3.6.79 \(\int \frac {a+b \text {csch}^2(x)}{c+d \sinh (x)} \, dx\) [579]

Optimal. Leaf size=69 \[ \frac {b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac {2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c^2 \sqrt {c^2+d^2}}-\frac {b \coth (x)}{c} \]

[Out]

b*d*arctanh(cosh(x))/c^2-b*coth(x)/c-2*(a*c^2+b*d^2)*arctanh((d-c*tanh(1/2*x))/(c^2+d^2)^(1/2))/c^2/(c^2+d^2)^
(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4318, 3135, 3080, 3855, 2739, 632, 212} \begin {gather*} -\frac {2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {c^2+d^2}}\right )}{c^2 \sqrt {c^2+d^2}}+\frac {b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac {b \coth (x)}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*ArcTanh[Cosh[x]])/c^2 - (2*(a*c^2 + b*d^2)*ArcTanh[(d - c*Tanh[x/2])/Sqrt[c^2 + d^2]])/(c^2*Sqrt[c^2 + d^
2]) - (b*Coth[x])/c

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 3080

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

Rule 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4318

Int[(csc[(a_.) + (b_.)*(x_)]^2*(C_.) + (A_))*(u_), x_Symbol] :> Int[ActivateTrig[u]*((C + A*Sin[a + b*x]^2)/Si
n[a + b*x]^2), x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 125, normalized size = 1.81 \begin {gather*} -\frac {\text {csch}\left (\frac {x}{2}\right ) \text {sech}\left (\frac {x}{2}\right ) \left (b c \sqrt {-c^2-d^2} \cosh (x)+\left (-2 \left (a c^2+b d^2\right ) \text {ArcTan}\left (\frac {d-c \tanh \left (\frac {x}{2}\right )}{\sqrt {-c^2-d^2}}\right )+b d \sqrt {-c^2-d^2} \log \left (\tanh \left (\frac {x}{2}\right )\right )\right ) \sinh (x)\right )}{2 c^2 \sqrt {-c^2-d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Csch[x/2]*Sech[x/2]*(b*c*Sqrt[-c^2 - d^2]*Cosh[x] + (-2*(a*c^2 + b*d^2)*ArcTan[(d - c*Tanh[x/2])/Sqrt[-c
^2 - d^2]] + b*d*Sqrt[-c^2 - d^2]*Log[Tanh[x/2]])*Sinh[x]))/(c^2*Sqrt[-c^2 - d^2])

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Maple [A]
time = 1.25, size = 86, normalized size = 1.25

