### 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[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]))

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Rubi [A]  time = 0.103961, antiderivative size = 89, normalized size of antiderivative = 1., 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 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]

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 12

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

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 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 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])

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*}

Mathematica [A]  time = 0.336885, 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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Maple [B]  time = 0.119, size = 469, normalized size = 5.3 \begin{align*} 2\,{\frac{{b}^{2} \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+2\,b \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,b\tanh \left ( 1/2\,dx+c/2 \right ) +a \right ) a \left ( 4\,{a}^{2}+{b}^{2} \right ) }}-8\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{d \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+2\,b \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,b\tanh \left ( 1/2\,dx+c/2 \right ) +a \right ) \left ( 4\,{a}^{2}+{b}^{2} \right ) }}+2\,{\frac{{b}^{2}\tanh \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+2\,b \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,b\tanh \left ( 1/2\,dx+c/2 \right ) +a \right ) a \left ( 4\,{a}^{2}+{b}^{2} \right ) }}+16\,{\frac{{a}^{3}\ln \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+\sqrt{4\,{a}^{2}+{b}^{2}}\tanh \left ( 1/2\,dx+c/2 \right ) +b\tanh \left ( 1/2\,dx+c/2 \right ) +a \right ) }{d \left ( 4\,{a}^{2}+{b}^{2} \right ) ^{5/2}}}+4\,{\frac{a\ln \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+\sqrt{4\,{a}^{2}+{b}^{2}}\tanh \left ( 1/2\,dx+c/2 \right ) +b\tanh \left ( 1/2\,dx+c/2 \right ) +a \right ){b}^{2}}{d \left ( 4\,{a}^{2}+{b}^{2} \right ) ^{5/2}}}-4\,{\frac{a\ln \left ( - \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+\sqrt{4\,{a}^{2}+{b}^{2}}\tanh \left ( 1/2\,dx+c/2 \right ) -b\tanh \left ( 1/2\,dx+c/2 \right ) -a \right ) }{d \left ( 4\,{a}^{2}+{b}^{2} \right ) ^{3/2}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.16257, size = 1895, normalized size = 21.29 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.19051, size = 208, normalized size = 2.34 \begin{align*} -\frac{4 \, 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} d + b^{2} d\right )} \sqrt{4 \, a^{2} + b^{2}}} + \frac{4 \,{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - b\right )}}{{\left (4 \, a^{2} d + b^{2} d\right )}{\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - b\right )}} \end{align*}

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