3.859 \(\int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac {8 a \tanh ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{d \left (4 a^2+b^2\right )^{3/2}}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))} \]

[Out]

-8*a*arctanh((b-2*a*tanh(d*x+c))/(4*a^2+b^2)^(1/2))/(4*a^2+b^2)^(3/2)/d-2*b*cosh(2*d*x+2*c)/(4*a^2+b^2)/d/(2*a
+b*sinh(2*d*x+2*c))

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Rubi [A]  time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2666, 2664, 12, 2660, 618, 204} \[ -\frac {8 a \tanh ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{d \left (4 a^2+b^2\right )^{3/2}}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-2),x]

[Out]

(-8*a*ArcTanh[(b - 2*a*Tanh[c + d*x])/Sqrt[4*a^2 + b^2]])/((4*a^2 + b^2)^(3/2)*d) - (2*b*Cosh[2*c + 2*d*x])/((
4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^2} \, dx\\ &=-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))}+\frac {4 \int \frac {a}{a+\frac {1}{2} b \sinh (2 c+2 d x)} \, dx}{4 a^2+b^2}\\ &=-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))}+\frac {(4 a) \int \frac {1}{a+\frac {1}{2} b \sinh (2 c+2 d x)} \, dx}{4 a^2+b^2}\\ &=-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))}-\frac {(4 i a) \operatorname {Subst}\left (\int \frac {1}{a-i b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (2 i c+2 i d x)\right )\right )}{\left (4 a^2+b^2\right ) d}\\ &=-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))}+\frac {(8 i a) \operatorname {Subst}\left (\int \frac {1}{-4 a^2-b^2-x^2} \, dx,x,-i b+2 a \tan \left (\frac {1}{2} (2 i c+2 i d x)\right )\right )}{\left (4 a^2+b^2\right ) d}\\ &=-\frac {8 a \tanh ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{\left (4 a^2+b^2\right )^{3/2} d}-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 90, normalized size = 1.01 \[ \frac {2 \left (-\frac {4 a \tan ^{-1}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {-4 a^2-b^2}}\right )}{\left (-4 a^2-b^2\right )^{3/2}}-\frac {b \cosh (2 (c+d x))}{\left (4 a^2+b^2\right ) (2 a+b \sinh (2 (c+d x)))}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-2),x]

[Out]

(2*((-4*a*ArcTan[(b - 2*a*Tanh[c + d*x])/Sqrt[-4*a^2 - b^2]])/(-4*a^2 - b^2)^(3/2) - (b*Cosh[2*(c + d*x)])/((4
*a^2 + b^2)*(2*a + b*Sinh[2*(c + d*x)]))))/d

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fricas [B]  time = 0.55, size = 765, normalized size = 8.60 \[ -\frac {4 \, {\left (4 \, a^{2} b + b^{3} - 2 \, {\left (4 \, a^{3} + a b^{2}\right )} \cosh \left (d x + c\right )^{2} - 4 \, {\left (4 \, a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - 2 \, {\left (4 \, a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )^{2} - {\left (a b \cosh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a b \sinh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a b \cosh \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sinh \left (d x + c\right )^{2} - a b + 4 \, {\left (a b \cosh \left (d x + c\right )^{3} + 2 \, a^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {4 \, a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + 8 \, a^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2 \, {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + 2 \, a\right )} \sqrt {4 \, a^{2} + b^{2}}}{b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cosh \left (d x + c\right )^{2} + 2 \, a\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b \cosh \left (d x + c\right )^{3} + 2 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - b}\right )\right )}}{{\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{4} + 4 \, {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{4} + 4 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )^{2} - {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d + 4 \, {\left ({\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{3} + 2 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

-4*(4*a^2*b + b^3 - 2*(4*a^3 + a*b^2)*cosh(d*x + c)^2 - 4*(4*a^3 + a*b^2)*cosh(d*x + c)*sinh(d*x + c) - 2*(4*a
^3 + a*b^2)*sinh(d*x + c)^2 - (a*b*cosh(d*x + c)^4 + 4*a*b*cosh(d*x + c)*sinh(d*x + c)^3 + a*b*sinh(d*x + c)^4
+ 4*a^2*cosh(d*x + c)^2 + 2*(3*a*b*cosh(d*x + c)^2 + 2*a^2)*sinh(d*x + c)^2 - a*b + 4*(a*b*cosh(d*x + c)^3 +
2*a^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(4*a^2 + b^2)*log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x
+ c)^3 + b^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b)*sinh(d*x + c)^2 + 8*a
^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + 2*a*b*cosh(d*x + c))*sinh(d*x + c) - 2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x +
c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a)*sqrt(4*a^2 + b^2))/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*
x + c)^3 + b*sinh(d*x + c)^4 + 4*a*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a)*sinh(d*x + c)^2 + 4*(b*cosh
(d*x + c)^3 + 2*a*cosh(d*x + c))*sinh(d*x + c) - b)))/((16*a^4*b + 8*a^2*b^3 + b^5)*d*cosh(d*x + c)^4 + 4*(16*
a^4*b + 8*a^2*b^3 + b^5)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (16*a^4*b + 8*a^2*b^3 + b^5)*d*sinh(d*x + c)^4 + 4*
(16*a^5 + 8*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + 2*(3*(16*a^4*b + 8*a^2*b^3 + b^5)*d*cosh(d*x + c)^2 + 2*(16*a
^5 + 8*a^3*b^2 + a*b^4)*d)*sinh(d*x + c)^2 - (16*a^4*b + 8*a^2*b^3 + b^5)*d + 4*((16*a^4*b + 8*a^2*b^3 + b^5)*
d*cosh(d*x + c)^3 + 2*(16*a^5 + 8*a^3*b^2 + a*b^4)*d*cosh(d*x + c))*sinh(d*x + c))

