\(\int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx\) [106]

Optimal result

Integrand size = 18, antiderivative size = 94 \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=-\frac {(3 A-5 B) \arctan \left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{16 \sqrt {2} a^{5/2}}-\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}-\frac {(3 A-5 B) \sinh (x)}{16 a (a-a \cosh (x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=\frac {\left (2 (3 A-5 B) \text {csch}^2\left (\frac {x}{4}\right )-(A+B) \text {csch}^4\left (\frac {x}{4}\right )-8 (3 A-5 B) \left (\log \left (\cosh \left (\frac {x}{4}\right )\right )-\log \left (\sinh \left (\frac {x}{4}\right )\right )\right )+2 (3 A-5 B) \text {sech}^2\left (\frac {x}{4}\right )+(A+B) \text {sech}^4\left (\frac {x}{4}\right )\right ) \sinh ^5\left (\frac {x}{2}\right )}{32 a^2 (-1+\cosh (x))^2 \sqrt {a-a \cosh (x)}} \] Input:

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

Output:

((2*(3*A - 5*B)*Csch[x/4]^2 - (A + B)*Csch[x/4]^4 - 8*(3*A - 5*B)*(Log[Cos 
h[x/4]] - Log[Sinh[x/4]]) + 2*(3*A - 5*B)*Sech[x/4]^2 + (A + B)*Sech[x/4]^ 
4)*Sinh[x/2]^5)/(32*a^2*(-1 + Cosh[x])^2*Sqrt[a - a*Cosh[x]])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 3229, 3042, 3129, 3042, 3128, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (\frac {\pi }{2}+i x\right )}{\left (a-a \sin \left (\frac {\pi }{2}+i x\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3229

\(\displaystyle \frac {(3 A-5 B) \int \frac {1}{(a-a \cosh (x))^{3/2}}dx}{8 a}-\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}+\frac {(3 A-5 B) \int \frac {1}{\left (a-a \sin \left (i x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{8 a}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {(3 A-5 B) \left (\frac {\int \frac {1}{\sqrt {a-a \cosh (x)}}dx}{4 a}-\frac {\sinh (x)}{2 (a-a \cosh (x))^{3/2}}\right )}{8 a}-\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}+\frac {(3 A-5 B) \left (-\frac {\sinh (x)}{2 (a-a \cosh (x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a-a \sin \left (i x+\frac {\pi }{2}\right )}}dx}{4 a}\right )}{8 a}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}+\frac {(3 A-5 B) \left (-\frac {\sinh (x)}{2 (a-a \cosh (x))^{3/2}}+\frac {i \int \frac {1}{\frac {a^2 \sinh ^2(x)}{a-a \cosh (x)}+2 a}d\frac {i a \sinh (x)}{\sqrt {a-a \cosh (x)}}}{2 a}\right )}{8 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(3 A-5 B) \left (-\frac {\arctan \left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{2 \sqrt {2} a^{3/2}}-\frac {\sinh (x)}{2 (a-a \cosh (x))^{3/2}}\right )}{8 a}-\frac {(A+B) \sinh (x)}{4 (a-a \cosh (x))^{5/2}}\)

Input:

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

Output:

-1/4*((A + B)*Sinh[x])/(a - a*Cosh[x])^(5/2) + ((3*A - 5*B)*(-1/2*ArcTan[( 
Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a - a*Cosh[x]])]/(Sqrt[2]*a^(3/2)) - Sinh[x 
]/(2*(a - a*Cosh[x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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

rule 3129

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] + Simp[(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 3229

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] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1))   I 
nt[(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)]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (-6 A +10 B \right ) \cosh \left (\frac {x}{2}\right ) \sinh \left (\frac {x}{2}\right )^{2}+\left (4 A +4 B \right ) \cosh \left (\frac {x}{2}\right )+\left (3 \ln \left (\cosh \left (\frac {x}{2}\right )+1\right ) A -3 \ln \left (\cosh \left (\frac {x}{2}\right )-1\right ) A -5 \ln \left (\cosh \left (\frac {x}{2}\right )+1\right ) B +5 \ln \left (\cosh \left (\frac {x}{2}\right )-1\right ) B \right ) \sinh \left (\frac {x}{2}\right )^{4}}{32 a^{2} \left (\cosh \left (\frac {x}{2}\right )+1\right ) \left (\cosh \left (\frac {x}{2}\right )-1\right ) \sinh \left (\frac {x}{2}\right ) \sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}\)	\(118\)
parts	\(-\frac {A \left (-6 \sinh \left (\frac {x}{2}\right )^{2} \cosh \left (\frac {x}{2}\right )+4 \cosh \left (\frac {x}{2}\right )+\left (3 \ln \left (\cosh \left (\frac {x}{2}\right )+1\right )-3 \ln \left (\cosh \left (\frac {x}{2}\right )-1\right )\right ) \sinh \left (\frac {x}{2}\right )^{4}\right )}{32 a^{2} \left (\cosh \left (\frac {x}{2}\right )+1\right ) \left (\cosh \left (\frac {x}{2}\right )-1\right ) \sinh \left (\frac {x}{2}\right ) \sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}+\frac {B \left (-10 \sinh \left (\frac {x}{2}\right )^{2} \cosh \left (\frac {x}{2}\right )-4 \cosh \left (\frac {x}{2}\right )+\left (5 \ln \left (\cosh \left (\frac {x}{2}\right )+1\right )-5 \ln \left (\cosh \left (\frac {x}{2}\right )-1\right )\right ) \sinh \left (\frac {x}{2}\right )^{4}\right )}{32 a^{2} \left (\cosh \left (\frac {x}{2}\right )+1\right ) \left (\cosh \left (\frac {x}{2}\right )-1\right ) \sinh \left (\frac {x}{2}\right ) \sqrt {-2 \sinh \left (\frac {x}{2}\right )^{2} a}}\)	\(170\)

