Optimal. Leaf size=123 \[ -\frac {b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d x^2\right )\right )}+\frac {x^2}{2 a^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.25, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5436, 3785, 3919, 3831, 2659, 208} \[ -\frac {b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d x^2\right )\right )}+\frac {x^2}{2 a^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 208
Rule 2659
Rule 3785
Rule 3831
Rule 3919
Rule 5436
Rubi steps
\begin {align*} \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx,x,x^2\right )\\ &=\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )}-\frac {\operatorname {Subst}\left (\int \frac {-a^2+b^2+a b \text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx,x,x^2\right )}{2 a \left (a^2-b^2\right )}\\ &=\frac {x^2}{2 a^2}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )}-\frac {\left (b \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {\text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {x^2}{2 a^2}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )}-\frac {\left (2 a^2-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a \cosh (c+d x)}{b}} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {x^2}{2 a^2}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )}+\frac {\left (i \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,i \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac {x^2}{2 a^2}-\frac {b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.46, size = 220, normalized size = 1.79 \[ \frac {b \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+a b \sqrt {a^2-b^2} \sinh \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right )+a \cosh \left (c+d x^2\right ) \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right )}{2 a^2 d (a-b) (a+b) \sqrt {a^2-b^2} \left (a \cosh \left (c+d x^2\right )+b\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.46, size = 1314, normalized size = 10.68 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.15, size = 148, normalized size = 1.20 \[ -\frac {{\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{\left (d x^{2} + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {a^{2} - b^{2}}} - \frac {b^{3} e^{\left (d x^{2} + c\right )} + a b^{2}}{{\left (a^{4} d - a^{2} b^{2} d\right )} {\left (a e^{\left (2 \, d x^{2} + 2 \, c\right )} + 2 \, b e^{\left (d x^{2} + c\right )} + a\right )}} + \frac {d x^{2} + c}{2 \, a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.27, size = 234, normalized size = 1.90 \[ -\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}+\frac {b^{2} \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tanh ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \,a^{2} \left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}} \]
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 = 1.76, size = 316, normalized size = 2.57 \[ \frac {\frac {b^2}{d\,\left (a\,b^2-a^3\right )}+\frac {b^3\,{\mathrm {e}}^{d\,x^2+c}}{a\,d\,\left (a\,b^2-a^3\right )}}{a+2\,b\,{\mathrm {e}}^{d\,x^2+c}+a\,{\mathrm {e}}^{2\,d\,x^2+2\,c}}+\frac {x^2}{2\,a^2}+\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2-b^2\right )}{a^3\,\left (a^2-b^2\right )}-\frac {2\,b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2+c}\right )\,\left (2\,a^2-b^2\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2-b^2\right )}{a^3\,\left (a^2-b^2\right )}+\frac {2\,b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2+c}\right )\,\left (2\,a^2-b^2\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {sech}{\left (c + d x^{2} \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________