3.36 \(\int \frac{\sinh (c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)
Optimal. Leaf size=84 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{2 a^{5/2} d}+\frac{3 \cosh (c+d x)}{2 a^2 d}-\frac{\cosh ^3(c+d x)}{2 a d \left (a \cosh ^2(c+d x)+b\right )} \]
[Out]
(-3*Sqrt[b]*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/(2*a^(5/2)*d) + (3*Cosh[c + d*x])/(2*a^2*d) - Cosh[c + d*
x]^3/(2*a*d*(b + a*Cosh[c + d*x]^2))
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Rubi [A] time = 0.0647587, antiderivative size = 84, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{4133, 288, 321, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{2 a^{5/2} d}+\frac{3 \cosh (c+d x)}{2 a^2 d}-\frac{\cosh ^3(c+d x)}{2 a d \left (a \cosh ^2(c+d x)+b\right )} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
(-3*Sqrt[b]*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/(2*a^(5/2)*d) + (3*Cosh[c + d*x])/(2*a^2*d) - Cosh[c + d*
x]^3/(2*a*d*(b + a*Cosh[c + d*x]^2))
Rule 4133
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*(ff*x)^n)^p)/(ff*x)^(n*
p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
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{\sinh (c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (b+a x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\cosh ^3(c+d x)}{2 a d \left (b+a \cosh ^2(c+d x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{b+a x^2} \, dx,x,\cosh (c+d x)\right )}{2 a d}\\ &=\frac{3 \cosh (c+d x)}{2 a^2 d}-\frac{\cosh ^3(c+d x)}{2 a d \left (b+a \cosh ^2(c+d x)\right )}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cosh (c+d x)\right )}{2 a^2 d}\\ &=-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{2 a^{5/2} d}+\frac{3 \cosh (c+d x)}{2 a^2 d}-\frac{\cosh ^3(c+d x)}{2 a d \left (b+a \cosh ^2(c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 2.70154, size = 479, normalized size = 5.7 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b)^2 \left (\frac{2 \left (-\left (a^2+24 b^2\right ) \tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )+a^2 \tan ^{-1}\left (\frac{\sqrt{a}-i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )+a^2 \tan ^{-1}\left (\frac{\sqrt{a}+i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )+a^2 \left (-\tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )\right )+16 \sqrt{a} b^{3/2} \sinh (c) \sinh (d x)-24 b^2 \tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )\right )}{a^{5/2} b^{3/2}}+\frac{32 b \cosh (c+d x)}{a^2 (a \cosh (2 (c+d x))+a+2 b)}+\frac{32 \cosh (c) \cosh (d x)}{a^2}\right )}{128 d \left (a+b \text{sech}^2(c+d x)\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*((32*Cosh[c]*Cosh[d*x])/a^2 + (32*b*Cosh[c + d*x])/(a^2*(a
+ 2*b + a*Cosh[2*(c + d*x)])) + (2*(-((a^2 + 24*b^2)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])
^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt
[b]]) - a^2*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqr
t[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] - 24*b^2*ArcTan[((Sqrt[a] + I*Sqrt[a
+ b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Si
nh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + a^2*ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + a^2*Arc
Tan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + 16*Sqrt[a]*b^(3/2)*Sinh[c]*Sinh[d*x]))/(a^(5/2)*b^(
3/2))))/(128*d*(a + b*Sech[c + d*x]^2)^2)
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Maple [A] time = 0.031, size = 74, normalized size = 0.9 \begin{align*}{\frac{b{\rm sech} \left (dx+c\right )}{2\,d{a}^{2} \left ( a+b \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) }}+{\frac{3\,b}{2\,d{a}^{2}}\arctan \left ({b{\rm sech} \left (dx+c\right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{d{a}^{2}{\rm sech} \left (dx+c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^2,x)
[Out]
1/2/d/a^2*b*sech(d*x+c)/(a+b*sech(d*x+c)^2)+3/2/d/a^2*b/(a*b)^(1/2)*arctan(sech(d*x+c)*b/(a*b)^(1/2))+1/d/a^2/
