### 3.725 $$\int \frac{A+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx$$

Optimal. Leaf size=82 $-\frac{-A b \sinh (x)-A c \cosh (x)+b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{c C \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0790252, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {3154, 3074, 206} $-\frac{-A b \sinh (x)-A c \cosh (x)+b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{c C \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3154

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> -Simp[(b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Co
s[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - 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, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 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 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])

Rubi steps

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

Mathematica [A]  time = 0.439, size = 155, normalized size = 1.89 $-\frac{\sinh (x) \left (2 b c^2 C \sqrt{b+c} \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )-A (b-c)^{3/2} (b+c)^2\right )+2 b^2 c C \sqrt{b+c} \cosh (x) \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )+b^2 C \sqrt{b-c} (b+c)}{b (b-c)^{3/2} (b+c)^2 (b \cosh (x)+c \sinh (x))}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.067, size = 115, normalized size = 1.4 \begin{align*} -2\,{\frac{1}{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+2\,c\tanh \left ( x/2 \right ) +b} \left ( -{\frac{ \left ( A{b}^{2}-A{c}^{2}-Ccb \right ) \tanh \left ( x/2 \right ) }{ \left ({b}^{2}-{c}^{2} \right ) b}}+{\frac{bC}{{b}^{2}-{c}^{2}}} \right ) }-2\,{\frac{Cc}{ \left ({b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,c}{\sqrt{{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-(A*b^2-A*c^2-C*b*c)/(b^2-c^2)/b*tanh(1/2*x)+b*C/(b^2-c^2))/(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b)-2*C*c/(b^2
-c^2)^(3/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.42513, size = 1648, normalized size = 20.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*A*b^3 - 2*A*b^2*c - 2*A*b*c^2 + 2*A*c^3 - (C*b*c - C*c^2 + (C*b*c + C*c^2)*cosh(x)^2 + 2*(C*b*c + C*c^2)*
cosh(x)*sinh(x) + (C*b*c + C*c^2)*sinh(x)^2)*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)*cosh(x)^2 + 2*(b + c)*cosh(x
)*sinh(x) + (b + c)*sinh(x)^2 + b - c)) + 2*(C*b^3 - C*b*c^2)*cosh(x) + 2*(C*b^3 - C*b*c^2)*sinh(x))/(b^5 - b^
4*c - 2*b^3*c^2 + 2*b^2*c^3 + b*c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^2 + 2*
(b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b
*c^4 + c^5)*sinh(x)^2), -2*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 - (C*b*c - C*c^2 + (C*b*c + C*c^2)*cosh(x)^2 + 2
*(C*b*c + C*c^2)*cosh(x)*sinh(x) + (C*b*c + C*c^2)*sinh(x)^2)*sqrt(b^2 - c^2)*arctan(sqrt(b^2 - c^2)/((b + c)*
cosh(x) + (b + c)*sinh(x))) + (C*b^3 - C*b*c^2)*cosh(x) + (C*b^3 - C*b*c^2)*sinh(x))/(b^5 - b^4*c - 2*b^3*c^2
+ 2*b^2*c^3 + b*c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^2 + 2*(b^5 + b^4*c - 2
*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*sinh
(x)^2)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14731, size = 112, normalized size = 1.37 \begin{align*} -\frac{2 \, C c \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{2 \,{\left (C b e^{x} + A b - A c\right )}}{{\left (b^{2} - c^{2}\right )}{\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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