3.111 \(\int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx\)

Optimal. Leaf size=82 \[ \frac {2 (a A-b B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))} \]

[Out]

2*(A*a-B*b)*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)-(A*b-B*a)*sinh(x)/(a^2-b^2)/(
a+b*cosh(x))

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Rubi [A]  time = 0.08, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2754, 12, 2659, 208} \[ \frac {2 (a A-b B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(a + b*Cosh[x])^2,x]

[Out]

(2*(a*A - b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) - ((A*b - a*B)*Sinh
[x])/((a^2 - b^2)*(a + b*Cosh[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 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 2754

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 81, normalized size = 0.99 \[ \frac {2 (a A-b B) \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {\sinh (x) (a B-A b)}{(a-b) (a+b) (a+b \cosh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/(a + b*Cosh[x])^2,x]

[Out]

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

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fricas [B]  time = 1.38, size = 828, normalized size = 10.10 \[ \left [-\frac {2 \, B a^{3} b - 2 \, A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4} - {\left (A a b^{2} - B b^{3} + {\left (A a b^{2} - B b^{3}\right )} \cosh \relax (x)^{2} + {\left (A a b^{2} - B b^{3}\right )} \sinh \relax (x)^{2} + 2 \, {\left (A a^{2} b - B a b^{2}\right )} \cosh \relax (x) + 2 \, {\left (A a^{2} b - B a b^{2} + {\left (A a b^{2} - B b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cosh \relax (x) + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \sinh \relax (x)}{a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)^{2} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \relax (x) + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}, -\frac {2 \, {\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A b^{4} + {\left (A a b^{2} - B b^{3} + {\left (A a b^{2} - B b^{3}\right )} \cosh \relax (x)^{2} + {\left (A a b^{2} - B b^{3}\right )} \sinh \relax (x)^{2} + 2 \, {\left (A a^{2} b - B a b^{2}\right )} \cosh \relax (x) + 2 \, {\left (A a^{2} b - B a b^{2} + {\left (A a b^{2} - B b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cosh \relax (x) + {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \sinh \relax (x)\right )}}{a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)^{2} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \relax (x) + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 107, normalized size = 1.30 \[ \frac {2 \, {\left (A a - B b\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (B a^{2} e^{x} - A a b e^{x} + B a b - A b^{2}\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 108, normalized size = 1.32 \[ \frac {2 \left (A b -a B \right ) \tanh \left (\frac {x}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}+\frac {2 \left (A a -B b \right ) \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x))/(a+b*cosh(x))^2,x)

[Out]

2*(A*b-B*a)/(a^2-b^2)*tanh(1/2*x)/(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)+2*(A*a-B*b)/(a+b)/(a-b)/((a+b)*(a-b))^
(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x))^2,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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.42, size = 246, normalized size = 3.00 \[ \frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{b\,\left (a^2\,b-b^3\right )}-\frac {2\,{\mathrm {e}}^x\,\left (B\,a^2\,b^2-A\,a\,b^3\right )}{b^2\,\left (a^2\,b-b^3\right )}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (A\,a-B\,b\right )}{b\,\left (a^2-b^2\right )}-\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )\,\left (A\,a-B\,b\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (A\,a-B\,b\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\ln \left (\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )\,\left (A\,a-B\,b\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,{\mathrm {e}}^x\,\left (A\,a-B\,b\right )}{b\,\left (a^2-b^2\right )}\right )\,\left (A\,a-B\,b\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x))/(a + b*cosh(x))^2,x)

[Out]

((2*(A*b^3 - B*a*b^2))/(b*(a^2*b - b^3)) - (2*exp(x)*(B*a^2*b^2 - A*a*b^3))/(b^2*(a^2*b - b^3)))/(b + 2*a*exp(
x) + b*exp(2*x)) + (log(- (2*exp(x)*(A*a - B*b))/(b*(a^2 - b^2)) - (2*(b + a*exp(x))*(A*a - B*b))/(b*(a + b)^(
3/2)*(a - b)^(3/2)))*(A*a - B*b))/((a + b)^(3/2)*(a - b)^(3/2)) - (log((2*(b + a*exp(x))*(A*a - B*b))/(b*(a +
b)^(3/2)*(a - b)^(3/2)) - (2*exp(x)*(A*a - B*b))/(b*(a^2 - b^2)))*(A*a - B*b))/((a + b)^(3/2)*(a - b)^(3/2))

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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((A+B*cosh(x))/(a+b*cosh(x))**2,x)

[Out]

Timed out

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