3.25 \(\int \frac{\sinh ^4(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)
Optimal. Leaf size=117 \[ \frac{x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}-\frac{\sqrt{b} (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 d}-\frac{(5 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]
[Out]
((3*a^2 + 12*a*b + 8*b^2)*x)/(8*a^3) - (Sqrt[b]*(a + b)^(3/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a
^3*d) - ((5*a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*a^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*a*d)
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Rubi [A] time = 0.195511, antiderivative size = 117, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used =
{4132, 470, 527, 522, 206, 208} \[ \frac{x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}-\frac{\sqrt{b} (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 d}-\frac{(5 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
[Out]
((3*a^2 + 12*a*b + 8*b^2)*x)/(8*a^3) - (Sqrt[b]*(a + b)^(3/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a
^3*d) - ((5*a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*a^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*a*d)
Rule 4132
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]
Rule 470
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 527
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, 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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sinh ^4(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}-\frac{\operatorname{Subst}\left (\int \frac{a+b+(4 a+3 b) x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{4 a d}\\ &=-\frac{(5 a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}-\frac{\operatorname{Subst}\left (\int \frac{-(a+b) (3 a+4 b)-b (5 a+4 b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 a^2 d}\\ &=-\frac{(5 a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}-\frac{\left (b (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{a^3 d}+\frac{\left (3 a^2+12 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^3 d}\\ &=\frac{\left (3 a^2+12 a b+8 b^2\right ) x}{8 a^3}-\frac{\sqrt{b} (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 d}-\frac{(5 a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}\\ \end{align*}
Mathematica [B] time = 2.40738, size = 294, normalized size = 2.51 \[ -\frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\sqrt{b} \left (34 a^2 b+3 a^3+64 a b^2+32 b^3\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )-\sqrt{b (\cosh (c)-\sinh (c))^4} \left (\sqrt{b} \sqrt{a+b} \left (a^2 \sinh (4 (c+d x))-2 a^2 c+12 a^2 d x-8 a (a+b) \sinh (2 (c+d x))+48 a b d x+32 b^2 d x\right )+a^2 (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )\right )\right )}{64 a^3 \sqrt{b} d \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4} \left (a+b \text{sech}^2(c+d x)\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
[Out]
-((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(Sqrt[b]*(3*a^3 + 34*a^2*b + 64*a*b^2 + 32*b^3)*ArcTanh[(Sec
h[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sin
h[c])^4])]*(Cosh[2*c] - Sinh[2*c]) - Sqrt[b*(Cosh[c] - Sinh[c])^4]*(a^2*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c +
d*x])/Sqrt[a + b]] + Sqrt[b]*Sqrt[a + b]*(-2*a^2*c + 12*a^2*d*x + 48*a*b*d*x + 32*b^2*d*x - 8*a*(a + b)*Sinh[2
*(c + d*x)] + a^2*Sinh[4*(c + d*x)]))))/(64*a^3*Sqrt[b]*Sqrt[a + b]*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c]
- Sinh[c])^4])
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Maple [B] time = 0.089, size = 708, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x)
[Out]
-1/4/d/a/(tanh(1/2*d*x+1/2*c)+1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)^3+1/8/d/a/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/a
^2/(tanh(1/2*d*x+1/2*c)+1)^2*b-3/8/d/a/(tanh(1/2*d*x+1/2*c)+1)-1/2/d/a^2/(tanh(1/2*d*x+1/2*c)+1)*b+3/8/d/a*ln(
tanh(1/2*d*x+1/2*c)+1)+3/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)*b+1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2-1/2/d*b^(1/
2)/a/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))-1/d*b^(3/2)/a
^2/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))-1/2/d*b^(5/2)/a
^3/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/2/d*b^(1/2)/a
/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/d*b^(3/2)/a^2/(
a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/2/d*b^(5/2)/a^3/(
a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/4/d/a/(tanh(1/2*d
*x+1/2*c)-1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)^3-1/8/d/a/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/d/a^2/(tanh(1/2*d*x+1/2
*c)-1)^2*b-3/8/d/a/(tanh(1/2*d*x+1/2*c)-1)-1/2/d/a^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/a*ln(tanh(1/2*d*x+1/2*c)-
1)-3/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)*b-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)-1)*b^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(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.96917, size = 4338, normalized size = 37.08 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/64*(a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 + 12*a*b + 8
*b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x
+ c)^6 + 8*(7*a^2*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4
+ 4*(3*a^2 + 12*a*b + 8*b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5
+ 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*
b)*cosh(d*x + c)^2 + 4*(7*a^2*cosh(d*x + c)^6 + 12*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*
b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*sinh(d*x + c)^2 + 32*((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)^3*
sinh(d*x + c) + 6*(a + b)*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*
sinh(d*x + c)^4)*sqrt(a*b + b^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x
+ c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*
a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*cosh(d*x + c)^2 + 2*a
*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x
+ c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*si
nh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - a^2 + 8*(a^2*cosh(d*x +
c)^7 + 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x
+ c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^3*d*cosh(d*x + c)^2
*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x + c)^4), 1/64*(a^2*cosh(d*x + c)^8 +
8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^4 -
8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7*a^2*cosh(d*x
+ c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 + 4*(3*a^2 + 12*a*b + 8*b^2)
*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 + 4*(3*a^2 + 12*a*b + 8*b^2)
*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cosh(d*x + c)^2 + 4*(7*a^
2*cosh(d*x + c)^6 + 12*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*b)*cosh(d*x + c)^4 + 2*a^2 +
2*a*b)*sinh(d*x + c)^2 - 64*((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a + b)*co
sh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4)*sqrt(-a*b -
b^2)*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-a*b
- b^2)/(a*b + b^2)) - a^2 + 8*(a^2*cosh(d*x + c)^7 + 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 - 6*(a^2
+ a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x
+ c)^3*sinh(d*x + c) + 6*a^3*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*
d*sinh(d*x + c)^4)]
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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(sinh(d*x+c)**4/(a+b*sech(d*x+c)**2),x)
[Out]
Timed out
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Giac [B] time = 1.17853, size = 312, normalized size = 2.67 \begin{align*} \frac{{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )}{\left (d x + c\right )}}{8 \, a^{3} d} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 72 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, a^{3} d} - \frac{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a^{3} d} + \frac{a d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, a^{2} d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="giac")
[Out]
1/8*(3*a^2 + 12*a*b + 8*b^2)*(d*x + c)/(a^3*d) - 1/64*(18*a^2*e^(4*d*x + 4*c) + 72*a*b*e^(4*d*x + 4*c) + 48*b^
2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) - 8*a*b*e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c)/(a^3*d) - (a^2*b + 2
*a*b^2 + b^3)*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^3*d) + 1/64*(a*d*
e^(4*d*x + 4*c) - 8*a*d*e^(2*d*x + 2*c) - 8*b*d*e^(2*d*x + 2*c))/(a^2*d^2)