3.726 \(\int \frac {A+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^3} \, dx\)
Optimal. Leaf size=123 \[ -\frac {-A b \sinh (x)-A c \cosh (x)+b C}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac {A \tan ^{-1}\left (\frac {b \sinh (x)+c \cosh (x)}{\sqrt {b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}-\frac {b c C \sinh (x)+c^2 C \cosh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))} \]
[Out]
1/2*A*arctan((c*cosh(x)+b*sinh(x))/(b^2-c^2)^(1/2))/(b^2-c^2)^(3/2)+1/2*(-b*C+A*c*cosh(x)+A*b*sinh(x))/(b^2-c^
2)/(b*cosh(x)+c*sinh(x))^2+(-c^2*C*cosh(x)-b*c*C*sinh(x))/(b^2-c^2)^2/(b*cosh(x)+c*sinh(x))
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Rubi [A] time = 0.13, antiderivative size = 123, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used =
{3157, 3153, 3074, 206} \[ -\frac {-A b \sinh (x)-A c \cosh (x)+b C}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac {A \tan ^{-1}\left (\frac {b \sinh (x)+c \cosh (x)}{\sqrt {b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}-\frac {b c C \sinh (x)+c^2 C \cosh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^3,x]
[Out]
(A*ArcTan[(c*Cosh[x] + b*Sinh[x])/Sqrt[b^2 - c^2]])/(2*(b^2 - c^2)^(3/2)) - (b*C - A*c*Cosh[x] - A*b*Sinh[x])/
(2*(b^2 - c^2)*(b*Cosh[x] + c*Sinh[x])^2) - (c^2*C*Cosh[x] + b*c*C*Sinh[x])/((b^2 - c^2)^2*(b*Cosh[x] + c*Sinh
[x]))
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3074
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[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]
Rule 3153
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] + Dist[(a*A - b*B - c*C)/(a^2 -
b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[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]
Rule 3157
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(d_.) + (e
_.)*(x_)]), x_Symbol] :> Simp[((b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin
[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] + Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a + b*Cos[d +
e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - c*C) - (n + 2)*b*A*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[
d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]
Rubi steps
\begin {align*} \int \frac {A+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^3} \, dx &=-\frac {b C-A c \cosh (x)-A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}+\frac {\int \frac {-2 c C+A b \cosh (x)+A c \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx}{2 \left (b^2-c^2\right )}\\ &=-\frac {b C-A c \cosh (x)-A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}-\frac {c^2 C \cosh (x)+b c C \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}+\frac {A \int \frac {1}{b \cosh (x)+c \sinh (x)} \, dx}{2 \left (b^2-c^2\right )}\\ &=-\frac {b C-A c \cosh (x)-A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}-\frac {c^2 C \cosh (x)+b c C \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}+\frac {(i A) \operatorname {Subst}\left (\int \frac {1}{b^2-c^2-x^2} \, dx,x,-i c \cosh (x)-i b \sinh (x)\right )}{2 \left (b^2-c^2\right )}\\ &=\frac {A \tan ^{-1}\left (\frac {c \cosh (x)+b \sinh (x)}{\sqrt {b^2-c^2}}\right )}{2 \left (b^2-c^2\right )^{3/2}}-\frac {b C-A c \cosh (x)-A b \sinh (x)}{2 \left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))^2}-\frac {c^2 C \cosh (x)+b c C \sinh (x)}{\left (b^2-c^2\right )^2 (b \cosh (x)+c \sinh (x))}\\ \end {align*}
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Mathematica [A] time = 1.19, size = 134, normalized size = 1.09 \[ \frac {1}{2} \left (\frac {A \left (b^2-c^2\right ) \sinh (x)-b^2 C}{b (b-c) (b+c) (b \cosh (x)+c \sinh (x))^2}+\frac {c (A-2 C \sinh (x))}{b (b-c) (b+c) (b \cosh (x)+c \sinh (x))}+\frac {2 A \tan ^{-1}\left (\frac {b \tanh \left (\frac {x}{2}\right )+c}{\sqrt {b-c} \sqrt {b+c}}\right )}{(b-c)^{3/2} (b+c)^{3/2}}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(A + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^3,x]