Input:

int((A+B*cosh(x))/(a-cosh(x)*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32/a^2*((-6*A+10*B)*cosh(1/2*x)*sinh(1/2*x)^2+(4*A+4*B)*cosh(1/2*x)+(3* 
ln(cosh(1/2*x)+1)*A-3*ln(cosh(1/2*x)-1)*A-5*ln(cosh(1/2*x)+1)*B+5*ln(cosh( 
1/2*x)-1)*B)*sinh(1/2*x)^4)/(cosh(1/2*x)+1)/(cosh(1/2*x)-1)/sinh(1/2*x)/(- 
2*sinh(1/2*x)^2*a)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (75) = 150\).

Time = 0.09 (sec) , antiderivative size = 548, normalized size of antiderivative = 5.83 \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/32*(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)*cosh(x) - 3*A + 5*B)*sinh(x) + 3*A - 5*B)* 
sqrt(-a)*log((2*sqrt(2)*sqrt(1/2)*sqrt(-a)*sqrt(-a/(cosh(x) + sinh(x)))*(c 
osh(x) + sinh(x)) - a*cosh(x) - a*sinh(x) - a)/(cosh(x) + sinh(x) - 1)) - 
4*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*cos 
h(x)^2 + 3*a^3*cosh(x) - a^3)*sinh(x))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cosh(x))/(a-a*cosh(x))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=\int { \frac {B \cosh \left (x\right ) + A}{{\left (-a \cosh \left (x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (75) = 150\).

Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.01 \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left (3 \, A - 5 \, B\right )} \arctan \left (\frac {\sqrt {-a e^{x}}}{\sqrt {a}}\right )}{16 \, a^{\frac {5}{2}} \mathrm {sgn}\left (-e^{x} + 1\right )} + \frac {\sqrt {2} {\left (3 \, \sqrt {-a e^{x}} A a^{3} e^{\left (3 \, x\right )} - 5 \, \sqrt {-a e^{x}} B a^{3} e^{\left (3 \, x\right )} - 11 \, \sqrt {-a e^{x}} A a^{3} e^{\left (2 \, x\right )} - 3 \, \sqrt {-a e^{x}} B a^{3} e^{\left (2 \, x\right )} - 11 \, \sqrt {-a e^{x}} A a^{3} e^{x} - 3 \, \sqrt {-a e^{x}} B a^{3} e^{x} + 3 \, \sqrt {-a e^{x}} A a^{3} - 5 \, \sqrt {-a e^{x}} B a^{3}\right )}}{16 \, {\left (a e^{x} - a\right )}^{4} a^{2} \mathrm {sgn}\left (-e^{x} + 1\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {cosh}\left (x\right )}{{\left (a-a\,\mathrm {cosh}\left (x\right )\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \cosh (x)}{(a-a \cosh (x))^{5/2}} \, dx=-\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {-\cosh \left (x \right )+1}}{\cosh \left (x \right )^{3}-3 \cosh \left (x \right )^{2}+3 \cosh \left (x \right )-1}d x \right ) a +\left (\int \frac {\sqrt {-\cosh \left (x \right )+1}\, \cosh \left (x \right )}{\cosh \left (x \right )^{3}-3 \cosh \left (x \right )^{2}+3 \cosh \left (x \right )-1}d x \right ) b \right )}{a^{3}} \] Input:

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

Output:

( - sqrt(a)*(int(sqrt( - cosh(x) + 1)/(cosh(x)**3 - 3*cosh(x)**2 + 3*cosh( 
x) - 1),x)*a + int((sqrt( - cosh(x) + 1)*cosh(x))/(cosh(x)**3 - 3*cosh(x)* 
*2 + 3*cosh(x) - 1),x)*b))/a**3