3.796 \(\int \frac {B \cosh (x)+C \sinh (x)}{(a+b \cosh (x)+c \sinh (x))^2} \, dx\)

Optimal. Leaf size=108 \[ \frac {2 (b B-c C) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {-a B \sinh (x)-a C \cosh (x)-b C+B c}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

[Out]

2*(B*b-C*c)*arctanh((c-(a-b)*tanh(1/2*x))/(a^2-b^2+c^2)^(1/2))/(a^2-b^2+c^2)^(3/2)+(-B*c+b*C+a*C*cosh(x)+a*B*s
inh(x))/(a^2-b^2+c^2)/(a+b*cosh(x)+c*sinh(x))

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Rubi [A]  time = 0.13, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3153, 3124, 618, 206} \[ \frac {2 (b B-c C) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {-a B \sinh (x)-a C \cosh (x)-b C+B c}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + c*Sinh[x])^2,x]

[Out]

(2*(b*B - c*C)*ArcTanh[(c - (a - b)*Tanh[x/2])/Sqrt[a^2 - b^2 + c^2]])/(a^2 - b^2 + c^2)^(3/2) - (B*c - b*C -
a*C*Cosh[x] - a*B*Sinh[x])/((a^2 - b^2 + c^2)*(a + 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 618

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 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^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]

Rubi steps

\begin {align*} \int \frac {B \cosh (x)+C \sinh (x)}{(a+b \cosh (x)+c \sinh (x))^2} \, dx &=-\frac {B c-b C-a C \cosh (x)-a B \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {(b B-c C) \int \frac {1}{a+b \cosh (x)+c \sinh (x)} \, dx}{a^2-b^2+c^2}\\ &=-\frac {B c-b C-a C \cosh (x)-a B \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}-\frac {(2 (b B-c C)) \operatorname {Subst}\left (\int \frac {1}{a+b+2 c x-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2+c^2}\\ &=-\frac {B c-b C-a C \cosh (x)-a B \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}+\frac {(4 (b B-c C)) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2-b^2+c^2\right )-x^2} \, dx,x,2 c+2 (-a+b) \tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2+c^2}\\ &=\frac {2 (b B-c C) \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (a^2-b^2+c^2\right )^{3/2}}-\frac {B c-b C-a C \cosh (x)-a B \sinh (x)}{\left (a^2-b^2+c^2\right ) (a+b \cosh (x)+c \sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 123, normalized size = 1.14 \[ \frac {2 (b B-c C) \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\left (-a^2+b^2-c^2\right )^{3/2}}+\frac {a^2 C+a \sinh (x) (c C-b B)-b^2 C+b B c}{b \left (-a^2+b^2-c^2\right ) (a+b \cosh (x)+c \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + c*Sinh[x])^2,x]

[Out]

(2*(b*B - c*C)*ArcTan[(c + (-a + b)*Tanh[x/2])/Sqrt[-a^2 + b^2 - c^2]])/(-a^2 + b^2 - c^2)^(3/2) + (b*B*c + a^
2*C - b^2*C + a*(-(b*B) + c*C)*Sinh[x])/(b*(-a^2 + b^2 - c^2)*(a + b*Cosh[x] + c*Sinh[x]))

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fricas [B]  time = 0.49, size = 2119, normalized size = 19.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="fricas")

[Out]

