3.1.36 \(\int \frac {\sinh (c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [36]
Optimal. Leaf size=84 \[ -\frac {3 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{2 a^{5/2} d}+\frac {3 \cosh (c+d x)}{2 a^2 d}-\frac {\cosh ^3(c+d x)}{2 a d \left (b+a \cosh ^2(c+d x)\right )} \]
[Out]
3/2*cosh(d*x+c)/a^2/d-1/2*cosh(d*x+c)^3/a/d/(b+a*cosh(d*x+c)^2)-3/2*arctan(cosh(d*x+c)*a^(1/2)/b^(1/2))*b^(1/2
)/a^(5/2)/d
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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4218, 294, 327,
211} \begin {gather*} -\frac {3 \sqrt {b} \text {ArcTan}\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 )} \end {gather*}
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 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 294
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 327
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 4218
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
]
Rubi steps
\begin {align*} \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx &=\frac {\text {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 \text {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) \text {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*}
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Mathematica [C] Result contains complex when optimal does not.
time = 1.87, size = 479, normalized size = 5.70 \begin {gather*} \frac {(a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}^4(c+d x) \left (\frac {32 \cosh (c) \cosh (d x)}{a^2}+\frac {32 b \cosh (c+d x)}{a^2 (a+2 b+a \cosh (2 (c+d x)))}+\frac {2 \left (-\left (\left (a^2+24 b^2\right ) \text {ArcTan}\left (\frac {\left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{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^2 \text {ArcTan}\left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{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 )-24 b^2 \text {ArcTan}\left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{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 \text {ArcTan}\left (\frac {\sqrt {a}-i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+a^2 \text {ArcTan}\left (\frac {\sqrt {a}+i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+16 \sqrt {a} b^{3/2} \sinh (c) \sinh (d x)\right )}{a^{5/2} b^{3/2}}\right )}{128 d \left (a+b \text {sech}^2(c+d x)\right )^2} \end {gather*}
Warning: Unable to verify antiderivative.
[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 = 0.94, size = 70, normalized size = 0.83
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(-\frac {-\frac {1}{a^{2} \mathrm {sech}\left (d x +c \right )}-\frac {b \left (\frac {\mathrm {sech}\left (d x +c \right )}{2 a +2 b \mathrm {sech}\left (d x +c \right )^{2}}+\frac {3 \arctan \left (\frac {b \,\mathrm {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{d}\) |
\(70\) |
| default |
\(-\frac {-\frac {1}{a^{2} \mathrm {sech}\left (d x +c \right )}-\frac {b \left (\frac {\mathrm {sech}\left (d x +c \right )}{2 a +2 b \mathrm {sech}\left (d x +c \right )^{2}}+\frac {3 \arctan \left (\frac {b \,\mathrm {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{d}\) |
\(70\) |
| risch |
\(\frac {{\mathrm e}^{d x +c}}{2 a^{2} d}+\frac {{\mathrm e}^{-d x -c}}{2 a^{2} d}+\frac {{\mathrm e}^{d x +c} b \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a^{2} d \left (a \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}\) |
\(183\) |
| | |
|
|
|
|
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,method=_RETURNVERBOSE)
[Out]
-1/d*(-1/a^2/sech(d*x+c)-b/a^2*(1/2*sech(d*x+c)/(a+b*sech(d*x+c)^2)+3/2/(a*b)^(1/2)*arctan(b*sech(d*x+c)/(a*b)
^(1/2))))
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
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] Leaf count of result is larger than twice the leaf count of optimal. 877 vs.
\(2 (70) = 140\).
time = 0.40, size = 1780, normalized size = 21.19 \begin {gather*} \text {Too large to display} \end {gather*}
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh {\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}
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]
Integral(sinh(c + d*x)/(a + b*sech(c + d*x)**2)**2, x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done
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Mupad [B]
time = 1.65, size = 71, normalized size = 0.85 \begin {gather*} \frac {b\,\mathrm {cosh}\left (c+d\,x\right )}{2\,\left (d\,a^3\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\,d\,a^2\right )}+\frac {\mathrm {cosh}\left (c+d\,x\right )}{a^2\,d}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {cosh}\left (c+d\,x\right )}{\sqrt {b}}\right )}{2\,a^{5/2}\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(c + d*x)/(a + b/cosh(c + d*x)^2)^2,x)
[Out]
(b*cosh(c + d*x))/(2*(a^3*d*cosh(c + d*x)^2 + a^2*b*d)) + cosh(c + d*x)/(a^2*d) - (3*b^(1/2)*atan((a^(1/2)*cos
h(c + d*x))/b^(1/2)))/(2*a^(5/2)*d)
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