Integrand size = 21, antiderivative size = 300 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {\sqrt {2} \left (i e-\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \arctan \left (\frac {2 i c-i b \tanh \left (\frac {x}{2}\right )+\sqrt {-b^2+4 a c} \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}+\frac {\sqrt {2} \left (i e+\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \arctan \left (\frac {2 i c-\left (i b+\sqrt {-b^2+4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}} \]
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))*2^(1/2)*(I*e+(-b*e+2*c*d)/(4*a*c-b^2)^(1 /2))/(b^2-2*(a-c)*c-I*b*(4*a*c-b^2)^(1/2))^(1/2)+arctan(1/2*(2*I*c-I*b*tan h(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))*2^(1/2)*(I*e+(b*e-2*c*d)/(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 = 1.73 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.86 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {\sqrt {2} \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {2 c+\left (-b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 (a-c) c+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c+b \sqrt {b^2-4 a c}}}+\frac {\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {2 c-\left (b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c}} \]
(Sqrt[2]*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*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]] + ((-2*c*d + (b + Sqrt[b ^2 - 4*a*c])*e)*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 = 0.66 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3773, 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.
\(\displaystyle \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {d-i e \sin (i x)}{a-i b \sin (i x)-c \sin (i x)^2}dx\) |
\(\Big \downarrow \) 3773 |
\(\displaystyle -\left (\left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-i b-2 i c \sinh (x)-\sqrt {4 a c-b^2}}dx\right )-\left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-i b-2 i c \sinh (x)+\sqrt {4 a c-b^2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\left (\left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-i b-2 c \sin (i x)-\sqrt {4 a c-b^2}}dx\right )-\left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-i b-2 c \sin (i x)+\sqrt {4 a c-b^2}}dx\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -2 \left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{\left (i b-\sqrt {4 a c-b^2}\right ) \tanh ^2\left (\frac {x}{2}\right )-4 i c \tanh \left (\frac {x}{2}\right )-i b+\sqrt {4 a c-b^2}}d\tanh \left (\frac {x}{2}\right )-2 \left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{\left (i b+\sqrt {4 a c-b^2}\right ) \tanh ^2\left (\frac {x}{2}\right )-4 i c \tanh \left (\frac {x}{2}\right )-i b-\sqrt {4 a c-b^2}}d\tanh \left (\frac {x}{2}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle 4 \left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-\left (2 \left (i b-\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )^2-8 \left (b^2+i \sqrt {4 a c-b^2} b-2 (a-c) c\right )}d\left (2 \left (i b-\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )+4 \left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \int \frac {1}{-\left (2 \left (i b+\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )^2-8 \left (b^2-i \sqrt {4 a c-b^2} b-2 (a-c) c\right )}d\left (2 \left (i b+\sqrt {4 a c-b^2}\right ) \tanh \left (\frac {x}{2}\right )-4 i c\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\sqrt {2} \left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \left (-\sqrt {4 a c-b^2}+i b\right )-4 i c}{2 \sqrt {2} \sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}-\frac {\sqrt {2} \left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \left (\sqrt {4 a c-b^2}+i b\right )-4 i c}{2 \sqrt {2} \sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\) |
-((Sqrt[2]*(I*e - (2*c*d - b*e)/Sqrt[-b^2 + 4*a*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 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]] ) - (Sqrt[2]*(I*e + (2*c*d - b*e)/Sqrt[-b^2 + 4*a*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 - 2*(a - c)*c - I*b*Sqrt[-b^2 + 4*a*c ]]
3.9.31.3.1 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[((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_)])/((a_.) + (b_.)*sin[(d_.) + (e_.) *(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(B + (b*B - 2*A*c)/q) Int[1/(b + q + 2*c*Sin[d + e*x]) , x], x] + Simp[(B - (b*B - 2*A*c)/q) Int[1/(b - q + 2*c*Sin[d + e*x]), x ], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.67 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.26
method | result | size |
default | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{4}-2 b \,\textit {\_Z}^{3}+\left (-2 a +4 c \right ) \textit {\_Z}^{2}+2 b \textit {\_Z} +a \right )}{\sum }\frac {\left (-\textit {\_R}^{2} d +2 \textit {\_R} e +d \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{2 \textit {\_R}^{3} a -3 \textit {\_R}^{2} b -2 \textit {\_R} a +4 \textit {\_R} c +b}\) | \(79\) |
risch | \(\text {Expression too large to display}\) | \(8284\) |
sum((-_R^2*d+2*_R*e+d)/(2*_R^3*a-3*_R^2*b-2*_R*a+4*_R*c+b)*ln(tanh(1/2*x)- _R),_R=RootOf(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. 6841 vs. \(2 (244) = 488\).
Time = 2.97 (sec) , antiderivative size = 6841, normalized size of antiderivative = 22.80 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\int { \frac {e \sinh \left (x\right ) + d}{c \sinh \left (x\right )^{2} + b \sinh \left (x\right ) + a} \,d x } \]
Time = 1.55 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=0 \]
Timed out. \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Hanged} \]