3.254 \(\int \frac{A+B \cosh (d+e x)+C \sinh (d+e x)}{(a+c \sinh (d+e x))^2} \, dx\)

Optimal. Leaf size=113 \[ -\frac{2 (a A+c C) \tanh ^{-1}\left (\frac{c-a \tanh \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2+c^2}}\right )}{e \left (a^2+c^2\right )^{3/2}}-\frac{(A c-a C) \cosh (d+e x)}{e \left (a^2+c^2\right ) (a+c \sinh (d+e x))}-\frac{B}{c e (a+c \sinh (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.169336, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {4376, 2754, 12, 2660, 618, 204, 2668, 32} \[ -\frac{2 (a A+c C) \tanh ^{-1}\left (\frac{c-a \tanh \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2+c^2}}\right )}{e \left (a^2+c^2\right )^{3/2}}-\frac{(A c-a C) \cosh (d+e x)}{e \left (a^2+c^2\right ) (a+c \sinh (d+e x))}-\frac{B}{c e (a+c \sinh (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4376

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Sin[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Cos[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Sin[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Cos] || EqQ[F, cos])

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

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 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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.52467, size = 113, normalized size = 1. \[ \frac{\frac{2 (a A+c C) \tan ^{-1}\left (\frac{c-a \tanh \left (\frac{1}{2} (d+e x)\right )}{\sqrt{-a^2-c^2}}\right )}{\sqrt{-a^2-c^2}}-\frac{B \left (a^2+c^2\right )+c (A c-a C) \cosh (d+e x)}{c (a+c \sinh (d+e x))}}{e \left (a^2+c^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 151, normalized size = 1.3 \begin{align*}{\frac{1}{e} \left ( -2\,{\frac{1}{a \left ( \tanh \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}-2\,c\tanh \left ( 1/2\,ex+d/2 \right ) -a} \left ( -{\frac{ \left ( A{c}^{2}-B{a}^{2}-B{c}^{2}-Cac \right ) \tanh \left ( 1/2\,ex+d/2 \right ) }{a \left ({a}^{2}+{c}^{2} \right ) }}-{\frac{Ac-Ca}{{a}^{2}+{c}^{2}}} \right ) }+2\,{\frac{Aa+Cc}{ \left ({a}^{2}+{c}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,ex+d/2 \right ) -2\,c}{\sqrt{{a}^{2}+{c}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(-2*(-(A*c^2-B*a^2-B*c^2-C*a*c)/a/(a^2+c^2)*tanh(1/2*e*x+1/2*d)-(A*c-C*a)/(a^2+c^2))/(a*tanh(1/2*e*x+1/2*d
)^2-2*c*tanh(1/2*e*x+1/2*d)-a)+2*(A*a+C*c)/(a^2+c^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*e*x+1/2*d)-2*c)/(a^2+c^2)
^(1/2)))