[Out]
((2*A*ArcTan[(c + b*Tanh[x/2])/(Sqrt[b - c]*Sqrt[b + c])])/((b - c)^(3/2)*(b + c)^(3/2)) + (-(b^2*C) + A*(b^2
- c^2)*Sinh[x])/(b*(b - c)*(b + c)*(b*Cosh[x] + c*Sinh[x])^2) + (c*(A - 2*C*Sinh[x]))/(b*(b - c)*(b + c)*(b*Co
sh[x] + c*Sinh[x])))/2
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fricas [B] time = 0.56, size = 1855, normalized size = 15.08 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")
[Out]
[1/2*(4*C*b^2*c - 8*C*b*c^2 + 4*C*c^3 + 2*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^3 + 2*(A*b^3 + A*b^2*c -
A*b*c^2 - A*c^3)*sinh(x)^3 - 4*(C*b^3 - C*b^2*c - C*b*c^2 + C*c^3)*cosh(x)^2 - 2*(2*C*b^3 - 2*C*b^2*c - 2*C*b
*c^2 + 2*C*c^3 - 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x))*sinh(x)^2 + ((A*b^2 + 2*A*b*c + A*c^2)*cosh(x)
^4 + 4*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)*sinh(x)^3 + (A*b^2 + 2*A*b*c + A*c^2)*sinh(x)^4 + A*b^2 - 2*A*b*c + A
*c^2 + 2*(A*b^2 - A*c^2)*cosh(x)^2 + 2*(A*b^2 - A*c^2 + 3*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^2)*sinh(x)^2 + 4*(
(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^3 + (A*b^2 - A*c^2)*cosh(x))*sinh(x))*sqrt(-b^2 + c^2)*log(((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)) - b + c)/((b + c)*c
osh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 + b - c)) - 2*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3)*cos
h(x) - 2*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 - 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^2 + 4*(C*b^3 - C*b
^2*c - C*b*c^2 + C*c^3)*cosh(x))*sinh(x))/(b^6 - 2*b^5*c - b^4*c^2 + 4*b^3*c^3 - b^2*c^4 - 2*b*c^5 + c^6 + (b^
6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^4 + 4*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^
3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)*sinh(x)^3 + (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^
6)*sinh(x)^4 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*cosh(x)^2 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6 + 3*(b^6
+ 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^2)*sinh(x)^2 + 4*((b^6 + 2*b^5*c - b^4*c^2
- 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)^3 + (b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*cosh(x))*sinh(x)), (2*
C*b^2*c - 4*C*b*c^2 + 2*C*c^3 + (A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^3 + (A*b^3 + A*b^2*c - A*b*c^2 - A
*c^3)*sinh(x)^3 - 2*(C*b^3 - C*b^2*c - C*b*c^2 + C*c^3)*cosh(x)^2 - (2*C*b^3 - 2*C*b^2*c - 2*C*b*c^2 + 2*C*c^3
- 3*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x))*sinh(x)^2 - ((A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^4 + 4*(A*b^2
+ 2*A*b*c + A*c^2)*cosh(x)*sinh(x)^3 + (A*b^2 + 2*A*b*c + A*c^2)*sinh(x)^4 + A*b^2 - 2*A*b*c + A*c^2 + 2*(A*b^
2 - A*c^2)*cosh(x)^2 + 2*(A*b^2 - A*c^2 + 3*(A*b^2 + 2*A*b*c + A*c^2)*cosh(x)^2)*sinh(x)^2 + 4*((A*b^2 + 2*A*b
*c + A*c^2)*cosh(x)^3 + (A*b^2 - A*c^2)*cosh(x))*sinh(x))*sqrt(b^2 - c^2)*arctan(sqrt(b^2 - c^2)/((b + c)*cosh
(x) + (b + c)*sinh(x))) - (A*b^3 - A*b^2*c - A*b*c^2 + A*c^3)*cosh(x) - (A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 - 3
*(A*b^3 + A*b^2*c - A*b*c^2 - A*c^3)*cosh(x)^2 + 4*(C*b^3 - C*b^2*c - C*b*c^2 + C*c^3)*cosh(x))*sinh(x))/(b^6
- 2*b^5*c - b^4*c^2 + 4*b^3*c^3 - b^2*c^4 - 2*b*c^5 + c^6 + (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2
*b*c^5 + c^6)*cosh(x)^4 + 4*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)*sinh(x)^3
+ (b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*sinh(x)^4 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 -
