3.103 \(\int \frac {A+B \cosh (x)}{(a+a \cosh (x))^{5/2}} \, dx\)

Optimal. Leaf size=93 \[ \frac {(3 A+5 B) \tan ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a \cosh (x)+a}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(3 A+5 B) \sinh (x)}{16 a (a \cosh (x)+a)^{3/2}}+\frac {(A-B) \sinh (x)}{4 (a \cosh (x)+a)^{5/2}} \]

[Out]

1/4*(A-B)*sinh(x)/(a+a*cosh(x))^(5/2)+1/16*(3*A+5*B)*sinh(x)/a/(a+a*cosh(x))^(3/2)+1/32*(3*A+5*B)*arctan(1/2*s
inh(x)*a^(1/2)*2^(1/2)/(a+a*cosh(x))^(1/2))/a^(5/2)*2^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2750, 2650, 2649, 206} \[ \frac {(3 A+5 B) \tan ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a \cosh (x)+a}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(3 A+5 B) \sinh (x)}{16 a (a \cosh (x)+a)^{3/2}}+\frac {(A-B) \sinh (x)}{4 (a \cosh (x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(a + a*Cosh[x])^(5/2),x]

[Out]

((3*A + 5*B)*ArcTan[(Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a + a*Cosh[x]])])/(16*Sqrt[2]*a^(5/2)) + ((A - B)*Sinh[x])
/(4*(a + a*Cosh[x])^(5/2)) + ((3*A + 5*B)*Sinh[x])/(16*a*(a + a*Cosh[x])^(3/2))

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])

Rule 2649

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

Rule 2650

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

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \cosh (x)}{(a+a \cosh (x))^{5/2}} \, dx &=\frac {(A-B) \sinh (x)}{4 (a+a \cosh (x))^{5/2}}+\frac {(3 A+5 B) \int \frac {1}{(a+a \cosh (x))^{3/2}} \, dx}{8 a}\\ &=\frac {(A-B) \sinh (x)}{4 (a+a \cosh (x))^{5/2}}+\frac {(3 A+5 B) \sinh (x)}{16 a (a+a \cosh (x))^{3/2}}+\frac {(3 A+5 B) \int \frac {1}{\sqrt {a+a \cosh (x)}} \, dx}{32 a^2}\\ &=\frac {(A-B) \sinh (x)}{4 (a+a \cosh (x))^{5/2}}+\frac {(3 A+5 B) \sinh (x)}{16 a (a+a \cosh (x))^{3/2}}+\frac {(i (3 A+5 B)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {i a \sinh (x)}{\sqrt {a+a \cosh (x)}}\right )}{16 a^2}\\ &=\frac {(3 A+5 B) \tan ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a+a \cosh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(A-B) \sinh (x)}{4 (a+a \cosh (x))^{5/2}}+\frac {(3 A+5 B) \sinh (x)}{16 a (a+a \cosh (x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 57, normalized size = 0.61 \[ \frac {\sinh (x) ((3 A+5 B) \cosh (x)+7 A+B)+4 (3 A+5 B) \cosh ^5\left (\frac {x}{2}\right ) \tan ^{-1}\left (\sinh \left (\frac {x}{2}\right )\right )}{16 (a (\cosh (x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/(a + a*Cosh[x])^(5/2),x]

[Out]

(4*(3*A + 5*B)*ArcTan[Sinh[x/2]]*Cosh[x/2]^5 + (7*A + B + (3*A + 5*B)*Cosh[x])*Sinh[x])/(16*(a*(1 + Cosh[x]))^
(5/2))