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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+B*cosh(e*x+d)+C*sinh(e*x+d))/(a+c*sinh(e*x+d))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.16489, size = 1339, normalized size = 11.85 \begin{align*} \frac{2 \, C a^{3} c - 2 \, A a^{2} c^{2} + 2 \, C a c^{3} - 2 \, A c^{4} -{\left (A a c^{2} + C c^{3} -{\left (A a c^{2} + C c^{3}\right )} \cosh \left (e x + d\right )^{2} -{\left (A a c^{2} + C c^{3}\right )} \sinh \left (e x + d\right )^{2} - 2 \,{\left (A a^{2} c + C a c^{2}\right )} \cosh \left (e x + d\right ) - 2 \,{\left (A a^{2} c + C a c^{2} +{\left (A a c^{2} + C c^{3}\right )} \cosh \left (e x + d\right )\right )} \sinh \left (e x + d\right )\right )} \sqrt{a^{2} + c^{2}} \log \left (\frac{c^{2} \cosh \left (e x + d\right )^{2} + c^{2} \sinh \left (e x + d\right )^{2} + 2 \, a c \cosh \left (e x + d\right ) + 2 \, a^{2} + c^{2} + 2 \,{\left (c^{2} \cosh \left (e x + d\right ) + a c\right )} \sinh \left (e x + d\right ) - 2 \, \sqrt{a^{2} + c^{2}}{\left (c \cosh \left (e x + d\right ) + c \sinh \left (e x + d\right ) + a\right )}}{c \cosh \left (e x + d\right )^{2} + c \sinh \left (e x + d\right )^{2} + 2 \, a \cosh \left (e x + d\right ) + 2 \,{\left (c \cosh \left (e x + d\right ) + a\right )} \sinh \left (e x + d\right ) - c}\right ) - 2 \,{\left ({\left (B + C\right )} a^{4} - A a^{3} c +{\left (2 \, B + C\right )} a^{2} c^{2} - A a c^{3} + B c^{4}\right )} \cosh \left (e x + d\right ) - 2 \,{\left ({\left (B + C\right )} a^{4} - A a^{3} c +{\left (2 \, B + C\right )} a^{2} c^{2} - A a c^{3} + B c^{4}\right )} \sinh \left (e x + d\right )}{{\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6}\right )} e \cosh \left (e x + d\right )^{2} +{\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6}\right )} e \sinh \left (e x + d\right )^{2} + 2 \,{\left (a^{5} c + 2 \, a^{3} c^{3} + a c^{5}\right )} e \cosh \left (e x + d\right ) -{\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6}\right )} e + 2 \,{\left ({\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6}\right )} e \cosh \left (e x + d\right ) +{\left (a^{5} c + 2 \, a^{3} c^{3} + a c^{5}\right )} e\right )} \sinh \left (e x + d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*a^3*c - 2*A*a^2*c^2 + 2*C*a*c^3 - 2*A*c^4 - (A*a*c^2 + C*c^3 - (A*a*c^2 + C*c^3)*cosh(e*x + d)^2 - (A*a*c
^2 + C*c^3)*sinh(e*x + d)^2 - 2*(A*a^2*c + C*a*c^2)*cosh(e*x + d) - 2*(A*a^2*c + C*a*c^2 + (A*a*c^2 + C*c^3)*c
osh(e*x + d))*sinh(e*x + d))*sqrt(a^2 + c^2)*log((c^2*cosh(e*x + d)^2 + c^2*sinh(e*x + d)^2 + 2*a*c*cosh(e*x +
d) + 2*a^2 + c^2 + 2*(c^2*cosh(e*x + d) + a*c)*sinh(e*x + d) - 2*sqrt(a^2 + c^2)*(c*cosh(e*x + d) + c*sinh(e*
x + d) + a))/(c*cosh(e*x + d)^2 + c*sinh(e*x + d)^2 + 2*a*cosh(e*x + d) + 2*(c*cosh(e*x + d) + a)*sinh(e*x + d
) - c)) - 2*((B + C)*a^4 - A*a^3*c + (2*B + C)*a^2*c^2 - A*a*c^3 + B*c^4)*cosh(e*x + d) - 2*((B + C)*a^4 - A*a
^3*c + (2*B + C)*a^2*c^2 - A*a*c^3 + B*c^4)*sinh(e*x + d))/((a^4*c^2 + 2*a^2*c^4 + c^6)*e*cosh(e*x + d)^2 + (a
^4*c^2 + 2*a^2*c^4 + c^6)*e*sinh(e*x + d)^2 + 2*(a^5*c + 2*a^3*c^3 + a*c^5)*e*cosh(e*x + d) - (a^4*c^2 + 2*a^2
*c^4 + c^6)*e + 2*((a^4*c^2 + 2*a^2*c^4 + c^6)*e*cosh(e*x + d) + (a^5*c + 2*a^3*c^3 + a*c^5)*e)*sinh(e*x + d))

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

[Out]

Timed out

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Giac [A]  time = 1.18165, size = 262, normalized size = 2.32 \begin{align*} \frac{{\left (A a + C c\right )} \log \left (\frac{{\left | 2 \, c e^{\left (x e + d\right )} + 2 \, a - 2 \, \sqrt{a^{2} + c^{2}} \right |}}{{\left | 2 \, c e^{\left (x e + d\right )} + 2 \, a + 2 \, \sqrt{a^{2} + c^{2}} \right |}}\right )}{{\left (a^{2} e + c^{2} e\right )} \sqrt{a^{2} + c^{2}}} - \frac{2 \,{\left (B a^{2} e^{\left (x e + d\right )} + C a^{2} e^{\left (x e + d\right )} - A a c e^{\left (x e + d\right )} + B c^{2} e^{\left (x e + d\right )} - C a c + A c^{2}\right )}}{{\left (a^{2} c e + c^{3} e\right )}{\left (c e^{\left (2 \, x e + 2 \, d\right )} + 2 \, a e^{\left (x e + d\right )} - c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a + C*c)*log(abs(2*c*e^(x*e + d) + 2*a - 2*sqrt(a^2 + c^2))/abs(2*c*e^(x*e + d) + 2*a + 2*sqrt(a^2 + c^2)))
/((a^2*e + c^2*e)*sqrt(a^2 + c^2)) - 2*(B*a^2*e^(x*e + d) + C*a^2*e^(x*e + d) - A*a*c*e^(x*e + d) + B*c^2*e^(x
*e + d) - C*a*c + A*c^2)/((a^2*c*e + c^3*e)*(c*e^(2*x*e + 2*d) + 2*a*e^(x*e + d) - c))