Integrand size = 28, antiderivative size = 159 \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt {2} \left (b^2-c^2\right )^{3/4}}-\frac {c \cosh (x)+b \sinh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \]
1/4*arctanh(1/2*(b^2-c^2)^(1/4)*sinh(x+I*arctan(b,-I*c))*2^(1/2)/(-(b^2-c^ 2)^(1/2)+cosh(x+I*arctan(b,-I*c))*(b^2-c^2)^(1/2))^(1/2))/(b^2-c^2)^(3/4)* 2^(1/2)+1/2*(-c*cosh(x)-b*sinh(x))/(b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^( 3/2)/(b^2-c^2)^(1/2)
Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {\$Aborted} \]
Time = 0.55 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
| \(\displaystyle \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}-\frac {i \int \frac {1}{\frac {\left (b^2-c^2\right ) \sinh ^2\left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )-\sqrt {b^2-c^2}}-2 \sqrt {b^2-c^2}}d\left (-\frac {i \sqrt {b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )-\sqrt {b^2-c^2}}}\right )}{2 \sqrt {b^2-c^2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {b \sinh (x)+c \cosh (x)}{2 \sqrt {b^2-c^2} \left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{2 \sqrt {2} \left (b^2-c^2\right )^{3/4}}\) |
ArcTanh[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[-S qrt[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))
3.8.78.3.1 Defintions of rubi rules used
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])
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(414\) vs. \(2(134)=268\).
Time = 0.23 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.61
| method | result | size |
| default | \(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cosh \left (x \right ) \sqrt {2}}{2}\right )}{2 \sqrt {b^{2}-c^{2}}\, \sqrt {-\frac {\sinh \left (x \right ) b^{2}-\sinh \left (x \right ) c^{2}+b^{2}-c^{2}}{\sqrt {b^{2}-c^{2}}}}}-\frac {\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right ) \sinh \left (x \right )^{2}}\, \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \left (\ln \left (-\frac {2 \left (\cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \sinh \left (x \right )-\sinh \left (x \right ) \sqrt {b^{2}-c^{2}}+\cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}-\sqrt {b^{2}-c^{2}}-\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right )}\, \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right ) \sinh \left (x \right )^{2}}\right )}{\cosh \left (x \right )-\sqrt {2}}\right )-\ln \left (\frac {2 \cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}\, \sinh \left (x \right )+2 \sinh \left (x \right ) \sqrt {b^{2}-c^{2}}+2 \cosh \left (x \right ) \sqrt {b^{2}-c^{2}}\, \sqrt {2}+2 \sqrt {b^{2}-c^{2}}+2 \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right )}\, \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right ) \sinh \left (x \right )^{2}}}{\cosh \left (x \right )+\sqrt {2}}\right )\right )}{4 \left (b -c \right ) \left (b +c \right ) \sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \left (x \right )+1\right )}\, \sinh \left (x \right ) \sqrt {-\frac {\sinh \left (x \right ) b^{2}-\sinh \left (x \right ) c^{2}+b^{2}-c^{2}}{\sqrt {b^{2}-c^{2}}}}}\) | \(415\) |
1/2/(b^2-c^2)^(1/2)/(-(sinh(x)*b^2-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)-sinh(x)*(b^2-c^2)^(1/2)+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)+sinh(x)*(b^2-c^2)^(1/2)+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)/(-(sinh(x)*b^2-sinh(x)*c^2+b^2-c^2)/(b^2-c^2) ^(1/2))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 2137 vs. \(2 (130) = 260\).
Time = 0.47 (sec) , antiderivative size = 2137, normalized size of antiderivative = 13.44 \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {Too large to display} \]
-1/4*((sqrt(2)*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^6 + 6*sqrt(2)*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)*sinh(x)^5 + sqrt(2)*(b^3 + 3*b^2*c + 3* b*c^2 + c^3)*sinh(x)^6 - 3*sqrt(2)*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^4 + 3*(5*sqrt(2)*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^2 - sqrt(2)*(b^3 + b ^2*c - b*c^2 - c^3))*sinh(x)^4 + 4*(5*sqrt(2)*(b^3 + 3*b^2*c + 3*b*c^2 + c ^3)*cosh(x)^3 - 3*sqrt(2)*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x))*sinh(x)^3 + 3*sqrt(2)*(b^3 - b^2*c - b*c^2 + c^3)*cosh(x)^2 + 3*(5*sqrt(2)*(b^3 + 3*b ^2*c + 3*b*c^2 + c^3)*cosh(x)^4 - 6*sqrt(2)*(b^3 + b^2*c - b*c^2 - c^3)*co sh(x)^2 + sqrt(2)*(b^3 - b^2*c - b*c^2 + c^3))*sinh(x)^2 + 6*(sqrt(2)*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^5 - 2*sqrt(2)*(b^3 + b^2*c - b*c^2 - c ^3)*cosh(x)^3 + sqrt(2)*(b^3 - b^2*c - b*c^2 + c^3)*cosh(x))*sinh(x) - sqr t(2)*(b^3 - 3*b^2*c + 3*b*c^2 - c^3))*(b^2 - c^2)^(1/4)*log(-((b^2 + 2*b*c + c^2)*cosh(x)^4 + 4*(b^2 + 2*b*c + c^2)*cosh(x)^3*sinh(x) + 6*(b^2 + 2*b *c + c^2)*cosh(x)^2*sinh(x)^2 + 4*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^3 + (b^2 + 2*b*c + c^2)*sinh(x)^4 - 2*sqrt(1/2)*(sqrt(2)*(b + c)*cosh(x)^3 + 3 *sqrt(2)*(b + c)*cosh(x)*sinh(x)^2 + sqrt(2)*(b + c)*sinh(x)^3 + sqrt(2)*( b - c)*cosh(x) + (3*sqrt(2)*(b + c)*cosh(x)^2 + sqrt(2)*(b - c))*sinh(x) + 2*(sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2)*sqr t(b^2 - c^2))*(b^2 - c^2)^(1/4)*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)) +...
\[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \cosh {\left (x \right )} + c \sinh {\left (x \right )} - \sqrt {b^{2} - c^{2}}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) - \sqrt {b^{2} - c^{2}}\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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%%%{%%%{8,[4,0]%%%}+%%%{16,[3,1]%%%}+%%%{-8,[3,0]%%%}+%%%{-8,[ 2,1]%%%}+
Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {cosh}\left (x\right )-\sqrt {b^2-c^2}+c\,\mathrm {sinh}\left (x\right )\right )}^{3/2}} \,d x \]