3.86 \(\int \frac{\text{sech}(c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)
Optimal. Leaf size=82 \[ \frac{(2 a+b) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{2 a^{3/2} d (a+b)^{3/2}}-\frac{b \sinh (c+d x)}{2 a d (a+b) \left (a \sinh ^2(c+d x)+a+b\right )} \]
[Out]
((2*a + b)*ArcTan[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b]])/(2*a^(3/2)*(a + b)^(3/2)*d) - (b*Sinh[c + d*x])/(2*a*(
a + b)*d*(a + b + a*Sinh[c + d*x]^2))
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Rubi [A] time = 0.0655658, antiderivative size = 82, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4147, 385, 205} \[ \frac{(2 a+b) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{2 a^{3/2} d (a+b)^{3/2}}-\frac{b \sinh (c+d x)}{2 a d (a+b) \left (a \sinh ^2(c+d x)+a+b\right )} \]
Antiderivative was successfully verified.
[In]
Int[Sech[c + d*x]/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((2*a + b)*ArcTan[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b]])/(2*a^(3/2)*(a + b)^(3/2)*d) - (b*Sinh[c + d*x])/(2*a*(
a + b)*d*(a + b + a*Sinh[c + d*x]^2))
Rule 4147
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\text{sech}(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (a+b+a x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=-\frac{b \sinh (c+d x)}{2 a (a+b) d \left (a+b+a \sinh ^2(c+d x)\right )}+\frac{(2 a+b) \operatorname{Subst}\left (\int \frac{1}{a+b+a x^2} \, dx,x,\sinh (c+d x)\right )}{2 a (a+b) d}\\ &=\frac{(2 a+b) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{2 a^{3/2} (a+b)^{3/2} d}-\frac{b \sinh (c+d x)}{2 a (a+b) d \left (a+b+a \sinh ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.301814, size = 124, normalized size = 1.51 \[ \frac{\left (2 a^2+3 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )-\sqrt{a} b \sqrt{a+b} \sinh (c+d x)+a (2 a+b) \sinh ^2(c+d x) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{a^{3/2} d (a+b)^{3/2} (a \cosh (2 (c+d x))+a+2 b)} \]
Antiderivative was successfully verified.
[In]
Integrate[Sech[c + d*x]/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((2*a^2 + 3*a*b + b^2)*ArcTan[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b]] - Sqrt[a]*b*Sqrt[a + b]*Sinh[c + d*x] + a*(
2*a + b)*ArcTan[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b]]*Sinh[c + d*x]^2)/(a^(3/2)*(a + b)^(3/2)*d*(a + 2*b + a*Co
sh[2*(c + d*x)]))
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Maple [B] time = 0.068, size = 332, normalized size = 4.1 \begin{align*}{\frac{b}{da \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}-{\frac{b}{da \left ( a+b \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}+{\frac{1}{d}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}+2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}}-{\frac{1}{d}\arctan \left ({\frac{1}{2} \left ( -2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}+2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}}+{\frac{b}{2\,d}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}+2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{3}{2}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{b}{2\,d}\arctan \left ({\frac{1}{2} \left ( -2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}+2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{3}{2}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(d*x+c)/(a+b*sech(d*x+c)^2)^2,x)
[Out]
1/d/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*
b/a/(a+b)*tanh(1/2*d*x+1/2*c)^3-1/d/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a
-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*b/a/(a+b)*tanh(1/2*d*x+1/2*c)+1/d/(a+b)^(3/2)/a^(1/2)*arctan(1/2*(2*tanh(1/2*d
*x+1/2*c)*(a+b)^(1/2)+2*b^(1/2))/a^(1/2))-1/d/(a+b)^(3/2)/a^(1/2)*arctan(1/2*(-2*tanh(1/2*d*x+1/2*c)*(a+b)^(1/
2)+2*b^(1/2))/a^(1/2))+1/2/d/a^(3/2)*b/(a+b)^(3/2)*arctan(1/2*(2*tanh(1/2*d*x+1/2*c)*(a+b)^(1/2)+2*b^(1/2))/a^
(1/2))-1/2/d/a^(3/2)*b/(a+b)^(3/2)*arctan(1/2*(-2*tanh(1/2*d*x+1/2*c)*(a+b)^(1/2)+2*b^(1/2))/a^(1/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}}{a^{3} d + a^{2} b d +{\left (a^{3} d e^{\left (4 \, c\right )} + a^{2} b d e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \,{\left (a^{3} d e^{\left (2 \, c\right )} + 3 \, a^{2} b d e^{\left (2 \, c\right )} + 2 \, a b^{2} d e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}} + 2 \, \int \frac{{\left (2 \, a e^{\left (3 \, c\right )} + b e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} +{\left (2 \, a e^{c} + b e^{c}\right )} e^{\left (d x\right )}}{2 \,{\left (a^{3} + a^{2} b +{\left (a^{3} e^{\left (4 \, c\right )} + a^{2} b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \,{\left (a^{3} e^{\left (2 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, c\right )} + 2 \, a b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