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fricas [B]  time = 0.85, size = 509, normalized size = 5.47 \[ -\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{4} + {\left (3 \, A + 5 \, B\right )} \sinh \relax (x)^{4} + 4 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{3} + 4 \, {\left ({\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + 3 \, A + 5 \, B\right )} \sinh \relax (x)^{3} + 6 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{2} + 6 \, {\left ({\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + 3 \, A + 5 \, B\right )} \sinh \relax (x)^{2} + 4 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + 4 \, {\left ({\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{3} + 3 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{2} + 3 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + 3 \, A + 5 \, B\right )} \sinh \relax (x) + 3 \, A + 5 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{\cosh \relax (x) + \sinh \relax (x)}}}{\sqrt {a}}\right ) - 2 \, \sqrt {\frac {1}{2}} {\left ({\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{4} + {\left (3 \, A + 5 \, B\right )} \sinh \relax (x)^{4} + {\left (11 \, A - 3 \, B\right )} \cosh \relax (x)^{3} + {\left (4 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + 11 \, A - 3 \, B\right )} \sinh \relax (x)^{3} - {\left (11 \, A - 3 \, B\right )} \cosh \relax (x)^{2} + {\left (6 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{2} + 3 \, {\left (11 \, A - 3 \, B\right )} \cosh \relax (x) - 11 \, A + 3 \, B\right )} \sinh \relax (x)^{2} - {\left (3 \, A + 5 \, B\right )} \cosh \relax (x) + {\left (4 \, {\left (3 \, A + 5 \, B\right )} \cosh \relax (x)^{3} + 3 \, {\left (11 \, A - 3 \, B\right )} \cosh \relax (x)^{2} - 2 \, {\left (11 \, A - 3 \, B\right )} \cosh \relax (x) - 3 \, A - 5 \, B\right )} \sinh \relax (x)\right )} \sqrt {\frac {a}{\cosh \relax (x) + \sinh \relax (x)}}}{16 \, {\left (a^{3} \cosh \relax (x)^{4} + a^{3} \sinh \relax (x)^{4} + 4 \, a^{3} \cosh \relax (x)^{3} + 6 \, a^{3} \cosh \relax (x)^{2} + 4 \, a^{3} \cosh \relax (x) + 4 \, {\left (a^{3} \cosh \relax (x) + a^{3}\right )} \sinh \relax (x)^{3} + a^{3} + 6 \, {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) + a^{3}\right )} \sinh \relax (x)^{2} + 4 \, {\left (a^{3} \cosh \relax (x)^{3} + 3 \, a^{3} \cosh \relax (x)^{2} + 3 \, a^{3} \cosh \relax (x) + a^{3}\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+a*cosh(x))^(5/2),x, algorithm="fricas")

[Out]

-1/16*(sqrt(2)*((3*A + 5*B)*cosh(x)^4 + (3*A + 5*B)*sinh(x)^4 + 4*(3*A + 5*B)*cosh(x)^3 + 4*((3*A + 5*B)*cosh(
x) + 3*A + 5*B)*sinh(x)^3 + 6*(3*A + 5*B)*cosh(x)^2 + 6*((3*A + 5*B)*cosh(x)^2 + 2*(3*A + 5*B)*cosh(x) + 3*A +
5*B)*sinh(x)^2 + 4*(3*A + 5*B)*cosh(x) + 4*((3*A + 5*B)*cosh(x)^3 + 3*(3*A + 5*B)*cosh(x)^2 + 3*(3*A + 5*B)*c
osh(x) + 3*A + 5*B)*sinh(x) + 3*A + 5*B)*sqrt(a)*arctan(sqrt(2)*sqrt(1/2)*sqrt(a/(cosh(x) + sinh(x)))/sqrt(a))
- 2*sqrt(1/2)*((3*A + 5*B)*cosh(x)^4 + (3*A + 5*B)*sinh(x)^4 + (11*A - 3*B)*cosh(x)^3 + (4*(3*A + 5*B)*cosh(x
) + 11*A - 3*B)*sinh(x)^3 - (11*A - 3*B)*cosh(x)^2 + (6*(3*A + 5*B)*cosh(x)^2 + 3*(11*A - 3*B)*cosh(x) - 11*A
+ 3*B)*sinh(x)^2 - (3*A + 5*B)*cosh(x) + (4*(3*A + 5*B)*cosh(x)^3 + 3*(11*A - 3*B)*cosh(x)^2 - 2*(11*A - 3*B)*
cosh(x) - 3*A - 5*B)*sinh(x))*sqrt(a/(cosh(x) + sinh(x))))/(a^3*cosh(x)^4 + a^3*sinh(x)^4 + 4*a^3*cosh(x)^3 +
6*a^3*cosh(x)^2 + 4*a^3*cosh(x) + 4*(a^3*cosh(x) + a^3)*sinh(x)^3 + a^3 + 6*(a^3*cosh(x)^2 + 2*a^3*cosh(x) + a
^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + 3*a^3*cosh(x)^2 + 3*a^3*cosh(x) + a^3)*sinh(x))

