\(\int \frac {1}{(-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x))^{5/2}} \, dx\) [539]

Optimal result

Integrand size = 28, antiderivative size = 211 \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{16 \sqrt {2} \left (b^2-c^2\right )^{5/4}}-\frac {c \cosh (x)+b \sinh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}+\frac {3 (c \cosh (x)+b \sinh (x))}{16 \left (b^2-c^2\right ) \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \] Output:

-3/32*arctanh(1/2*(b^2-c^2)^(1/4)*sinh(x+I*arctan(-I*c,b))*2^(1/2)/(-(b^2- 
c^2)^(1/2)+(b^2-c^2)^(1/2)*cosh(x+I*arctan(-I*c,b)))^(1/2))*2^(1/2)/(b^2-c 
^2)^(5/4)-1/4*(c*cosh(x)+b*sinh(x))/(b^2-c^2)^(1/2)/(-(b^2-c^2)^(1/2)+b*co 
sh(x)+c*sinh(x))^(5/2)+3/16*(c*cosh(x)+b*sinh(x))/(b^2-c^2)/(-(b^2-c^2)^(1 
/2)+b*cosh(x)+c*sinh(x))^(3/2)
 

Mathematica [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=\text {\$Aborted} \] Input:

Integrate[(-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-5/2),x]
 

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 3595, 3042, 3595, 3042, 3594, 3042, 3128, 217}

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.

\(\displaystyle \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cos (i x)-i c \sin (i x)\right )^{5/2}}dx\)

\(\Big \downarrow \) 3595

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}-\frac {3 \int \frac {1}{\left (b \cos (i x)-i c \sin (i x)-\sqrt {b^2-c^2}\right )^{3/2}}dx}{8 \sqrt {b^2-c^2}}\)

\(\Big \downarrow \) 3595

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}-\frac {3 \left (-\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {b \cos (i x)-i c \sin (i x)-\sqrt {b^2-c^2}}}dx}{4 \sqrt {b^2-c^2}}\right )}{8 \sqrt {b^2-c^2}}\)

\(\Big \downarrow \) 3594

\(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}-\frac {3 \left (-\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )-\sqrt {b^2-c^2}}}dx}{4 \sqrt {b^2-c^2}}\right )}{8 \sqrt {b^2-c^2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}-\frac {3 \left (-\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}-\frac {i \int \frac {1}{\frac {\left (b^2-c^2\right ) \sinh ^2\left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )-\sqrt {b^2-c^2}}-2 \sqrt {b^2-c^2}}d\left (-\frac {i \sqrt {b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )-\sqrt {b^2-c^2}}}\right )}{2 \sqrt {b^2-c^2}}\right )}{8 \sqrt {b^2-c^2}}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{4 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}}-\frac {3 \left (-\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt {2} \left (b^2-c^2\right )^{3/4}}\right )}{8 \sqrt {b^2-c^2}}\)

Input:

Int[(-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-5/2),x]
 

Output:

-1/4*(c*Cosh[x] + b*Sinh[x])/(Sqrt[b^2 - c^2]*(-Sqrt[b^2 - c^2] + b*Cosh[x 
] + c*Sinh[x])^(5/2)) - (3*(ArcTanh[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b 
, (-I)*c]])/(Sqrt[2]*Sqrt[-Sqrt[b^2 - c^2] + Sqrt[b^2 - c^2]*Cosh[x + I*Ar 
cTan[b, (-I)*c]]])]/(2*Sqrt[2]*(b^2 - c^2)^(3/4)) - (c*Cosh[x] + b*Sinh[x] 
)/(2*Sqrt[b^2 - c^2]*(-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(3/2))))/( 
8*Sqrt[b^2 - c^2])
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[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 3594

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

rule 3595

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(816\) vs. \(2(178)=356\).