[-(2*B*a^3*b - 2*B*a*b^3 + 2*B*a*b*c^2 - 2*C*a*c^3 + (B*b^3 - C*b^2*c - B*b*c^2 + C*c^3 + (B*b^3 + (2*B - C)*b
^2*c + (B - 2*C)*b*c^2 - C*c^3)*cosh(x)^2 + (B*b^3 + (2*B - C)*b^2*c + (B - 2*C)*b*c^2 - C*c^3)*sinh(x)^2 + 2*
(B*a*b^2 + (B - C)*a*b*c - C*a*c^2)*cosh(x) + 2*(B*a*b^2 + (B - C)*a*b*c - C*a*c^2 + (B*b^3 + (2*B - C)*b^2*c
+ (B - 2*C)*b*c^2 - C*c^3)*cosh(x))*sinh(x))*sqrt(a^2 - b^2 + c^2)*log(((b^2 + 2*b*c + c^2)*cosh(x)^2 + (b^2 +
2*b*c + c^2)*sinh(x)^2 + 2*a^2 - b^2 + c^2 + 2*(a*b + a*c)*cosh(x) + 2*(a*b + a*c + (b^2 + 2*b*c + c^2)*cosh(
x))*sinh(x) - 2*sqrt(a^2 - b^2 + c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a))/((b + c)*cosh(x)^2 + (b + c)*si
nh(x)^2 + 2*a*cosh(x) + 2*((b + c)*cosh(x) + a)*sinh(x) + b - c)) - 2*(C*a^3 - C*a*b^2)*c + 2*((B + C)*a^4 - (
B + 2*C)*a^2*b^2 + C*b^4 + (B - C)*b*c^3 + B*c^4 + ((2*B + C)*a^2 - (B + C)*b^2)*c^2 + ((B - C)*a^2*b - (B - C
)*b^3)*c)*cosh(x) + 2*((B + C)*a^4 - (B + 2*C)*a^2*b^2 + C*b^4 + (B - C)*b*c^3 + B*c^4 + ((2*B + C)*a^2 - (B +
C)*b^2)*c^2 + ((B - C)*a^2*b - (B - C)*b^3)*c)*sinh(x))/(a^4*b^2 - 2*a^2*b^4 + b^6 - c^6 - (2*a^2 - 3*b^2)*c^
4 - (a^4 - 4*a^2*b^2 + 3*b^4)*c^2 + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b
- b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*cosh(x)^2 + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5
+ c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*sinh(x)^2
+ 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5 + 2*(a^3 - a*b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*
b^2 + a*b^4)*c)*cosh(x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5 + 2*(a^3 - a*b^2)*c^3 + 2*(a^3*b - a*
b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a
^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*cosh(x))*sinh(x)), -2*(B*a^3*b - B*a*b^3 +
B*a*b*c^2 - C*a*c^3 + (B*b^3 - C*b^2*c - B*b*c^2 + C*c^3 + (B*b^3 + (2*B - C)*b^2*c + (B - 2*C)*b*c^2 - C*c^3)
*cosh(x)^2 + (B*b^3 + (2*B - C)*b^2*c + (B - 2*C)*b*c^2 - C*c^3)*sinh(x)^2 + 2*(B*a*b^2 + (B - C)*a*b*c - C*a*
c^2)*cosh(x) + 2*(B*a*b^2 + (B - C)*a*b*c - C*a*c^2 + (B*b^3 + (2*B - C)*b^2*c + (B - 2*C)*b*c^2 - C*c^3)*cosh
(x))*sinh(x))*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a)/(a^
2 - b^2 + c^2)) - (C*a^3 - C*a*b^2)*c + ((B + C)*a^4 - (B + 2*C)*a^2*b^2 + C*b^4 + (B - C)*b*c^3 + B*c^4 + ((2
*B + C)*a^2 - (B + C)*b^2)*c^2 + ((B - C)*a^2*b - (B - C)*b^3)*c)*cosh(x) + ((B + C)*a^4 - (B + 2*C)*a^2*b^2 +
C*b^4 + (B - C)*b*c^3 + B*c^4 + ((2*B + C)*a^2 - (B + C)*b^2)*c^2 + ((B - C)*a^2*b - (B - C)*b^3)*c)*sinh(x))
/(a^4*b^2 - 2*a^2*b^4 + b^6 - c^6 - (2*a^2 - 3*b^2)*c^4 - (a^4 - 4*a^2*b^2 + 3*b^4)*c^2 + (a^4*b^2 - 2*a^2*b^4
+ b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^
5)*c)*cosh(x)^2 + (a^4*b^2 - 2*a^2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4
- b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c)*sinh(x)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + a*b*c^4 + a*c^5 + 2*(a^
3 - a*b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c)*cosh(x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 +
a*b*c^4 + a*c^5 + 2*(a^3 - a*b^2)*c^3 + 2*(a^3*b - a*b^3)*c^2 + (a^5 - 2*a^3*b^2 + a*b^4)*c + (a^4*b^2 - 2*a^
2*b^4 + b^6 + 2*b*c^5 + c^6 + (2*a^2 - b^2)*c^4 + 4*(a^2*b - b^3)*c^3 + (a^4 - b^4)*c^2 + 2*(a^4*b - 2*a^2*b^3
+ b^5)*c)*cosh(x))*sinh(x))]

