### 3.587 $$\int \frac{1}{(a \cosh (x)+b \sinh (x))^3} \, dx$$

Optimal. Leaf size=77 $\frac{a \sinh (x)+b \cosh (x)}{2 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^2}+\frac{\tan ^{-1}\left (\frac{a \sinh (x)+b \cosh (x)}{\sqrt{a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0466193, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {3076, 3074, 206} $\frac{a \sinh (x)+b \cosh (x)}{2 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^2}+\frac{\tan ^{-1}\left (\frac{a \sinh (x)+b \cosh (x)}{\sqrt{a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-3),x]

[Out]

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

Rule 3076

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

Rule 3074

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.478082, size = 96, normalized size = 1.25 $\frac{1}{2} \left (\frac{2 \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac{b}{a (a-b) (a+b) (a \cosh (x)+b \sinh (x))}+\frac{\sinh (x)}{a (a \cosh (x)+b \sinh (x))^2}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^(-3),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.066, size = 167, normalized size = 2.2 \begin{align*} 2\,{\frac{1}{ \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) ^{2}} \left ( -1/2\,{\frac{ \left ({a}^{2}-2\,{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{ \left ({a}^{2}-{b}^{2} \right ) a}}+1/2\,{\frac{b \left ({a}^{2}+2\,{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{ \left ({a}^{2}-{b}^{2} \right ){a}^{2}}}+1/2\,{\frac{ \left ({a}^{2}+2\,{b}^{2} \right ) \tanh \left ( x/2 \right ) }{ \left ({a}^{2}-{b}^{2} \right ) a}}+1/2\,{\frac{b}{{a}^{2}-{b}^{2}}} \right ) }+{\arctan \left ({\frac{1}{2} \left ( 2\,a\tanh \left ( x/2 \right ) +2\,b \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)+b*sinh(x))^3,x)

[Out]

2*(-1/2*(a^2-2*b^2)/(a^2-b^2)/a*tanh(1/2*x)^3+1/2*b*(a^2+2*b^2)/(a^2-b^2)/a^2*tanh(1/2*x)^2+1/2*(a^2+2*b^2)/(a
^2-b^2)/a*tanh(1/2*x)+1/2*b/(a^2-b^2))/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2+1/(a^2-b^2)^(3/2)*arctan(1/2*(2*a
*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))

________________________________________________________________________________________

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*cosh(x)+b*sinh(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.6482, size = 3501, normalized size = 45.47 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^3,x, algorithm="fricas")

[Out]

[1/2*(2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^3 + 6*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)*sinh(x)^2 + 2*(a^3 + a^2
*b - a*b^2 - b^3)*sinh(x)^3 + ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2
+ 2*a*b + b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2
+ a^2 - 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*log((
(a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + 2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x)) - a
+ b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) - 2*(a^3 - a^2*b - a*b^2 +
b^3)*cosh(x) - 2*(a^3 - a^2*b - a*b^2 + b^3 - 3*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2)*sinh(x))/(a^6 - 2*a^5*b
- a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 + (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 +
b^6)*cosh(x)^4 + 4*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)*sinh(x)^3 + (a^6 +
2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*sinh(x)^4 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos
h(x)^2 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 3*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b
^6)*cosh(x)^2)*sinh(x)^2 + 4*((a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)^3 + (a^6
- 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x)), ((a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^3 + 3*(a^3 + a^2*b - a
*b^2 - b^3)*cosh(x)*sinh(x)^2 + (a^3 + a^2*b - a*b^2 - b^3)*sinh(x)^3 - ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^
2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a
*b + b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 + (a^2 - b^2
)*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) - (a^3 - a^2*b
- a*b^2 + b^3)*cosh(x) - (a^3 - a^2*b - a*b^2 + b^3 - 3*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2)*sinh(x))/(a^6
- 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 + (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2
*a*b^5 + b^6)*cosh(x)^4 + 4*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)*sinh(x)^3
+ (a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*sinh(x)^4 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 -
b^6)*cosh(x)^2 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + 3*(a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*
a*b^5 + b^6)*cosh(x)^2)*sinh(x)^2 + 4*((a^6 + 2*a^5*b - a^4*b^2 - 4*a^3*b^3 - a^2*b^4 + 2*a*b^5 + b^6)*cosh(x)
^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x))*sinh(x))]

________________________________________________________________________________________

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*cosh(x)+b*sinh(x))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.147, size = 119, normalized size = 1.55 \begin{align*} \frac{\arctan \left (\frac{a e^{x} + b e^{x}}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} - a e^{x} + b e^{x}}{{\left (a^{2} - b^{2}\right )}{\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^3,x, algorithm="giac")

[Out]

arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/(a^2 - b^2)^(3/2) + (a*e^(3*x) + b*e^(3*x) - a*e^x + b*e^x)/((a^2 - b^
2)*(a*e^(2*x) + b*e^(2*x) + a - b)^2)