\(\int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx\) [831]

Optimal result

Integrand size = 21, antiderivative size = 300 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {\sqrt {2} \left (i e-\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \arctan \left (\frac {2 i c-i b \tanh \left (\frac {x}{2}\right )+\sqrt {-b^2+4 a c} \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}}}+\frac {\sqrt {2} \left (i e+\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \arctan \left (\frac {2 i c-\left (i b+\sqrt {-b^2+4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}}} \]

[Out]

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*t
anh(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)

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3373, 2739, 632, 210} \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {\sqrt {2} \left (-\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \arctan \left (\frac {\tanh \left (\frac {x}{2}\right ) \sqrt {4 a c-b^2}-i b \tanh \left (\frac {x}{2}\right )+2 i c}{\sqrt {2} \sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}+\frac {\sqrt {2} \left (\frac {2 c d-b e}{\sqrt {4 a c-b^2}}+i e\right ) \arctan \left (\frac {2 i c-\tanh \left (\frac {x}{2}\right ) \left (\sqrt {4 a c-b^2}+i b\right )}{\sqrt {2} \sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt {-i b \sqrt {4 a c-b^2}-2 c (a-c)+b^2}} \]

[In]

Int[(d + e*Sinh[x])/(a + b*Sinh[x] + c*Sinh[x]^2),x]

[Out]

(Sqrt[2]*(I*e - (2*c*d - b*e)/Sqrt[-b^2 + 4*a*c])*ArcTan[((2*I)*c - I*b*Tanh[x/2] + Sqrt[-b^2 + 4*a*c]*Tanh[x/
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[((2*I)*c - (I*b + Sqrt[-b^2 + 4*a*c])*Tanh[x/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]
]

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 3373

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]}, Dist[B + (b*B - 2*A*c)/q, Int[1/(b + q + 2*c*Sin[d + e*x
]), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \left (-i e-\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \int \frac {1}{-i b-\sqrt {-b^2+4 a c}-2 i c \sinh (x)} \, dx+\left (-i e+\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right ) \int \frac {1}{-i b+\sqrt {-b^2+4 a c}-2 i c \sinh (x)} \, dx \\ & = -\left (\left (2 \left (i e-\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right )\right ) \text {Subst}\left (\int \frac {1}{-i b+\sqrt {-b^2+4 a c}-4 i c x-\left (-i b+\sqrt {-b^2+4 a c}\right ) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\right )-\left (2 \left (i e+\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right )\right ) \text {Subst}\left (\int \frac {1}{-i b-\sqrt {-b^2+4 a c}-4 i c x-\left (-i b-\sqrt {-b^2+4 a c}\right ) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right ) \\ & = \left (4 \left (i e-\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right )\right ) \text {Subst}\left (\int \frac {1}{-8 \left (b^2-2 (a-c) c+i b \sqrt {-b^2+4 a c}\right )-x^2} \, dx,x,-4 i c+2 \left (i b-\sqrt {-b^2+4 a c}\right ) \tanh \left (\frac {x}{2}\right )\right )+\left (4 \left (i e+\frac {2 c d-b e}{\sqrt {-b^2+4 a c}}\right )\right ) \text {Subst}\left (\int \frac {1}{-8 \left (b^2-2 (a-c) c-i b \sqrt {-b^2+4 a c}\right )-x^2} \, dx,x,-4 i c+2 \left (i b+\sqrt {-b^2+4 a c}\right ) \tanh \left (\frac {x}{2}\right )\right ) \\ & = \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}}}+\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}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.86 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\frac {\sqrt {2} \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {2 c+\left (-b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 (a-c) c+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c+b \sqrt {b^2-4 a c}}}+\frac {\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \arctan \left (\frac {2 c-\left (b+\sqrt {b^2-4 a c}\right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {-b^2+2 (a-c) c-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c}} \]

[In]

Integrate[(d + e*Sinh[x])/(a + b*Sinh[x] + c*Sinh[x]^2),x]

[Out]

(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]

Maple [C] (verified)

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	\(\text {Expression too large to display}\)	\(8284\)

[In]

int((d+e*sinh(x))/(a+b*sinh(x)+c*sinh(x)^2),x,method=_RETURNVERBOSE)

[Out]

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))

Fricas [B] (verification not implemented)

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 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Too large to display} \]

[In]

integrate((d+e*sinh(x))/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Timed out} \]

[In]

integrate((d+e*sinh(x))/(a+b*sinh(x)+c*sinh(x)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\int { \frac {e \sinh \left (x\right ) + d}{c \sinh \left (x\right )^{2} + b \sinh \left (x\right ) + a} \,d x } \]

[In]

integrate((d+e*sinh(x))/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="maxima")

[Out]

integrate((e*sinh(x) + d)/(c*sinh(x)^2 + b*sinh(x) + a), x)

Giac [A] (verification not implemented)

none

Time = 1.55 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.00 \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=0 \]

[In]

integrate((d+e*sinh(x))/(a+b*sinh(x)+c*sinh(x)^2),x, algorithm="giac")

[Out]

0

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e \sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx=\text {Hanged} \]

[In]

int((d + e*sinh(x))/(a + c*sinh(x)^2 + b*sinh(x)),x)

[Out]

\text{Hanged}