\(\int \frac {1}{(\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x))^{3/2}} \, dx\) [532]

Optimal result

Integrand size = 26, antiderivative size = 155 \[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt {2} \left (b^2-c^2\right )^{3/4}}+\frac {c \cosh (x)+b \sinh (x)}{2 \sqrt {b^2-c^2} \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \] Output:

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

Mathematica [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {\$Aborted} \] Input:

Integrate[(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-3/2),x]
 

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3595, 3042, 3594, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3595

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

Int[(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x])^(-3/2),x]
 

Output:

ArcTan[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[Sqr 
t[b^2 - c^2] + Sqrt[b^2 - c^2]*Cosh[x + I*ArcTan[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))
 

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 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. \(416\) vs. \(2(128)=256\).

Time = 0.23 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.69

method	result	size
default	\(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cosh \left (x \right ) \sqrt {2}}{2}\right )}{2 \sqrt {b^{2}-c^{2}}\, \sqrt {-\frac {b^{2} \sinh \left (x \right )-\sinh \left (x \right ) c^{2}-b^{2}+c^{2}}{\sqrt {b^{2}-c^{2}}}}}+\frac {\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right ) \sinh \left (x \right )^{2}}\, \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \left (\ln \left (-\frac {2 \left (\cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \sinh \left (x \right )-\sqrt {b^{2}-c^{2}}\, \sinh \left (x \right )-\cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}+\sqrt {b^{2}-c^{2}}-\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right )}\, \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right ) \sinh \left (x \right )^{2}}\right )}{\cosh \left (x \right )-\sqrt {2}}\right )-\ln \left (\frac {2 \cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \sinh \left (x \right )+2 \sqrt {b^{2}-c^{2}}\, \sinh \left (x \right )-2 \cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}-2 \sqrt {b^{2}-c^{2}}+2 \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right )}\, \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right ) \sinh \left (x \right )^{2}}}{\cosh \left (x \right )+\sqrt {2}}\right )\right )}{4 \left (b -c \right ) \left (b +c \right ) \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )-1\right )}\, \sinh \left (x \right ) \sqrt {-\frac {b^{2} \sinh \left (x \right )-\sinh \left (x \right ) c^{2}-b^{2}+c^{2}}{\sqrt {b^{2}-c^{2}}}}}\)	\(417\)

Input:

int(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/(b^2-c^2)^(1/2)/(-(b^2*sinh(x)-sinh(x)*c^2-b^2+c^2)/(b^2-c^2)^(1/2))^( 
1/2)*2^(1/2)*arctanh(1/2*cosh(x)*2^(1/2))+1/4*(-(b^2-c^2)^(1/2)*(sinh(x)-1 
)*sinh(x)^2)^(1/2)*(b^2-c^2)^(1/2)*2^(1/2)*(ln(-2*(cosh(x)*(b^2-c^2)^(1/2) 
*2^(1/2)*sinh(x)-(b^2-c^2)^(1/2)*sinh(x)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)+( 
b^2-c^2)^(1/2)-(-(b^2-c^2)^(1/2)*(sinh(x)-1))^(1/2)*(-(b^2-c^2)^(1/2)*(sin 
h(x)-1)*sinh(x)^2)^(1/2))/(cosh(x)-2^(1/2)))-ln(2*(cosh(x)*(b^2-c^2)^(1/2) 
*2^(1/2)*sinh(x)+(b^2-c^2)^(1/2)*sinh(x)-cosh(x)*(b^2-c^2)^(1/2)*2^(1/2)-( 
b^2-c^2)^(1/2)+(-(b^2-c^2)^(1/2)*(sinh(x)-1))^(1/2)*(-(b^2-c^2)^(1/2)*(sin 
h(x)-1)*sinh(x)^2)^(1/2))/(cosh(x)+2^(1/2))))/(b-c)/(b+c)/(-(b^2-c^2)^(1/2 
)*(sinh(x)-1))^(1/2)/sinh(x)/(-(b^2*sinh(x)-sinh(x)*c^2-b^2+c^2)/(b^2-c^2) 
^(1/2))^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1735 vs. \(2 (122) = 244\).