Time = 0.27 (sec) , antiderivative size = 817, normalized size of antiderivative = 3.87

method	result	size
default	\(\text {Expression too large to display}\)	\(817\)

Input:

int(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x,method=_RETURNVERBOSE 
)
 

Output:

2*(-b^2+c^2)/(-(b^2*sinh(x)-sinh(x)*c^2+b^2-c^2)/(b^2-c^2)^(1/2))^(1/2)/(b 
^4-2*b^2*c^2+c^4)*(-1/4*cosh(x)/(cosh(x)^2-2)+1/8*2^(1/2)*arctanh(1/2*cosh 
(x)*2^(1/2)))+(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)*(b^2-c^2)*(1/ 
4/(b-c)^2/(b+c)^2*(1/(b^2-c^2)^(1/2)/(sinh(x)+1)/(cosh(x)-2^(1/2))*(-(b^2- 
c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)+2^(1/2)/(-(b^2-c^2)^(1/2)*(sinh(x) 
+1))^(1/2)*ln((-2*(b^2-c^2)^(1/2)*(sinh(x)+1)-2*(sinh(x)+1)*2^(1/2)*(b^2-c 
^2)^(1/2)*(cosh(x)-2^(1/2))+2*(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/ 
2)*(-(b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2))/(cosh(x)-2^(1/2))))+1/4/(b-c)^2/( 
b+c)^2*(1/(b^2-c^2)^(1/2)/(sinh(x)+1)/(cosh(x)+2^(1/2))*(-(b^2-c^2)^(1/2)* 
(sinh(x)+1)*sinh(x)^2)^(1/2)-2^(1/2)/(-(b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2)* 
ln((-2*(b^2-c^2)^(1/2)*(sinh(x)+1)+2*(sinh(x)+1)*2^(1/2)*(b^2-c^2)^(1/2)*( 
cosh(x)+2^(1/2))+2*(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)*(-(b^2-c 
^2)^(1/2)*(sinh(x)+1))^(1/2))/(cosh(x)+2^(1/2))))-1/8/(b-c)^2/(b+c)^2*2^(1 
/2)/(-(b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2)*ln((-2*(b^2-c^2)^(1/2)*(sinh(x)+1 
)-2*(sinh(x)+1)*2^(1/2)*(b^2-c^2)^(1/2)*(cosh(x)-2^(1/2))+2*(-(b^2-c^2)^(1 
/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)*(-(b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2))/(co 
sh(x)-2^(1/2)))+1/8/(b-c)^2/(b+c)^2*2^(1/2)/(-(b^2-c^2)^(1/2)*(sinh(x)+1)) 
^(1/2)*ln((-2*(b^2-c^2)^(1/2)*(sinh(x)+1)+2*(sinh(x)+1)*2^(1/2)*(b^2-c^2)^ 
(1/2)*(cosh(x)+2^(1/2))+2*(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)*( 
-(b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2))/(cosh(x)+2^(1/2))))/sinh(x)/(-(b^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5675 vs. \(2 (172) = 344\).

Time = 1.66 (sec) , antiderivative size = 5675, normalized size of antiderivative = 26.90 \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x, algorithm="fri 
cas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \cosh {\left (x \right )} + c \sinh {\left (x \right )} - \sqrt {b^{2} - c^{2}}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(-(b**2-c**2)**(1/2)+b*cosh(x)+c*sinh(x))**(5/2),x)
 

Output:

Integral((b*cosh(x) + c*sinh(x) - sqrt(b**2 - c**2))**(-5/2), x)
 

Maxima [F]

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

integrate(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x, algorithm="max 
ima")
 

Output:

integrate((b*cosh(x) + c*sinh(x) - sqrt(b^2 - c^2))^(-5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x, algorithm="gia 
c")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{-1,[1,0]%%%}+%%%{-1,[0,1]%%%},[0,5,0]%%%}+%%%{%%{[1,0] 
:[1,0,%%%
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {cosh}\left (x\right )-\sqrt {b^2-c^2}+c\,\mathrm {sinh}\left (x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(b*cosh(x) - (b^2 - c^2)^(1/2) + c*sinh(x))^(5/2),x)
 

Output:

int(1/(b*cosh(x) - (b^2 - c^2)^(1/2) + c*sinh(x))^(5/2), x)
 

Reduce [F]

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

int(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x)
 

Output:

int(1/(-(b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(5/2),x)