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giac [A]  time = 0.12, size = 179, normalized size = 1.66 \[ -\frac {2 \, {\left (B b - C c\right )} \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{{\left (a^{2} - b^{2} + c^{2}\right )} \sqrt {-a^{2} + b^{2} - c^{2}}} - \frac {2 \, {\left (B a^{2} e^{x} + C a^{2} e^{x} - C b^{2} e^{x} + B b c e^{x} - C b c e^{x} + B c^{2} e^{x} + B a b - C a c\right )}}{{\left (a^{2} b - b^{3} + a^{2} c - b^{2} c + b c^{2} + c^{3}\right )} {\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x))^2,x, algorithm="giac")

[Out]

-2*(B*b - C*c)*arctan((b*e^x + c*e^x + a)/sqrt(-a^2 + b^2 - c^2))/((a^2 - b^2 + c^2)*sqrt(-a^2 + b^2 - c^2)) -
2*(B*a^2*e^x + C*a^2*e^x - C*b^2*e^x + B*b*c*e^x - C*b*c*e^x + B*c^2*e^x + B*a*b - C*a*c)/((a^2*b - b^3 + a^2
*c - b^2*c + b*c^2 + c^3)*(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c))

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maple [B]  time = 0.27, size = 287, normalized size = 2.66 \[ \frac {-\frac {2 \left (B \,a^{2}-b B a +B \,c^{2}+a c C -C c b \right ) \tanh \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}-\frac {2 \left (b B c +a^{2} C -C \,b^{2}\right )}{a^{3}-a^{2} b -a \,b^{2}+a \,c^{2}+b^{3}-b \,c^{2}}}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) B b}{\left (a^{2}-b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) C c}{\left (a^{2}-b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}-c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x))^2,x)

[Out]

2*(-(B*a^2-B*a*b+B*c^2+C*a*c-C*b*c)/(a^3-a^2*b-a*b^2+a*c^2+b^3-b*c^2)*tanh(1/2*x)-(B*b*c+C*a^2-C*b^2)/(a^3-a^2
*b-a*b^2+a*c^2+b^3-b*c^2))/(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-2*c*tanh(1/2*x)-a-b)+2/(a^2-b^2+c^2)/(-a^2+b^2-c^2
)^(1/2)*arctan(1/2*(2*(a-b)*tanh(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2))*B*b-2/(a^2-b^2+c^2)/(-a^2+b^2-c^2)^(1/2)*ar
ctan(1/2*(2*(a-b)*tanh(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2))*C*c

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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((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x))^2,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(c^2-b^2+a^2>0)', see `assume?`
for more details)Is c^2-b^2+a^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {B\,\mathrm {cosh}\relax (x)+C\,\mathrm {sinh}\relax (x)}{{\left (a+b\,\mathrm {cosh}\relax (x)+c\,\mathrm {sinh}\relax (x)\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cosh(x) + C*sinh(x))/(a + b*cosh(x) + c*sinh(x))^2,x)

[Out]

int((B*cosh(x) + C*sinh(x))/(a + b*cosh(x) + c*sinh(x))^2, x)

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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((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x))**2,x)

[Out]

Timed out

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