Integrand size = 14, antiderivative size = 271 \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=-\frac {2 \sqrt {2} c \arctan \left (\frac {2 i c-i b \tanh \left (\frac {x}{2}\right )+\sqrt {-b^2+4 a c} \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-b^2+4 a c} \sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}+\frac {2 \sqrt {2} c \arctan \left (\frac {2 i c-\left (i b+\sqrt {-b^2+4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-b^2+4 a c} \sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}} \] Output:
-2*2^(1/2)*c*arctan(1/2*(2*I*c-I*b*tanh(1/2*x)+(4*a*c-b^2)^(1/2)*tanh(1/2* x))*2^(1/2)/(b^2-2*(a-c)*c+I*b*(4*a*c-b^2)^(1/2))^(1/2))/(4*a*c-b^2)^(1/2) /(b^2-2*(a-c)*c+I*b*(4*a*c-b^2)^(1/2))^(1/2)+2*2^(1/2)*c*arctan(1/2*(2*I*c -(I*b+(4*a*c-b^2)^(1/2))*tanh(1/2*x))*2^(1/2)/(b^2-2*(a-c)*c-I*b*(4*a*c-b^ 2)^(1/2))^(1/2))/(4*a*c-b^2)^(1/2)/(b^2-2*(a-c)*c-I*b*(4*a*c-b^2)^(1/2))^( 1/2)
Time = 1.20 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.80 \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {2 \sqrt {2} c \left (\frac {\arctan \left (\frac {2 c+\left (-b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 (a-c) c+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c+b \sqrt {b^2-4 a c}}}-\frac {\arctan \left (\frac {2 c-\left (b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c}} \] Input:
Integrate[(a + b*Sinh[x] + c*Sinh[x]^2)^(-1),x]
Output:
(2*Sqrt[2]*c*(ArcTan[(2*c + (-b + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/Sqrt[-2*b^ 2 + 4*(a - c)*c + 2*b*Sqrt[b^2 - 4*a*c]]]/Sqrt[-b^2 + 2*(a - c)*c + b*Sqrt [b^2 - 4*a*c]] - ArcTan[(2*c - (b + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/(Sqrt[2] *Sqrt[-b^2 + 2*(a - c)*c - b*Sqrt[b^2 - 4*a*c]])]/Sqrt[-b^2 + 2*(a - c)*c - b*Sqrt[b^2 - 4*a*c]]))/Sqrt[b^2 - 4*a*c]
Time = 1.12 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3729, 3042, 3139, 1083, 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.
\(\displaystyle \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-i b \sin (i x)-c \sin (i x)^2}dx\) |
\(\Big \downarrow \) 3729 |
\(\displaystyle \frac {2 c \int \frac {1}{-i b-2 i c \sinh (x)+\sqrt {4 a c-b^2}}dx}{\sqrt {4 a c-b^2}}-\frac {2 c \int \frac {1}{-i b-2 i c \sinh (x)-\sqrt {4 a c-b^2}}dx}{\sqrt {4 a c-b^2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c \int \frac {1}{-i b-2 c \sin (i x)+\sqrt {4 a c-b^2}}dx}{\sqrt {4 a c-b^2}}-\frac {2 c \int \frac {1}{-i b-2 c \sin (i x)-\sqrt {4 a c-b^2}}dx}{\sqrt {4 a c-b^2}}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {4 c \int \frac {1}{\left (i b-\sqrt {4 a c-b^2}\right ) \tanh ^2\left (\frac {x}{2}\right )-4 i c \tanh \left (\frac {x}{2}\right )-i b+\sqrt {4 a c-b^2}}d\tanh \left (\frac {x}{2}\right )}{\sqrt {4 a c-b^2}}-\frac {4 c \int \frac {1}{\left (i b+\sqrt {4 a c-b^2}\right ) \tanh ^2\left (\frac {x}{2}\right )-4 i c \tanh \left (\frac {x}{2}\right )-i b-\sqrt {4 a c-b^2}}d\tanh \left (\frac {x}{2}\right )}{\sqrt {4 a c-b^2}}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {8 c \int \frac {1}{-\left (2 \left (i b+\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )^2-8 \left (b^2-i \sqrt {4 a c-b^2} b-2 (a-c) c\right )}d\left (2 \left (i b+\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )}{\sqrt {4 a c-b^2}}-\frac {8 c \int \frac {1}{-\left (2 \left (i b-\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )^2-8 \left (b^2+i \sqrt {4 a c-b^2} b-2 (a-c) c\right )}d\left (2 \left (i b-\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )}{\sqrt {4 a c-b^2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \sqrt {2} c \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \left (-\sqrt {4 a c-b^2}+i b\right )-4 i c}{2 \sqrt {2} \sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {4 a c-b^2} \sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}-\frac {2 \sqrt {2} c \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \left (\sqrt {4 a c-b^2}+i b\right )-4 i c}{2 \sqrt {2} \sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {4 a c-b^2} \sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\) |