sech(d*x+c)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 \,{\left (a e^{\left (4 \, c\right )} + 2 \, b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \,{\left (a e^{\left (2 \, c\right )} + 2 \, b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + a e^{\left (6 \, d x + 6 \, c\right )} + a}{2 \,{\left (a^{3} d e^{\left (5 \, d x + 5 \, c\right )} + a^{3} d e^{\left (d x + c\right )} + 2 \,{\left (a^{3} d e^{\left (3 \, c\right )} + 2 \, a^{2} b d e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )}\right )}} - \frac{1}{2} \, \int \frac{6 \,{\left (b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}\right )}}{a^{3} e^{\left (4 \, d x + 4 \, c\right )} + a^{3} + 2 \,{\left (a^{3} e^{\left (2 \, c\right )} + 2 \, a^{2} b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
1/2*(3*(a*e^(4*c) + 2*b*e^(4*c))*e^(4*d*x) + 3*(a*e^(2*c) + 2*b*e^(2*c))*e^(2*d*x) + a*e^(6*d*x + 6*c) + a)/(a
^3*d*e^(5*d*x + 5*c) + a^3*d*e^(d*x + c) + 2*(a^3*d*e^(3*c) + 2*a^2*b*d*e^(3*c))*e^(3*d*x)) - 1/2*integrate(6*
(b*e^(3*d*x + 3*c) - b*e^(d*x + c))/(a^3*e^(4*d*x + 4*c) + a^3 + 2*(a^3*e^(2*c) + 2*a^2*b*e^(2*c))*e^(2*d*x)),
x)
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Fricas [B] time = 2.87968, size = 4732, normalized size = 56.33 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
[1/4*(2*a*cosh(d*x + c)^6 + 12*a*cosh(d*x + c)*sinh(d*x + c)^5 + 2*a*sinh(d*x + c)^6 + 6*(a + 2*b)*cosh(d*x +
c)^4 + 6*(5*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^4 + 8*(5*a*cosh(d*x + c)^3 + 3*(a + 2*b)*cosh(d*x + c))
*sinh(d*x + c)^3 + 6*(a + 2*b)*cosh(d*x + c)^2 + 6*(5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)^2 + a + 2*
b)*sinh(d*x + c)^2 + 3*(a*cosh(d*x + c)^5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 + 2*(a + 2*b
)*cosh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^3 + 3*(a + 2*b)*c
osh(d*x + c))*sinh(d*x + c)^2 + a*cosh(d*x + c) + (5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)^2 + a)*sinh
(d*x + c))*sqrt(-b/a)*log((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)*co
sh(d*x + c))*sinh(d*x + c) - 4*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3 + a*
cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))*sqrt(-b/a) + 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)) + 12*(a*cosh(d*x + c)^5 + 2*(a
+ 2*b)*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + 2*a)/(a^3*d*cosh(d*x + c)^5 + 5*a^3*d*cosh(
d*x + c)*sinh(d*x + c)^4 + a^3*d*sinh(d*x + c)^5 + a^3*d*cosh(d*x + c) + 2*(a^3 + 2*a^2*b)*d*cosh(d*x + c)^3 +
2*(5*a^3*d*cosh(d*x + c)^2 + (a^3 + 2*a^2*b)*d)*sinh(d*x + c)^3 + 2*(5*a^3*d*cosh(d*x + c)^3 + 3*(a^3 + 2*a^2
*b)*d*cosh(d*x + c))*sinh(d*x + c)^2 + (5*a^3*d*cosh(d*x + c)^4 + a^3*d + 6*(a^3 + 2*a^2*b)*d*cosh(d*x + c)^2)
*sinh(d*x + c)), 1/2*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*sinh(d*x + c)^6 + 3*(a + 2*b)*
cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3 + 3*(a + 2*b)*cos
h(d*x + c))*sinh(d*x + c)^3 + 3*(a + 2*b)*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)
^2 + a + 2*b)*sinh(d*x + c)^2 + 3*(a*cosh(d*x + c)^5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 +
2*(a + 2*b)*cosh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^3 + 3*
(a + 2*b)*cosh(d*x + c))*sinh(d*x + c)^2 + a*cosh(d*x + c) + (5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)^
2 + a)*sinh(d*x + c))*sqrt(b/a)*arctan(1/2*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x
+ c)^3 + (a + 4*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a + 4*b)*sinh(d*x + c))*sqrt(b/a)/b) - 3*(a*cosh(d*
x + c)^5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 + 2*(a + 2*b)*cosh(d*x + c)^3 + 2*(5*a*cosh(d
*x + c)^2 + a + 2*b)*sinh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^3 + 3*(a + 2*b)*cosh(d*x + c))*sinh(d*x + c)^2 + a
*cosh(d*x + c) + (5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)^2 + a)*sinh(d*x + c))*sqrt(b/a)*arctan(1/2*(
a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt(b/a)/b) + 6*(a*cosh(d*x + c)^5 + 2*(a + 2*b)*cosh(d*x + c)^3 + (a + 2*
b)*cosh(d*x + c))*sinh(d*x + c) + a)/(a^3*d*cosh(d*x + c)^5 + 5*a^3*d*cosh(d*x + c)*sinh(d*x + c)^4 + a^3*d*si
nh(d*x + c)^5 + a^3*d*cosh(d*x + c) + 2*(a^3 + 2*a^2*b)*d*cosh(d*x + c)^3 + 2*(5*a^3*d*cosh(d*x + c)^2 + (a^3
+ 2*a^2*b)*d)*sinh(d*x + c)^3 + 2*(5*a^3*d*cosh(d*x + c)^3 + 3*(a^3 + 2*a^2*b)*d*cosh(d*x + c))*sinh(d*x + c)^
2 + (5*a^3*d*cosh(d*x + c)^4 + a^3*d + 6*(a^3 + 2*a^2*b)*d*cosh(d*x + c)^2)*sinh(d*x + c))]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)**2)**2,x)
[Out]
Timed out
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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(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
[Out]
Exception raised: TypeError