method	result	size
default	\(-\frac {b \tanh \left (\frac {x}{2}\right )}{2 c}-\frac {\left (-4 a \,c^{2}-4 d^{2} b \right ) \arctanh \left (\frac {2 c \tanh \left (\frac {x}{2}\right )-2 d}{2 \sqrt {c^{2}+d^{2}}}\right )}{2 c^{2} \sqrt {c^{2}+d^{2}}}-\frac {b}{2 c \tanh \left (\frac {x}{2}\right )}-\frac {b d \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{c^{2}}\)	\(86\)
risch	\(-\frac {2 b}{c \left ({\mathrm e}^{2 x}-1\right )}+\frac {b d \ln \left ({\mathrm e}^{x}+1\right )}{c^{2}}-\frac {b d \ln \left ({\mathrm e}^{x}-1\right )}{c^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c -c^{2}-d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) d^{2} b}{\sqrt {c^{2}+d^{2}}\, c^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) a}{\sqrt {c^{2}+d^{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {\sqrt {c^{2}+d^{2}}\, c +c^{2}+d^{2}}{\sqrt {c^{2}+d^{2}}\, d}\right ) d^{2} b}{\sqrt {c^{2}+d^{2}}\, c^{2}}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/c*tanh(1/2*x)-1/2/c^2*(-4*a*c^2-4*b*d^2)/(c^2+d^2)^(1/2)*arctanh(1/2*(2*c*tanh(1/2*x)-2*d)/(c^2+d^2)^(1
/2))-1/2*b/c/tanh(1/2*x)-1/c^2*b*d*ln(tanh(1/2*x))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (65) = 130\).
time = 0.48, size = 158, normalized size = 2.29 \begin {gather*} b {\left (\frac {d^{2} \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}} c^{2}} + \frac {d \log \left (e^{\left (-x\right )} + 1\right )}{c^{2}} - \frac {d \log \left (e^{\left (-x\right )} - 1\right )}{c^{2}} + \frac {2}{c e^{\left (-2 \, x\right )} - c}\right )} + \frac {a \log \left (\frac {d e^{\left (-x\right )} - c - \sqrt {c^{2} + d^{2}}}{d e^{\left (-x\right )} - c + \sqrt {c^{2} + d^{2}}}\right )}{\sqrt {c^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(d^2*log((d*e^(-x) - c - sqrt(c^2 + d^2))/(d*e^(-x) - c + sqrt(c^2 + d^2)))/(sqrt(c^2 + d^2)*c^2) + d*log(e^
(-x) + 1)/c^2 - d*log(e^(-x) - 1)/c^2 + 2/(c*e^(-2*x) - c)) + a*log((d*e^(-x) - c - sqrt(c^2 + d^2))/(d*e^(-x)
- c + sqrt(c^2 + d^2)))/sqrt(c^2 + d^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (65) = 130\).
time = 0.59, size = 401, normalized size = 5.81 \begin {gather*} \frac {2 \, b c^{3} + 2 \, b c d^{2} + {\left (a c^{2} + b d^{2} - {\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a c^{2} + b d^{2}\right )} \sinh \left (x\right )^{2}\right )} \sqrt {c^{2} + d^{2}} \log \left (\frac {d^{2} \cosh \left (x\right )^{2} + d^{2} \sinh \left (x\right )^{2} + 2 \, c d \cosh \left (x\right ) + 2 \, c^{2} + d^{2} + 2 \, {\left (d^{2} \cosh \left (x\right ) + c d\right )} \sinh \left (x\right ) - 2 \, \sqrt {c^{2} + d^{2}} {\left (d \cosh \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{d \cosh \left (x\right )^{2} + d \sinh \left (x\right )^{2} + 2 \, c \cosh \left (x\right ) + 2 \, {\left (d \cosh \left (x\right ) + c\right )} \sinh \left (x\right ) - d}\right ) + {\left (b c^{2} d + b d^{3} - {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (b c^{2} d + b d^{3} - {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{c^{4} + c^{2} d^{2} - {\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (c^{4} + c^{2} d^{2}\right )} \sinh \left (x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*c^3 + 2*b*c*d^2 + (a*c^2 + b*d^2 - (a*c^2 + b*d^2)*cosh(x)^2 - 2*(a*c^2 + b*d^2)*cosh(x)*sinh(x) - (a*c^2
+ b*d^2)*sinh(x)^2)*sqrt(c^2 + d^2)*log((d^2*cosh(x)^2 + d^2*sinh(x)^2 + 2*c*d*cosh(x) + 2*c^2 + d^2 + 2*(d^2
*cosh(x) + c*d)*sinh(x) - 2*sqrt(c^2 + d^2)*(d*cosh(x) + d*sinh(x) + c))/(d*cosh(x)^2 + d*sinh(x)^2 + 2*c*cosh
(x) + 2*(d*cosh(x) + c)*sinh(x) - d)) + (b*c^2*d + b*d^3 - (b*c^2*d + b*d^3)*cosh(x)^2 - 2*(b*c^2*d + b*d^3)*c
osh(x)*sinh(x) - (b*c^2*d + b*d^3)*sinh(x)^2)*log(cosh(x) + sinh(x) + 1) - (b*c^2*d + b*d^3 - (b*c^2*d + b*d^3
)*cosh(x)^2 - 2*(b*c^2*d + b*d^3)*cosh(x)*sinh(x) - (b*c^2*d + b*d^3)*sinh(x)^2)*log(cosh(x) + sinh(x) - 1))/(
c^4 + c^2*d^2 - (c^4 + c^2*d^2)*cosh(x)^2 - 2*(c^4 + c^2*d^2)*cosh(x)*sinh(x) - (c^4 + c^2*d^2)*sinh(x)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {csch}^{2}{\left (x \right )}}{c + d \sinh {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*csch(x)**2)/(c + d*sinh(x)), x)

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Giac [A]
time = 0.40, size = 109, normalized size = 1.58 \begin {gather*} \frac {b d \log \left (e^{x} + 1\right )}{c^{2}} - \frac {b d \log \left ({\left | e^{x} - 1 \right |}\right )}{c^{2}} + \frac {{\left (a c^{2} + b d^{2}\right )} \log \left (\frac {{\left | 2 \, d e^{x} + 2 \, c - 2 \, \sqrt {c^{2} + d^{2}} \right |}}{{\left | 2 \, d e^{x} + 2 \, c + 2 \, \sqrt {c^{2} + d^{2}} \right |}}\right )}{\sqrt {c^{2} + d^{2}} c^{2}} - \frac {2 \, b}{c {\left (e^{\left (2 \, x\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d*log(e^x + 1)/c^2 - b*d*log(abs(e^x - 1))/c^2 + (a*c^2 + b*d^2)*log(abs(2*d*e^x + 2*c - 2*sqrt(c^2 + d^2))/
abs(2*d*e^x + 2*c + 2*sqrt(c^2 + d^2)))/(sqrt(c^2 + d^2)*c^2) - 2*b/(c*(e^(2*x) - 1))