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giac [A]  time = 0.51, size = 140, normalized size = 1.57 \[ \frac {4 \, {\left (\frac {a \log \left (\frac {{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}\right )}{{\left (4 \, a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, a e^{\left (2 \, d x + 2 \, c\right )} - b}{{\left (4 \, a^{2} + b^{2}\right )} {\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - b\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

4*(a*log(abs(2*b*e^(2*d*x + 2*c) + 4*a - 2*sqrt(4*a^2 + b^2))/abs(2*b*e^(2*d*x + 2*c) + 4*a + 2*sqrt(4*a^2 + b
^2)))/(4*a^2 + b^2)^(3/2) + (2*a*e^(2*d*x + 2*c) - b)/((4*a^2 + b^2)*(b*e^(4*d*x + 4*c) + 4*a*e^(2*d*x + 2*c)
- b)))/d

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maple [B]  time = 0.80, size = 469, normalized size = 5.27 \[ \frac {2 b^{2} \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a \left (4 a^{2}+b^{2}\right )}-\frac {8 b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \left (4 a^{2}+b^{2}\right )}+\frac {2 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a \left (4 a^{2}+b^{2}\right )}-\frac {4 a \ln \left (-\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {16 a^{3} \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}{d \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {4 a \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b^{2}}{d \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^2,x)

[Out]

2/d/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^3*b-2*tanh(1/2*d*x+1/2*c)^2*a+2*tanh(1/2*d*x+1/2*c)*b+a)*b^
2/a/(4*a^2+b^2)*tanh(1/2*d*x+1/2*c)^3-8/d/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^3*b-2*tanh(1/2*d*x+1/
2*c)^2*a+2*tanh(1/2*d*x+1/2*c)*b+a)*b/(4*a^2+b^2)*tanh(1/2*d*x+1/2*c)^2+2/d/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/
2*d*x+1/2*c)^3*b-2*tanh(1/2*d*x+1/2*c)^2*a+2*tanh(1/2*d*x+1/2*c)*b+a)*b^2/a/(4*a^2+b^2)*tanh(1/2*d*x+1/2*c)-4/
d*a/(4*a^2+b^2)^(3/2)*ln(-tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)-tanh(1/2*d*x+1/2*c)*b-
a)+16/d*a^3/(4*a^2+b^2)^(5/2)*ln(tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)+tanh(1/2*d*x+1/
2*c)*b+a)+4/d*a/(4*a^2+b^2)^(5/2)*ln(tanh(1/2*d*x+1/2*c)^2*a+(4*a^2+b^2)^(1/2)*tanh(1/2*d*x+1/2*c)+tanh(1/2*d*
x+1/2*c)*b+a)*b^2

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maxima [A]  time = 0.43, size = 150, normalized size = 1.69 \[ \frac {4 \, a \log \left (\frac {b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a - \sqrt {4 \, a^{2} + b^{2}}}{b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a + \sqrt {4 \, a^{2} + b^{2}}}\right )}{{\left (4 \, a^{2} + b^{2}\right )}^{\frac {3}{2}} d} - \frac {4 \, {\left (2 \, a e^{\left (-2 \, d x - 2 \, c\right )} + b\right )}}{{\left (4 \, a^{2} b + b^{3} + 4 \, {\left (4 \, a^{3} + a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

4*a*log((b*e^(-2*d*x - 2*c) - 2*a - sqrt(4*a^2 + b^2))/(b*e^(-2*d*x - 2*c) - 2*a + sqrt(4*a^2 + b^2)))/((4*a^2
+ b^2)^(3/2)*d) - 4*(2*a*e^(-2*d*x - 2*c) + b)/((4*a^2*b + b^3 + 4*(4*a^3 + a*b^2)*e^(-2*d*x - 2*c) - (4*a^2*
b + b^3)*e^(-4*d*x - 4*c))*d)

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mupad [B]  time = 2.11, size = 229, normalized size = 2.57 \[ \frac {4\,a\,\ln \left (\frac {16\,a\,\left (b-2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {16\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{4\,a^2\,b+b^3}\right )}{d\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {4\,a\,\ln \left (-\frac {16\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{4\,a^2\,b+b^3}-\frac {16\,a\,\left (b-2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b\,{\left (4\,a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {\frac {4\,b^2}{d\,\left (4\,a^2\,b+b^3\right )}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d\,\left (4\,a^2\,b+b^3\right )}}{4\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}-b+b\,{\mathrm {e}}^{4\,c+4\,d\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(c + d*x)*sinh(c + d*x))^2,x)

[Out]

(4*a*log((16*a*(b - 2*a*exp(2*c + 2*d*x)))/(b*(4*a^2 + b^2)^(3/2)) - (16*a*exp(2*c + 2*d*x))/(4*a^2*b + b^3)))
/(d*(4*a^2 + b^2)^(3/2)) - (4*a*log(- (16*a*exp(2*c + 2*d*x))/(4*a^2*b + b^3) - (16*a*(b - 2*a*exp(2*c + 2*d*x
)))/(b*(4*a^2 + b^2)^(3/2))))/(d*(4*a^2 + b^2)^(3/2)) - ((4*b^2)/(d*(4*a^2*b + b^3)) - (8*a*b*exp(2*c + 2*d*x)
)/(d*(4*a^2*b + b^3)))/(4*a*exp(2*c + 2*d*x) - b + b*exp(4*c + 4*d*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))**2,x)

[Out]

Timed out

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