{4\,x}\,\left (a-1\right )+1\right )}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}-\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}+\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{x}}{\left (a - \tanh {\left (2 x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________