\(\int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx\) [589]

Optimal result

Integrand size = 11, antiderivative size = 112 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\frac {3 \arctan \left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{8 \left (a^2-b^2\right )^{5/2}}+\frac {b \cosh (x)+a \sinh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4}+\frac {3 (b \cosh (x)+a \sinh (x))}{8 \left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))^2} \]

[Out]

3/8*arctan((b*cosh(x)+a*sinh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)+1/4*(b*cosh(x)+a*sinh(x))/(a^2-b^2)/(a*cosh(
x)+b*sinh(x))^4+3/8*(b*cosh(x)+a*sinh(x))/(a^2-b^2)^2/(a*cosh(x)+b*sinh(x))^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3155, 3153, 212} \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\frac {3 \arctan \left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{8 \left (a^2-b^2\right )^{5/2}}+\frac {3 (a \sinh (x)+b \cosh (x))}{8 \left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))^2}+\frac {a \sinh (x)+b \cosh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4} \]

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-5),x]

[Out]

(3*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(8*(a^2 - b^2)^(5/2)) + (b*Cosh[x] + a*Sinh[x])/(4*(a^2 -
b^2)*(a*Cosh[x] + b*Sinh[x])^4) + (3*(b*Cosh[x] + a*Sinh[x]))/(8*(a^2 - b^2)^2*(a*Cosh[x] + b*Sinh[x])^2)

Rule 212

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

Rule 3153

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 3155

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]

Rubi steps \begin{align*} \text {integral}& = \frac {b \cosh (x)+a \sinh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4}+\frac {3 \int \frac {1}{(a \cosh (x)+b \sinh (x))^3} \, dx}{4 \left (a^2-b^2\right )} \\ & = \frac {b \cosh (x)+a \sinh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4}+\frac {3 (b \cosh (x)+a \sinh (x))}{8 \left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))^2}+\frac {3 \int \frac {1}{a \cosh (x)+b \sinh (x)} \, dx}{8 \left (a^2-b^2\right )^2} \\ & = \frac {b \cosh (x)+a \sinh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4}+\frac {3 (b \cosh (x)+a \sinh (x))}{8 \left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))^2}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{8 \left (a^2-b^2\right )^2} \\ & = \frac {3 \arctan \left (\frac {b \cosh (x)+a \sinh (x)}{\sqrt {a^2-b^2}}\right )}{8 \left (a^2-b^2\right )^{5/2}}+\frac {b \cosh (x)+a \sinh (x)}{4 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^4}+\frac {3 (b \cosh (x)+a \sinh (x))}{8 \left (a^2-b^2\right )^2 (a \cosh (x)+b \sinh (x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\frac {1}{8} \left (\frac {6 \arctan \left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}+\frac {b \left (2 (a-b) (a+b)+3 (a \cosh (x)+b \sinh (x))^2\right )}{a (a-b)^2 (a+b)^2 (a \cosh (x)+b \sinh (x))^3}+\frac {\sinh (x) \left (2+\frac {3 (a \cosh (x)+b \sinh (x))^2}{(a-b) (a+b)}\right )}{a (a \cosh (x)+b \sinh (x))^4}\right ) \]

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^(-5),x]

[Out]

((6*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])])/((a - b)^(5/2)*(a + b)^(5/2)) + (b*(2*(a - b)*(a + b)
+ 3*(a*Cosh[x] + b*Sinh[x])^2))/(a*(a - b)^2*(a + b)^2*(a*Cosh[x] + b*Sinh[x])^3) + (Sinh[x]*(2 + (3*(a*Cosh[
x] + b*Sinh[x])^2)/((a - b)*(a + b))))/(a*(a*Cosh[x] + b*Sinh[x])^4))/8

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(461\) vs. \(2(102)=204\).