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giac [A]  time = 0.22, size = 118, normalized size = 1.27 \[ \frac {\sqrt {2} {\left (3 \, A + 5 \, B\right )} \arctan \left (e^{\left (\frac {1}{2} \, x\right )}\right )}{16 \, a^{\frac {5}{2}}} + \frac {\sqrt {2} {\left (3 \, A a^{\frac {7}{2}} e^{\left (\frac {7}{2} \, x\right )} + 5 \, B a^{\frac {7}{2}} e^{\left (\frac {7}{2} \, x\right )} + 11 \, A a^{\frac {7}{2}} e^{\left (\frac {5}{2} \, x\right )} - 3 \, B a^{\frac {7}{2}} e^{\left (\frac {5}{2} \, x\right )} - 11 \, A a^{\frac {7}{2}} e^{\left (\frac {3}{2} \, x\right )} + 3 \, B a^{\frac {7}{2}} e^{\left (\frac {3}{2} \, x\right )} - 3 \, A a^{\frac {7}{2}} e^{\left (\frac {1}{2} \, x\right )} - 5 \, B a^{\frac {7}{2}} e^{\left (\frac {1}{2} \, x\right )}\right )}}{16 \, {\left (a e^{x} + a\right )}^{4} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+a*cosh(x))^(5/2),x, algorithm="giac")

[Out]

1/16*sqrt(2)*(3*A + 5*B)*arctan(e^(1/2*x))/a^(5/2) + 1/16*sqrt(2)*(3*A*a^(7/2)*e^(7/2*x) + 5*B*a^(7/2)*e^(7/2*
x) + 11*A*a^(7/2)*e^(5/2*x) - 3*B*a^(7/2)*e^(5/2*x) - 11*A*a^(7/2)*e^(3/2*x) + 3*B*a^(7/2)*e^(3/2*x) - 3*A*a^(
7/2)*e^(1/2*x) - 5*B*a^(7/2)*e^(1/2*x))/((a*e^x + a)^4*a^2)

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maple [B]  time = 0.41, size = 209, normalized size = 2.25 \[ -\frac {\sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \left (3 A \ln \left (\frac {2 \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}-2 a}{\cosh \left (\frac {x}{2}\right )}\right ) \left (\cosh ^{4}\left (\frac {x}{2}\right )\right ) a +5 B \ln \left (\frac {2 \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}-2 a}{\cosh \left (\frac {x}{2}\right )}\right ) \left (\cosh ^{4}\left (\frac {x}{2}\right )\right ) a -3 A \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}\, \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-5 B \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}\, \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )-2 A \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}+2 B \sqrt {a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {-a}\right ) \sqrt {2}}{32 \cosh \left (\frac {x}{2}\right )^{3} a^{3} \sqrt {-a}\, \sinh \left (\frac {x}{2}\right ) \sqrt {a \left (\cosh ^{2}\left (\frac {x}{2}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x))/(a+a*cosh(x))^(5/2),x)

[Out]

-1/32*(a*sinh(1/2*x)^2)^(1/2)*(3*A*ln(2/cosh(1/2*x)*((a*sinh(1/2*x)^2)^(1/2)*(-a)^(1/2)-a))*cosh(1/2*x)^4*a+5*
B*ln(2/cosh(1/2*x)*((a*sinh(1/2*x)^2)^(1/2)*(-a)^(1/2)-a))*cosh(1/2*x)^4*a-3*A*(a*sinh(1/2*x)^2)^(1/2)*(-a)^(1
/2)*cosh(1/2*x)^2-5*B*(a*sinh(1/2*x)^2)^(1/2)*(-a)^(1/2)*cosh(1/2*x)^2-2*A*(a*sinh(1/2*x)^2)^(1/2)*(-a)^(1/2)+
2*B*(a*sinh(1/2*x)^2)^(1/2)*(-a)^(1/2))/cosh(1/2*x)^3/a^3/(-a)^(1/2)/sinh(1/2*x)*2^(1/2)/(a*cosh(1/2*x)^2)^(1/
2)

