Integrand size = 21, antiderivative size = 78 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\frac {(b B-c C) \arctan \left (\frac {c \cosh (x)+b \sinh (x)}{\sqrt {b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}+\frac {B c-b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))} \] Output:
(B*b-C*c)*arctan((c*cosh(x)+b*sinh(x))/(b^2-c^2)^(1/2))/(b^2-c^2)^(3/2)+(B *c-C*b)/(b^2-c^2)/(b*cosh(x)+c*sinh(x))
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\frac {2 (b B-c C) \arctan \left (\frac {c+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-c} \sqrt {b+c}}\right )}{(b-c)^{3/2} (b+c)^{3/2}}+\frac {B c-b C}{(b-c) (b+c) (b \cosh (x)+c \sinh (x))} \] Input:
Integrate[(B*Cosh[x] + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^2,x]
Output:
(2*(b*B - c*C)*ArcTan[(c + b*Tanh[x/2])/(Sqrt[b - c]*Sqrt[b + c])])/((b - c)^(3/2)*(b + c)^(3/2)) + (B*c - b*C)/((b - c)*(b + c)*(b*Cosh[x] + c*Sinh [x]))
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3632, 3042, 3553, 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.
\(\displaystyle \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \cos (i x)-i C \sin (i x)}{(b \cos (i x)-i c \sin (i x))^2}dx\) |
\(\Big \downarrow \) 3632 |
\(\displaystyle \frac {(b B-c C) \int \frac {1}{b \cosh (x)+c \sinh (x)}dx}{b^2-c^2}+\frac {B c-b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B c-b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac {(b B-c C) \int \frac {1}{b \cos (i x)-i c \sin (i x)}dx}{b^2-c^2}\) |
\(\Big \downarrow \) 3553 |
\(\displaystyle \frac {B c-b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac {i (b B-c C) \int \frac {1}{b^2-c^2-(-i c \cosh (x)-i b \sinh (x))^2}d(-i c \cosh (x)-i b \sinh (x))}{b^2-c^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B c-b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}+\frac {i (b B-c C) \text {arctanh}\left (\frac {-i b \sinh (x)-i c \cosh (x)}{\sqrt {b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}\) |
Input:
Int[(B*Cosh[x] + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^2,x]
Output:
(I*(b*B - c*C)*ArcTanh[((-I)*c*Cosh[x] - I*b*Sinh[x])/Sqrt[b^2 - c^2]])/(b ^2 - c^2)^(3/2) + (B*c - b*C)/((b^2 - c^2)*(b*Cosh[x] + c*Sinh[x]))
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])
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x _Symbol] :> Simp[-d^(-1) Subst[Int[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)*Sin[ d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2) Int[1/(a + b*Cos[d + e*x] + c*S in[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]
Time = 1.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.51
method | result | size |
default | \(\frac {\frac {2 c \left (B c -b C \right ) \tanh \left (\frac {x}{2}\right )}{\left (b^{2}-c^{2}\right ) b}+\frac {2 \left (B c -b C \right )}{b^{2}-c^{2}}}{b \tanh \left (\frac {x}{2}\right )^{2}+2 c \tanh \left (\frac {x}{2}\right )+b}+\frac {2 \left (b B -C c \right ) \arctan \left (\frac {2 b \tanh \left (\frac {x}{2}\right )+2 c}{2 \sqrt {b^{2}-c^{2}}}\right )}{\left (b^{2}-c^{2}\right )^{\frac {3}{2}}}\) | \(118\) |
risch | \(\frac {2 \,{\mathrm e}^{x} \left (B c -b C \right )}{\left (b -c \right ) \left (b +c \right ) \left (b \,{\mathrm e}^{2 x}+c \,{\mathrm e}^{2 x}+b -c \right )}-\frac {b B \ln \left ({\mathrm e}^{x}-\frac {b -c}{\sqrt {-b^{2}+c^{2}}}\right )}{\sqrt {-b^{2}+c^{2}}\, \left (b +c \right ) \left (b -c \right )}+\frac {c C \ln \left ({\mathrm e}^{x}-\frac {b -c}{\sqrt {-b^{2}+c^{2}}}\right )}{\sqrt {-b^{2}+c^{2}}\, \left (b +c \right ) \left (b -c \right )}+\frac {b B \ln \left ({\mathrm e}^{x}+\frac {b -c}{\sqrt {-b^{2}+c^{2}}}\right )}{\sqrt {-b^{2}+c^{2}}\, \left (b +c \right ) \left (b -c \right )}-\frac {c C \ln \left ({\mathrm e}^{x}+\frac {b -c}{\sqrt {-b^{2}+c^{2}}}\right )}{\sqrt {-b^{2}+c^{2}}\, \left (b +c \right ) \left (b -c \right )}\) | \(237\) |
Input:
int((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x,method=_RETURNVERBOSE)
Output:
2*(c*(B*c-C*b)/(b^2-c^2)/b*tanh(1/2*x)+(B*c-C*b)/(b^2-c^2))/(b*tanh(1/2*x) ^2+2*c*tanh(1/2*x)+b)+2*(B*b-C*c)/(b^2-c^2)^(3/2)*arctan(1/2*(2*b*tanh(1/2 *x)+2*c)/(b^2-c^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (75) = 150\).