Time = 0.27 (sec) , antiderivative size = 1735, normalized size of antiderivative = 11.19 \[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x, algorithm="fric 
as")
 

Output:

(sqrt(1/2)*((b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^6 + 6*(b^3 + 3*b^2*c + 
 3*b*c^2 + c^3)*cosh(x)*sinh(x)^5 + (b^3 + 3*b^2*c + 3*b*c^2 + c^3)*sinh(x 
)^6 - 3*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^4 - 3*(b^3 + b^2*c - b*c^2 - c 
^3 - 5*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(b^3 + 
3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^3 - 3*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x) 
)*sinh(x)^3 - b^3 + 3*b^2*c - 3*b*c^2 + c^3 + 3*(b^3 - b^2*c - b*c^2 + c^3 
)*cosh(x)^2 + 3*(5*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^4 + b^3 - b^2*c 
 - b*c^2 + c^3 - 6*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^2)*sinh(x)^2 + 6*(( 
b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^5 - 2*(b^3 + b^2*c - b*c^2 - c^3)*c 
osh(x)^3 + (b^3 - b^2*c - b*c^2 + c^3)*cosh(x))*sinh(x))*(b^2 - c^2)^(1/4) 
*arctan(-(b^2 - c^2)^(1/4)*((b + c)*cosh(x) + (b + c)*sinh(x) - sqrt(b^2 - 
 c^2))*sqrt(((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh( 
x)^2 + 2*sqrt(b^2 - c^2)*(cosh(x) + sinh(x)) + b - c)/(cosh(x) + sinh(x))) 
/((b^2 + 2*b*c + c^2)*cosh(x)^2 + 2*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x) + 
(b^2 + 2*b*c + c^2)*sinh(x)^2 - b^2 + c^2)) - sqrt(1/2)*(4*(b^3 + b^2*c - 
b*c^2 - c^3)*cosh(x)^4 + 16*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)*sinh(x)^3 
+ 4*(b^3 + b^2*c - b*c^2 - c^3)*sinh(x)^4 + 4*(b^3 - b^2*c - b*c^2 + c^3)* 
cosh(x)^2 + 4*(b^3 - b^2*c - b*c^2 + c^3 + 6*(b^3 + b^2*c - b*c^2 - c^3)*c 
osh(x)^2)*sinh(x)^2 + 8*(2*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^3 + (b^3 - 
b^2*c - b*c^2 + c^3)*cosh(x))*sinh(x) - ((b^2 + 2*b*c + c^2)*cosh(x)^5 ...
 

Sympy [F]

\[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \cosh {\left (x \right )} + c \sinh {\left (x \right )} + \sqrt {b^{2} - c^{2}}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/((b**2-c**2)**(1/2)+b*cosh(x)+c*sinh(x))**(3/2),x)
 

Output:

Integral((b*cosh(x) + c*sinh(x) + sqrt(b**2 - c**2))**(-3/2), x)
 

Maxima [F]

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

integrate(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x, algorithm="maxi 
ma")
 

Output:

integrate((b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^(-3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x, algorithm="giac 
")
 

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]%%%},[0,3,0]%%%}+%%%{%%{[1,0]:[ 
1,0,%%%{-
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {cosh}\left (x\right )+\sqrt {b^2-c^2}+c\,\mathrm {sinh}\left (x\right )\right )}^{3/2}} \,d x \] Input:

int(1/(b*cosh(x) + (b^2 - c^2)^(1/2) + c*sinh(x))^(3/2),x)
 

Output:

int(1/(b*cosh(x) + (b^2 - c^2)^(1/2) + c*sinh(x))^(3/2), x)
 

Reduce [F]

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

int(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x)
 

Output:

int(1/((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^(3/2),x)