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Mupad [B]
time = 3.77, size = 613, normalized size = 8.88 \begin {gather*} \frac {b\,d\,\ln \left ({\mathrm {e}}^x+1\right )}{c^2}-\frac {b\,d\,\ln \left ({\mathrm {e}}^x-1\right )}{c^2}-\frac {2\,b}{c\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {\ln \left (\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d+2\,b^2\,c^2\,d^2-3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}-\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (4\,a\,c^3\,{\mathrm {e}}^x-2\,b\,d^3-2\,a\,c^2\,d+a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}+\frac {32\,\left (a\,c^2+b\,d^2\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x-4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}\right )\,\left (a\,c^2+b\,d^2\right )\,\sqrt {c^2+d^2}}{c^4+c^2\,d^2}+\frac {\ln \left (-\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,\left (a^2\,c^4+2\,a\,b\,c^2\,d^2-4\,{\mathrm {e}}^x\,b^2\,c^3\,d+2\,b^2\,c^2\,d^2-3\,{\mathrm {e}}^x\,b^2\,c\,d^3+2\,b^2\,d^4\right )}{c^2\,d^4}+\frac {\left (a\,c^2+b\,d^2\right )\,\left (\frac {32\,c\,\left (4\,a\,c^3\,{\mathrm {e}}^x-2\,b\,d^3-2\,a\,c^2\,d+a\,c\,d^2\,{\mathrm {e}}^x+3\,b\,c\,d^2\,{\mathrm {e}}^x\right )}{d^5}-\frac {32\,\left (a\,c^2+b\,d^2\right )\,\left (-4\,{\mathrm {e}}^x\,c^3+3\,c^2\,d-3\,{\mathrm {e}}^x\,c\,d^2+2\,d^3\right )}{d^5\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}\right )}{c^2\,\sqrt {c^2+d^2}}-\frac {32\,b\,\left (a\,c^2+b\,d^2\right )\,\left (2\,b\,d+a\,c\,{\mathrm {e}}^x-4\,b\,c\,{\mathrm {e}}^x\right )}{c^3\,d^3}\right )\,\left (a\,c^2+b\,d^2\right )\,\sqrt {c^2+d^2}}{c^4+c^2\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*log(exp(x) + 1))/c^2 - (b*d*log(exp(x) - 1))/c^2 - (2*b)/(c*(exp(2*x) - 1)) - (log(((a*c^2 + b*d^2)*((32*
(a^2*c^4 + 2*b^2*d^4 + 2*b^2*c^2*d^2 - 3*b^2*c*d^3*exp(x) - 4*b^2*c^3*d*exp(x) + 2*a*b*c^2*d^2))/(c^2*d^4) - (
(a*c^2 + b*d^2)*((32*c*(4*a*c^3*exp(x) - 2*b*d^3 - 2*a*c^2*d + a*c*d^2*exp(x) + 3*b*c*d^2*exp(x)))/d^5 + (32*(
a*c^2 + b*d^2)*(3*c^2*d + 2*d^3 - 4*c^3*exp(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 + d^2)^(1/2))))/(c^2*(c^2 + d^2)^(
1/2))))/(c^2*(c^2 + d^2)^(1/2)) - (32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) - 4*b*c*exp(x)))/(c^3*d^3))*(a*c^2
+ b*d^2)*(c^2 + d^2)^(1/2))/(c^4 + c^2*d^2) + (log(- ((a*c^2 + b*d^2)*((32*(a^2*c^4 + 2*b^2*d^4 + 2*b^2*c^2*d
^2 - 3*b^2*c*d^3*exp(x) - 4*b^2*c^3*d*exp(x) + 2*a*b*c^2*d^2))/(c^2*d^4) + ((a*c^2 + b*d^2)*((32*c*(4*a*c^3*ex
p(x) - 2*b*d^3 - 2*a*c^2*d + a*c*d^2*exp(x) + 3*b*c*d^2*exp(x)))/d^5 - (32*(a*c^2 + b*d^2)*(3*c^2*d + 2*d^3 -
4*c^3*exp(x) - 3*c*d^2*exp(x)))/(d^5*(c^2 + d^2)^(1/2))))/(c^2*(c^2 + d^2)^(1/2))))/(c^2*(c^2 + d^2)^(1/2)) -
(32*b*(a*c^2 + b*d^2)*(2*b*d + a*c*exp(x) - 4*b*c*exp(x)))/(c^3*d^3))*(a*c^2 + b*d^2)*(c^2 + d^2)^(1/2))/(c^4
+ c^2*d^2)

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