\(\int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [231]

Optimal result

Integrand size = 21, antiderivative size = 71 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a x}{b^2}-\frac {2 a^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2} d}+\frac {\cosh (c+d x)}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2825, 12, 2814, 2739, 632, 210} \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 a^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a x}{b^2}+\frac {\cosh (c+d x)}{b d} \]

[In]

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

[Out]

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

Rule 12

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 2814

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

Rule 2825

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

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

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-a \left (c+d x-\frac {2 a \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}\right )+b \cosh (c+d x)}{b^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.70

method	result	size
derivativedivides	\(\frac {\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {2 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\)	\(121\)
default	\(\frac {\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}+\frac {2 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\)	\(121\)
risch	\(-\frac {a x}{b^{2}}+\frac {{\mathrm e}^{d x +c}}{2 b d}+\frac {{\mathrm e}^{-d x -c}}{2 b d}+\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,b^{2}}\)	\(161\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (68) = 136\).

Time = 0.26 (sec) , antiderivative size = 331, normalized size of antiderivative = 4.66 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right ) - a^{2} b - b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a^{2} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \, {\left ({\left (a^{3} + a b^{2}\right )} d x - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1748 vs. \(2 (61) = 122\).

Time = 176.26 (sec) , antiderivative size = 1748, normalized size of antiderivative = 24.62 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*x*sinh(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (cosh(c + d*x)/(b*d), Eq(a, 0)), (-b*d*x*tanh(c/2 +
d*x/2)**3/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 - b*d*sqrt(-b**2)*tanh
(c/2 + d*x/2)) + b*d*x*tanh(c/2 + d*x/2)/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d + b*d*sqrt(-b**2)*tanh(c/2 + d*
x/2)**3 - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) + 2*b*tanh(c/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d
+ b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) - 4*b/(b**2*d*tanh(c/2 + d*x/2)**2
- b**2*d + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) + d*x*sqrt(-b**2)*tanh(c
/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 - b*d*sqrt(-b**2)*
tanh(c/2 + d*x/2)) - d*x*sqrt(-b**2)/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)
**3 - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) - 2*sqrt(-b**2)*tanh(c/2 + d*x/2)/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2
*d + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)), Eq(a, -sqrt(-b**2))), (-b*d*x*
tanh(c/2 + d*x/2)**3/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 + b*d*sqrt(-
b**2)*tanh(c/2 + d*x/2)) + b*d*x*tanh(c/2 + d*x/2)/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d - b*d*sqrt(-b**2)*tan
h(c/2 + d*x/2)**3 + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) + 2*b*tanh(c/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*x/2)**2
- b**2*d - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) - 4*b/(b**2*d*tanh(c/2 +
d*x/2)**2 - b**2*d - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) - d*x*sqrt(-b*
*2)*tanh(c/2 + d*x/2)**2/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 + b*d*sq
rt(-b**2)*tanh(c/2 + d*x/2)) + d*x*sqrt(-b**2)/(b**2*d*tanh(c/2 + d*x/2)**2 - b**2*d - b*d*sqrt(-b**2)*tanh(c/
2 + d*x/2)**3 + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)) + 2*sqrt(-b**2)*tanh(c/2 + d*x/2)/(b**2*d*tanh(c/2 + d*x/2)
**2 - b**2*d - b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**3 + b*d*sqrt(-b**2)*tanh(c/2 + d*x/2)), Eq(a, sqrt(-b**2))),
((x*sinh(c + d*x)**2/2 - x*cosh(c + d*x)**2/2 + sinh(c + d*x)*cosh(c + d*x)/(2*d))/a, Eq(b, 0)), (x*sinh(c)**
2/(a + b*sinh(c)), Eq(d, 0)), (-a**2*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)*tanh(c/2 + d*x/2)**2/(
b**2*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)) + a**2*log(tanh(c/2 + d*x/2) - b/a -
sqrt(a**2 + b**2)/a)/(b**2*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)) + a**2*log(ta
nh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)*tanh(c/2 + d*x/2)**2/(b**2*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)*
*2 - b**2*d*sqrt(a**2 + b**2)) - a**2*log(tanh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/(b**2*d*sqrt(a**2 + b
**2)*tanh(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)) - a*d*x*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2/(b**2*d*s
qrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)) + a*d*x*sqrt(a**2 + b**2)/(b**2*d*sqrt(a**2
+ b**2)*tanh(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)) - 2*b*sqrt(a**2 + b**2)/(b**2*d*sqrt(a**2 + b**2)*tan
h(c/2 + d*x/2)**2 - b**2*d*sqrt(a**2 + b**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.68 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2} d} - \frac {{\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{2 \, b d} + \frac {e^{\left (-d x - c\right )}}{2 \, b d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.56 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {2 \, {\left (d x + c\right )} a}{b^{2}} + \frac {e^{\left (d x + c\right )}}{b} + \frac {e^{\left (-d x - c\right )}}{b}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.34 \[ \int \frac {\sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}+\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}-\frac {a\,x}{b^2}-\frac {a^2\,\ln \left (-\frac {2\,a^2\,{\mathrm {e}}^{c+d\,x}}{b^3}-\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,d\,\sqrt {a^2+b^2}}+\frac {a^2\,\ln \left (\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3\,\sqrt {a^2+b^2}}-\frac {2\,a^2\,{\mathrm {e}}^{c+d\,x}}{b^3}\right )}{b^2\,d\,\sqrt {a^2+b^2}} \]

[In]

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

[Out]

exp(c + d*x)/(2*b*d) + exp(- c - d*x)/(2*b*d) - (a*x)/b^2 - (a^2*log(- (2*a^2*exp(c + d*x))/b^3 - (2*a^2*(b -
a*exp(c + d*x)))/(b^3*(a^2 + b^2)^(1/2))))/(b^2*d*(a^2 + b^2)^(1/2)) + (a^2*log((2*a^2*(b - a*exp(c + d*x)))/(
b^3*(a^2 + b^2)^(1/2)) - (2*a^2*exp(c + d*x))/b^3))/(b^2*d*(a^2 + b^2)^(1/2))