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maxima [B]  time = 0.93, size = 427, normalized size = 4.59 \[ \frac {1}{80} \, {\left (\sqrt {2} {\left (\frac {15 \, e^{\left (\frac {9}{2} \, x\right )} + 70 \, e^{\left (\frac {7}{2} \, x\right )} + 128 \, e^{\left (\frac {5}{2} \, x\right )} - 70 \, e^{\left (\frac {3}{2} \, x\right )} - 15 \, e^{\left (\frac {1}{2} \, x\right )}}{a^{\frac {5}{2}} e^{\left (5 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{x} + a^{\frac {5}{2}}} + \frac {15 \, \arctan \left (e^{\left (\frac {1}{2} \, x\right )}\right )}{a^{\frac {5}{2}}}\right )} - \frac {128 \, \sqrt {2} e^{\left (\frac {5}{2} \, x\right )}}{a^{\frac {5}{2}} e^{\left (5 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{x} + a^{\frac {5}{2}}}\right )} A + \frac {1}{672} \, {\left (\sqrt {2} {\left (\frac {105 \, e^{\left (\frac {9}{2} \, x\right )} + 490 \, e^{\left (\frac {7}{2} \, x\right )} + 896 \, e^{\left (\frac {5}{2} \, x\right )} + 790 \, e^{\left (\frac {3}{2} \, x\right )} - 105 \, e^{\left (\frac {1}{2} \, x\right )}}{a^{\frac {5}{2}} e^{\left (5 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{x} + a^{\frac {5}{2}}} + \frac {105 \, \arctan \left (e^{\left (\frac {1}{2} \, x\right )}\right )}{a^{\frac {5}{2}}}\right )} + 7 \, \sqrt {2} {\left (\frac {15 \, e^{\left (\frac {9}{2} \, x\right )} + 70 \, e^{\left (\frac {7}{2} \, x\right )} - 128 \, e^{\left (\frac {5}{2} \, x\right )} - 70 \, e^{\left (\frac {3}{2} \, x\right )} - 15 \, e^{\left (\frac {1}{2} \, x\right )}}{a^{\frac {5}{2}} e^{\left (5 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} + 10 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + 5 \, a^{\frac {5}{2}} e^{x} + a^{\frac {5}{2}}} + \frac {15 \, \arctan \left (e^{\left (\frac {1}{2} \, x\right )}\right )}{a^{\frac {5}{2}}}\right )} - \frac {128 \, {\left (7 \, \sqrt {2} \sqrt {a} e^{\left (\frac {7}{2} \, x\right )} + 3 \, \sqrt {2} \sqrt {a} e^{\left (\frac {3}{2} \, x\right )}\right )}}{a^{3} e^{\left (5 \, x\right )} + 5 \, a^{3} e^{\left (4 \, x\right )} + 10 \, a^{3} e^{\left (3 \, x\right )} + 10 \, a^{3} e^{\left (2 \, x\right )} + 5 \, a^{3} e^{x} + a^{3}}\right )} B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+a*cosh(x))^(5/2),x, algorithm="maxima")

[Out]

1/80*(sqrt(2)*((15*e^(9/2*x) + 70*e^(7/2*x) + 128*e^(5/2*x) - 70*e^(3/2*x) - 15*e^(1/2*x))/(a^(5/2)*e^(5*x) +
5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5*a^(5/2)*e^x + a^(5/2)) + 15*arctan(e^(1/2*x))/
a^(5/2)) - 128*sqrt(2)*e^(5/2*x)/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x
) + 5*a^(5/2)*e^x + a^(5/2)))*A + 1/672*(sqrt(2)*((105*e^(9/2*x) + 490*e^(7/2*x) + 896*e^(5/2*x) + 790*e^(3/2*
x) - 105*e^(1/2*x))/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5*a^(5/2)
*e^x + a^(5/2)) + 105*arctan(e^(1/2*x))/a^(5/2)) + 7*sqrt(2)*((15*e^(9/2*x) + 70*e^(7/2*x) - 128*e^(5/2*x) - 7
0*e^(3/2*x) - 15*e^(1/2*x))/(a^(5/2)*e^(5*x) + 5*a^(5/2)*e^(4*x) + 10*a^(5/2)*e^(3*x) + 10*a^(5/2)*e^(2*x) + 5
*a^(5/2)*e^x + a^(5/2)) + 15*arctan(e^(1/2*x))/a^(5/2)) - 128*(7*sqrt(2)*sqrt(a)*e^(7/2*x) + 3*sqrt(2)*sqrt(a)
*e^(3/2*x))/(a^3*e^(5*x) + 5*a^3*e^(4*x) + 10*a^3*e^(3*x) + 10*a^3*e^(2*x) + 5*a^3*e^x + a^3))*B

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {cosh}\relax (x)}{{\left (a+a\,\mathrm {cosh}\relax (x)\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x))/(a + a*cosh(x))^(5/2),x)

[Out]

int((A + B*cosh(x))/(a + a*cosh(x))^(5/2), 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((A+B*cosh(x))/(a+a*cosh(x))**(5/2),x)

[Out]

Timed out

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