c^6)*cosh(x)^2 + 2*(b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6 + 3*(b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*
b*c^5 + c^6)*cosh(x)^2)*sinh(x)^2 + 4*((b^6 + 2*b^5*c - b^4*c^2 - 4*b^3*c^3 - b^2*c^4 + 2*b*c^5 + c^6)*cosh(x)
^3 + (b^6 - 3*b^4*c^2 + 3*b^2*c^4 - c^6)*cosh(x))*sinh(x))]
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giac [A] time = 0.13, size = 152, normalized size = 1.24 \[ \frac {A \arctan \left (\frac {b e^{x} + c e^{x}}{\sqrt {b^{2} - c^{2}}}\right )}{{\left (b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {A b^{2} e^{\left (3 \, x\right )} + 2 \, A b c e^{\left (3 \, x\right )} + A c^{2} e^{\left (3 \, x\right )} - 2 \, C b^{2} e^{\left (2 \, x\right )} + 2 \, C c^{2} e^{\left (2 \, x\right )} - A b^{2} e^{x} + A c^{2} e^{x} + 2 \, C b c - 2 \, C c^{2}}{{\left (b^{3} + b^{2} c - b c^{2} - c^{3}\right )} {\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="giac")
[Out]
A*arctan((b*e^x + c*e^x)/sqrt(b^2 - c^2))/(b^2 - c^2)^(3/2) + (A*b^2*e^(3*x) + 2*A*b*c*e^(3*x) + A*c^2*e^(3*x)
- 2*C*b^2*e^(2*x) + 2*C*c^2*e^(2*x) - A*b^2*e^x + A*c^2*e^x + 2*C*b*c - 2*C*c^2)/((b^3 + b^2*c - b*c^2 - c^3)
*(b*e^(2*x) + c*e^(2*x) + b - c)^2)
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maple [A] time = 0.28, size = 187, normalized size = 1.52 \[ \frac {-\frac {A \left (b^{2}-2 c^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{\left (b^{2}-c^{2}\right ) b}+\frac {\left (A \,b^{2} c +2 A \,c^{3}+2 C \,b^{3}-2 C b \,c^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{\left (b^{2}-c^{2}\right ) b^{2}}+\frac {A \left (b^{2}+2 c^{2}\right ) \tanh \left (\frac {x}{2}\right )}{\left (b^{2}-c^{2}\right ) b}+\frac {2 A c}{2 b^{2}-2 c^{2}}}{\left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 c \tanh \left (\frac {x}{2}\right )+b \right )^{2}}+\frac {A \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 c}{2 \sqrt {b^{2}-c^{2}}}\right )}{\left (b^{2}-c^{2}\right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x)
[Out]
2*(-1/2*A*(b^2-2*c^2)/(b^2-c^2)/b*tanh(1/2*x)^3+1/2*(A*b^2*c+2*A*c^3+2*C*b^3-2*C*b*c^2)/(b^2-c^2)/b^2*tanh(1/2
*x)^2+1/2*A*(b^2+2*c^2)/(b^2-c^2)/b*tanh(1/2*x)+1/2*A*c/(b^2-c^2))/(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b)^2+A/(b^
2-c^2)^(3/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*b^2>0)', see `assume?`
for more details)Is 4*c^2-4*b^2 positive or negative?
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mupad [B] time = 1.69, size = 217, normalized size = 1.76 \[ \frac {\mathrm {atan}\left (\frac {A\,{\mathrm {e}}^x\,\sqrt {b^6-3\,b^4\,c^2+3\,b^2\,c^4-c^6}}{b^3\,\sqrt {A^2}+c^3\,\sqrt {A^2}-b\,c^2\,\sqrt {A^2}-b^2\,c\,\sqrt {A^2}}\right )\,\sqrt {A^2}}{\sqrt {b^6-3\,b^4\,c^2+3\,b^2\,c^4-c^6}}-\frac {\frac {C}{{\left (b+c\right )}^2}-\frac {A\,{\mathrm {e}}^x}{\left (b+c\right )\,\left (b-c\right )}}{b-c+{\mathrm {e}}^{2\,x}\,\left (b+c\right )}-\frac {\frac {2\,A\,{\mathrm {e}}^x}{b+c}-\frac {C}{b+c}+\frac {C\,{\mathrm {e}}^{2\,x}}{b+c}}{{\mathrm {e}}^{4\,x}\,{\left (b+c\right )}^2+{\left (b-c\right )}^2+2\,{\mathrm {e}}^{2\,x}\,\left (b+c\right )\,\left (b-c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + C*sinh(x))/(b*cosh(x) + c*sinh(x))^3,x)
[Out]
(atan((A*exp(x)*(b^6 - c^6 + 3*b^2*c^4 - 3*b^4*c^2)^(1/2))/(b^3*(A^2)^(1/2) + c^3*(A^2)^(1/2) - b*c^2*(A^2)^(1
/2) - b^2*c*(A^2)^(1/2)))*(A^2)^(1/2))/(b^6 - c^6 + 3*b^2*c^4 - 3*b^4*c^2)^(1/2) - (C/(b + c)^2 - (A*exp(x))/(
(b + c)*(b - c)))/(b - c + exp(2*x)*(b + c)) - ((2*A*exp(x))/(b + c) - C/(b + c) + (C*exp(2*x))/(b + c))/(exp(
4*x)*(b + c)^2 + (b - c)^2 + 2*exp(2*x)*(b + c)*(b - c))
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))**3,x)
[Out]
Timed out
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