-(b*e^(3*d*x + 3*c) - b*e^(d*x + c))/(a^3*d + a^2*b*d + (a^3*d*e^(4*c) + a^2*b*d*e^(4*c))*e^(4*d*x) + 2*(a^3*d
*e^(2*c) + 3*a^2*b*d*e^(2*c) + 2*a*b^2*d*e^(2*c))*e^(2*d*x)) + 2*integrate(1/2*((2*a*e^(3*c) + b*e^(3*c))*e^(3
*d*x) + (2*a*e^c + b*e^c)*e^(d*x))/(a^3 + a^2*b + (a^3*e^(4*c) + a^2*b*e^(4*c))*e^(4*d*x) + 2*(a^3*e^(2*c) + 3
*a^2*b*e^(2*c) + 2*a*b^2*e^(2*c))*e^(2*d*x)), x)
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Fricas [B] time = 2.39493, size = 4533, normalized size = 55.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
[-1/4*(4*(a^2*b + a*b^2)*cosh(d*x + c)^3 + 12*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + 4*(a^2*b + a*b^2
)*sinh(d*x + c)^3 + ((2*a^2 + a*b)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2 +
a*b)*sinh(d*x + c)^4 + 2*(2*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 + a*b)*cosh(d*x + c)^2 + 2*a^2
+ 5*a*b + 2*b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b + 4*((2*a^2 + a*b)*cosh(d*x + c)^3 + (2*a^2 + 5*a*b + 2*b^2)*co
sh(d*x + c))*sinh(d*x + c))*sqrt(-a^2 - a*b)*log((a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*si
nh(d*x + c)^4 - 2*(3*a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 - 3*a - 2*b)*sinh(d*x + c)^2 + 4*(a*cos
h(d*x + c)^3 - (3*a + 2*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2
+ sinh(d*x + c)^3 + (3*cosh(d*x + c)^2 - 1)*sinh(d*x + c) - cosh(d*x + c))*sqrt(-a^2 - a*b) + a)/(a*cosh(d*x
+ c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x
+ c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - 4*(
a^2*b + a*b^2)*cosh(d*x + c) - 4*(a^2*b + a*b^2 - 3*(a^2*b + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c))/((a^5 + 2*
a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 2*a^
4*b + a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^5 + 4*a^4*b + 5*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + 2
*a^4*b + a^3*b^2)*d*cosh(d*x + c)^2 + (a^5 + 4*a^4*b + 5*a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + (a^5 + 2*a^
4*b + a^3*b^2)*d + 4*((a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^5 + 4*a^4*b + 5*a^3*b^2 + 2*a^2*b^3)*d*
cosh(d*x + c))*sinh(d*x + c)), -1/2*(2*(a^2*b + a*b^2)*cosh(d*x + c)^3 + 6*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(
d*x + c)^2 + 2*(a^2*b + a*b^2)*sinh(d*x + c)^3 - ((2*a^2 + a*b)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b)*cosh(d*x + c
)*sinh(d*x + c)^3 + (2*a^2 + a*b)*sinh(d*x + c)^4 + 2*(2*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 +
a*b)*cosh(d*x + c)^2 + 2*a^2 + 5*a*b + 2*b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b + 4*((2*a^2 + a*b)*cosh(d*x + c)^3
+ (2*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 + a*b)*arctan(1/2*(a*cosh(d*x + c)^3 + 3*a*c
osh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3 + (3*a + 4*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + 3*a + 4*
b)*sinh(d*x + c))/sqrt(a^2 + a*b)) - ((2*a^2 + a*b)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b)*cosh(d*x + c)*sinh(d*x +
c)^3 + (2*a^2 + a*b)*sinh(d*x + c)^4 + 2*(2*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 + a*b)*cosh(d*
x + c)^2 + 2*a^2 + 5*a*b + 2*b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b + 4*((2*a^2 + a*b)*cosh(d*x + c)^3 + (2*a^2 +
5*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 + a*b)*arctan(1/2*sqrt(a^2 + a*b)*(cosh(d*x + c) + sinh(
d*x + c))/(a + b)) - 2*(a^2*b + a*b^2)*cosh(d*x + c) - 2*(a^2*b + a*b^2 - 3*(a^2*b + a*b^2)*cosh(d*x + c)^2)*s
inh(d*x + c))/((a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(
d*x + c)^3 + (a^5 + 2*a^4*b + a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^5 + 4*a^4*b + 5*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*
x + c)^2 + 2*(3*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^2 + (a^5 + 4*a^4*b + 5*a^3*b^2 + 2*a^2*b^3)*d)*sinh(
d*x + c)^2 + (a^5 + 2*a^4*b + a^3*b^2)*d + 4*((a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^5 + 4*a^4*b + 5
*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}{\left (c + d x \right )}}{\left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)/(a+b*sech(d*x+c)**2)**2,x)
[Out]
Integral(sech(c + d*x)/(a + b*sech(c + d*x)**2)**2, x)
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
[Out]
Exception raised: TypeError