Time = 0.29 (sec) , antiderivative size = 462, normalized size of antiderivative = 4.12

\[\frac {-\frac {\left (5 a^{4}-16 a^{2} b^{2}+8 b^{4}\right ) \tanh \left (\frac {x}{2}\right )^{7}}{4 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {3 b \left (a^{4}-16 a^{2} b^{2}+8 b^{4}\right ) \tanh \left (\frac {x}{2}\right )^{6}}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {\left (3 a^{6}+36 a^{4} b^{2}+56 a^{2} b^{4}-32 b^{6}\right ) \tanh \left (\frac {x}{2}\right )^{5}}{4 a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (15 a^{6}+114 a^{4} b^{2}-8 a^{2} b^{4}-16 b^{6}\right ) \tanh \left (\frac {x}{2}\right )^{4}}{4 a^{4} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (3 a^{6}-84 a^{4} b^{2}-56 a^{2} b^{4}+32 b^{6}\right ) \tanh \left (\frac {x}{2}\right )^{3}}{4 a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (23 a^{4}+64 a^{2} b^{2}-24 b^{4}\right ) \tanh \left (\frac {x}{2}\right )^{2}}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {\left (5 a^{4}+24 a^{2} b^{2}-8 b^{4}\right ) \tanh \left (\frac {x}{2}\right )}{4 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 \left (5 a^{2}-2 b^{2}\right ) b}{8 a^{4}-16 a^{2} b^{2}+8 b^{4}}}{\left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )^{4}}+\frac {3 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\]

[In]

int(1/(a*cosh(x)+b*sinh(x))^5,x)

[Out]

2*(-1/8*(5*a^4-16*a^2*b^2+8*b^4)/a/(a^4-2*a^2*b^2+b^4)*tanh(1/2*x)^7-3/8*b*(a^4-16*a^2*b^2+8*b^4)/(a^4-2*a^2*b
^2+b^4)/a^2*tanh(1/2*x)^6+1/8/a^3*(3*a^6+36*a^4*b^2+56*a^2*b^4-32*b^6)/(a^4-2*a^2*b^2+b^4)*tanh(1/2*x)^5+1/8/a
^4*b*(15*a^6+114*a^4*b^2-8*a^2*b^4-16*b^6)/(a^4-2*a^2*b^2+b^4)*tanh(1/2*x)^4-1/8/a^3*(3*a^6-84*a^4*b^2-56*a^2*
b^4+32*b^6)/(a^4-2*a^2*b^2+b^4)*tanh(1/2*x)^3+1/8*b*(23*a^4+64*a^2*b^2-24*b^4)/(a^4-2*a^2*b^2+b^4)/a^2*tanh(1/
2*x)^2+1/8*(5*a^4+24*a^2*b^2-8*b^4)/a/(a^4-2*a^2*b^2+b^4)*tanh(1/2*x)+1/8*(5*a^2-2*b^2)*b/(a^4-2*a^2*b^2+b^4))
/(tanh(1/2*x)^2*a+2*b*tanh(1/2*x)+a)^4+3/4/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b
)/(a^2-b^2)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3408 vs. \(2 (102) = 204\).

Time = 0.35 (sec) , antiderivative size = 6874, normalized size of antiderivative = 61.38 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^5,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\text {Timed out} \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))**5,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (102) = 204\).

Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.11 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\frac {3 \, \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} e^{\left (7 \, x\right )} + 9 \, a^{2} b e^{\left (7 \, x\right )} + 9 \, a b^{2} e^{\left (7 \, x\right )} + 3 \, b^{3} e^{\left (7 \, x\right )} + 11 \, a^{3} e^{\left (5 \, x\right )} + 11 \, a^{2} b e^{\left (5 \, x\right )} - 11 \, a b^{2} e^{\left (5 \, x\right )} - 11 \, b^{3} e^{\left (5 \, x\right )} - 11 \, a^{3} e^{\left (3 \, x\right )} + 11 \, a^{2} b e^{\left (3 \, x\right )} + 11 \, a b^{2} e^{\left (3 \, x\right )} - 11 \, b^{3} e^{\left (3 \, x\right )} - 3 \, a^{3} e^{x} + 9 \, a^{2} b e^{x} - 9 \, a b^{2} e^{x} + 3 \, b^{3} e^{x}}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{4}} \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^5,x, algorithm="giac")