Input:
Int[(a + b*Sinh[x] + c*Sinh[x]^2)^(-1),x]
Output:
(2*Sqrt[2]*c*ArcTan[((-4*I)*c + 2*(I*b - Sqrt[-b^2 + 4*a*c])*Tanh[x/2])/(2 *Sqrt[2]*Sqrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[-b^2 + 4*a*c]*Sqrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]]) - (2*Sqrt[2]*c*Ar cTan[((-4*I)*c + 2*(I*b + Sqrt[-b^2 + 4*a*c])*Tanh[x/2])/(2*Sqrt[2]*Sqrt[b ^2 - 2*(a - c)*c - I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[-b^2 + 4*a*c]*Sqrt[b^2 - 2*(a - c)*c - I*b*Sqrt[-b^2 + 4*a*c]])
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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)* (x_)]^(n2_.))^(-1), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c /q) Int[1/(b - q + 2*c*Sin[d + e*x]^n), x], x] - Simp[2*(c/q) Int[1/(b + q + 2*c*Sin[d + e*x]^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.27
method | result | size |
default | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{4}-2 b \,\textit {\_Z}^{3}+\left (-2 a +4 c \right ) \textit {\_Z}^{2}+2 b \textit {\_Z} +a \right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3} a -3 \textit {\_R}^{2} b -2 \textit {\_R} a +4 \textit {\_R} c +b}\) | \(74\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{4} c^{2}-8 a^{3} b^{2} c -32 a^{3} c^{3}+a^{2} b^{4}+32 a^{2} b^{2} c^{2}+16 a^{2} c^{4}-10 a \,b^{4} c -8 a \,b^{2} c^{3}+b^{6}+c^{2} b^{4}\right ) \textit {\_Z}^{4}+\left (-8 a^{2} c^{2}+6 a \,b^{2} c +8 a \,c^{3}-b^{4}-2 b^{2} c^{2}\right ) \textit {\_Z}^{2}+c^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\left (-3 b^{3}+\frac {8 c^{3} a}{b}+\frac {6 b \,a^{3}}{c}+\frac {24 c \,a^{3}}{b}-\frac {b^{3} a^{2}}{c^{2}}-\frac {24 c^{2} a^{2}}{b}+\frac {8 b^{3} a}{c}+18 a b c -\frac {8 a^{4}}{b}-22 a^{2} b -\frac {b^{5}}{c^{2}}-2 b \,c^{2}\right ) \textit {\_R}^{3}+\left (-\frac {b \,a^{2}}{c}-\frac {8 c \,a^{2}}{b}+\frac {4 c^{2} a}{b}-\frac {b^{3}}{c}-b c +\frac {4 a^{3}}{b}+6 a b \right ) \textit {\_R}^{2}+\left (2 b -\frac {4 b a}{c}-\frac {4 c a}{b}+\frac {b^{3}}{c^{2}}+\frac {2 c^{2}}{b}+\frac {2 a^{2}}{b}\right ) \textit {\_R} +\frac {b}{c}-\frac {a}{b}+\frac {c}{b}\right )\) | \(352\) |
Input:
int(1/(a+b*sinh(x)+c*sinh(x)^2),x,method=_RETURNVERBOSE)
Output:
sum((-_R^2+1)/(2*_R^3*a-3*_R^2*b-2*_R*a+4*_R*c+b)*ln(tanh(1/2*x)-_R),_R=Ro otOf(a*_Z^4-2*b*_Z^3+(-2*a+4*c)*_Z^2+2*b*_Z+a))
Leaf count of result is larger than twice the leaf count of optimal. 3313 vs. \(2 (219) = 438\).
Time = 0.21 (sec) , antiderivative size = 3313, normalized size of antiderivative = 12.23 \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sinh(x)+c*sinh(x)**2),x)
Output:
Timed out
\[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\int { \frac {1}{c \sinh \left (x\right )^{2} + b \sinh \left (x\right ) + a} \,d x } \] Input:
integrate(1/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="maxima")
Output:
integrate(1/(c*sinh(x)^2 + b*sinh(x) + a), x)
Time = 62.76 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=0 \] Input:
integrate(1/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="giac")
Output:
0
Timed out. \[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Hanged} \] Input:
int(1/(a + c*sinh(x)^2 + b*sinh(x)),x)
Output:
\text{Hanged}
\[ \int \frac {1}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\int \frac {1}{\sinh \left (x \right )^{2} c +\sinh \left (x \right ) b +a}d x \] Input:
int(1/(a+b*sinh(x)+c*sinh(x)^2),x)
Output:
int(1/(sinh(x)**2*c + sinh(x)*b + a),x)