Time = 0.12 (sec) , antiderivative size = 749, normalized size of antiderivative = 9.60 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx =\text {Too large to display} \] Input:
integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="fric as")
Output:
[-((B*b^2 - (B + C)*b*c + C*c^2 + (B*b^2 + (B - C)*b*c - C*c^2)*cosh(x)^2 + 2*(B*b^2 + (B - C)*b*c - C*c^2)*cosh(x)*sinh(x) + (B*b^2 + (B - C)*b*c - C*c^2)*sinh(x)^2)*sqrt(-b^2 + c^2)*log(((b + c)*cosh(x)^2 + 2*(b + c)*cos h(x)*sinh(x) + (b + c)*sinh(x)^2 - 2*sqrt(-b^2 + c^2)*(cosh(x) + sinh(x)) - b + c)/((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^ 2 + b - c)) + 2*(C*b^3 - B*b^2*c - C*b*c^2 + B*c^3)*cosh(x) + 2*(C*b^3 - B *b^2*c - C*b*c^2 + B*c^3)*sinh(x))/(b^5 - b^4*c - 2*b^3*c^2 + 2*b^2*c^3 + b*c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^ 2 + 2*(b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*sinh(x)^2), -2*((B*b ^2 - (B + C)*b*c + C*c^2 + (B*b^2 + (B - C)*b*c - C*c^2)*cosh(x)^2 + 2*(B* b^2 + (B - C)*b*c - C*c^2)*cosh(x)*sinh(x) + (B*b^2 + (B - C)*b*c - C*c^2) *sinh(x)^2)*sqrt(b^2 - c^2)*arctan(sqrt(b^2 - c^2)/((b + c)*cosh(x) + (b + c)*sinh(x))) + (C*b^3 - B*b^2*c - C*b*c^2 + B*c^3)*cosh(x) + (C*b^3 - B*b ^2*c - C*b*c^2 + B*c^3)*sinh(x))/(b^5 - b^4*c - 2*b^3*c^2 + 2*b^2*c^3 + b* c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^2 + 2*(b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*sinh(x)^2)]
Timed out. \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\text {Timed out} \] Input:
integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="maxi ma")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*b^2>0)', see `assume?` f or more de
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.13 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\frac {2 \, {\left (B b - C c\right )} \arctan \left (\frac {b e^{x} + c e^{x}}{\sqrt {b^{2} - c^{2}}}\right )}{{\left (b^{2} - c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (C b e^{x} - B c e^{x}\right )}}{{\left (b^{2} - c^{2}\right )} {\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}} \] Input:
integrate((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="giac ")
Output:
2*(B*b - C*c)*arctan((b*e^x + c*e^x)/sqrt(b^2 - c^2))/(b^2 - c^2)^(3/2) - 2*(C*b*e^x - B*c*e^x)/((b^2 - c^2)*(b*e^(2*x) + c*e^(2*x) + b - c))
Time = 1.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.55 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\frac {\ln \left (\frac {2\,\left (B\,b-C\,c\right )}{{\left (b+c\right )}^{5/2}\,\sqrt {c-b}}+\frac {2\,{\mathrm {e}}^x\,\left (B\,b-C\,c\right )}{-b^3-b^2\,c+b\,c^2+c^3}\right )\,\left (B\,b-C\,c\right )}{{\left (b+c\right )}^{3/2}\,{\left (c-b\right )}^{3/2}}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (B\,b-C\,c\right )}{-b^3-b^2\,c+b\,c^2+c^3}-\frac {2\,\left (B\,b-C\,c\right )}{{\left (b+c\right )}^{5/2}\,\sqrt {c-b}}\right )\,\left (B\,b-C\,c\right )}{{\left (b+c\right )}^{3/2}\,{\left (c-b\right )}^{3/2}}+\frac {2\,{\mathrm {e}}^x\,\left (B\,c-C\,b\right )}{\left (b+c\right )\,\left (b-c\right )\,\left (b-c+{\mathrm {e}}^{2\,x}\,\left (b+c\right )\right )} \] Input:
int((B*cosh(x) + C*sinh(x))/(b*cosh(x) + c*sinh(x))^2,x)
Output:
(log((2*(B*b - C*c))/((b + c)^(5/2)*(c - b)^(1/2)) + (2*exp(x)*(B*b - C*c) )/(b*c^2 - b^2*c - b^3 + c^3))*(B*b - C*c))/((b + c)^(3/2)*(c - b)^(3/2)) - (log((2*exp(x)*(B*b - C*c))/(b*c^2 - b^2*c - b^3 + c^3) - (2*(B*b - C*c) )/((b + c)^(5/2)*(c - b)^(1/2)))*(B*b - C*c))/((b + c)^(3/2)*(c - b)^(3/2) ) + (2*exp(x)*(B*c - C*b))/((b + c)*(b - c)*(b - c + exp(2*x)*(b + c)))
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {B \cosh (x)+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx=\frac {2 \sqrt {b^{2}-c^{2}}\, \mathit {atan} \left (\frac {e^{x} b +e^{x} c}{\sqrt {b^{2}-c^{2}}}\right )}{b^{2}-c^{2}} \] Input:
int((B*cosh(x)+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x)
Output:
(2*sqrt(b**2 - c**2)*atan((e**x*b + e**x*c)/sqrt(b**2 - c**2)))/(b**2 - c* *2)