[Out]

3/4*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + 1/4*(3*a^3*e^(7*x) + 9
*a^2*b*e^(7*x) + 9*a*b^2*e^(7*x) + 3*b^3*e^(7*x) + 11*a^3*e^(5*x) + 11*a^2*b*e^(5*x) - 11*a*b^2*e^(5*x) - 11*b
^3*e^(5*x) - 11*a^3*e^(3*x) + 11*a^2*b*e^(3*x) + 11*a*b^2*e^(3*x) - 11*b^3*e^(3*x) - 3*a^3*e^x + 9*a^2*b*e^x -
9*a*b^2*e^x + 3*b^3*e^x)/((a^4 - 2*a^2*b^2 + b^4)*(a*e^(2*x) + b*e^(2*x) + a - b)^4)

Mupad [B] (verification not implemented)

Time = 2.41 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.16 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^5} \, dx=\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}{a^5-a^4\,b-2\,a^3\,b^2+2\,a^2\,b^3+a\,b^4-b^5}\right )}{4\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}-\frac {4\,{\mathrm {e}}^{3\,x}}{\left (a+b\right )\,\left ({\mathrm {e}}^{8\,x}\,{\left (a+b\right )}^4+{\left (a-b\right )}^4+4\,{\mathrm {e}}^{2\,x}\,\left (a+b\right )\,{\left (a-b\right )}^3+4\,{\mathrm {e}}^{6\,x}\,{\left (a+b\right )}^3\,\left (a-b\right )+6\,{\mathrm {e}}^{4\,x}\,{\left (a+b\right )}^2\,{\left (a-b\right )}^2\right )}-\frac {2\,{\mathrm {e}}^x}{{\left (a+b\right )}^2\,\left ({\mathrm {e}}^{6\,x}\,{\left (a+b\right )}^3+{\left (a-b\right )}^3+3\,{\mathrm {e}}^{2\,x}\,\left (a+b\right )\,{\left (a-b\right )}^2+3\,{\mathrm {e}}^{4\,x}\,{\left (a+b\right )}^2\,\left (a-b\right )\right )}+\frac {3\,{\mathrm {e}}^x}{4\,{\left (a+b\right )}^2\,{\left (a-b\right )}^2\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )}+\frac {{\mathrm {e}}^x}{2\,{\left (a+b\right )}^2\,\left (a-b\right )\,\left ({\mathrm {e}}^{4\,x}\,{\left (a+b\right )}^2+{\left (a-b\right )}^2+2\,{\mathrm {e}}^{2\,x}\,\left (a+b\right )\,\left (a-b\right )\right )} \]

[In]

int(1/(a*cosh(x) + b*sinh(x))^5,x)

[Out]

(3*atan((exp(x)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/(a*b^4 - a^4*b + a^5 -
b^5 + 2*a^2*b^3 - 2*a^3*b^2)))/(4*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2)) - (4*
exp(3*x))/((a + b)*(exp(8*x)*(a + b)^4 + (a - b)^4 + 4*exp(2*x)*(a + b)*(a - b)^3 + 4*exp(6*x)*(a + b)^3*(a -
b) + 6*exp(4*x)*(a + b)^2*(a - b)^2)) - (2*exp(x))/((a + b)^2*(exp(6*x)*(a + b)^3 + (a - b)^3 + 3*exp(2*x)*(a
+ b)*(a - b)^2 + 3*exp(4*x)*(a + b)^2*(a - b))) + (3*exp(x))/(4*(a + b)^2*(a - b)^2*(a - b + exp(2*x)*(a + b))
) + exp(x)/(2*(a + b)^2*(a - b)*(exp(4*x)*(a + b)^2 + (a - b)^2 + 2*exp(2*x)*(a + b)*(a - b)))