3.370 \(\int \frac{\cosh ^2(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=141 \[ \frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{a \sinh (c+d x) \cosh (c+d x)}{2 b^2 d}+\frac{\sinh ^2(c+d x) \cosh (c+d x)}{3 b d} \]

[Out]

-(a*(2*a^2 + b^2)*x)/(2*b^4) - (2*a^2*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^4
*d) + ((3*a^2 + b^2)*Cosh[c + d*x])/(3*b^3*d) - (a*Cosh[c + d*x]*Sinh[c + d*x])/(2*b^2*d) + (Cosh[c + d*x]*Sin
h[c + d*x]^2)/(3*b*d)

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Rubi [A]  time = 0.498199, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{a \sinh (c+d x) \cosh (c+d x)}{2 b^2 d}+\frac{\sinh ^2(c+d x) \cosh (c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2*a^2 + b^2)*x)/(2*b^4) - (2*a^2*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^4
*d) + ((3*a^2 + b^2)*Cosh[c + d*x])/(3*b^3*d) - (a*Cosh[c + d*x]*Sinh[c + d*x])/(2*b^2*d) + (Cosh[c + d*x]*Sin
h[c + d*x]^2)/(3*b*d)

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3050

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

Rule 3049

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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 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])

Rubi steps

\begin{align*} \int \frac{\cosh ^2(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=\int \frac{\sinh ^2(c+d x) \left (1+\sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx\\ &=\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}+\frac{\int \frac{\sinh (c+d x) \left (-2 a+b \sinh (c+d x)-3 a \sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx}{3 b}\\ &=-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}+\frac{\int \frac{3 a^2-a b \sinh (c+d x)+2 \left (3 a^2+b^2\right ) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx}{6 b^2}\\ &=\frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}+\frac{i \int \frac{-3 i a^2 b+3 i a \left (2 a^2+b^2\right ) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx}{6 b^3}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}+\frac{\left (a^2 \left (a^2+b^2\right )\right ) \int \frac{1}{a+b \sinh (c+d x)} \, dx}{b^4}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}-\frac{\left (2 i a^2 \left (a^2+b^2\right )\right ) \operatorname{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^4 d}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}+\frac{\left (4 i a^2 \left (a^2+b^2\right )\right ) \operatorname{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^4 d}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}-\frac{2 a^2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b^4 d}+\frac{\left (3 a^2+b^2\right ) \cosh (c+d x)}{3 b^3 d}-\frac{a \cosh (c+d x) \sinh (c+d x)}{2 b^2 d}+\frac{\cosh (c+d x) \sinh ^2(c+d x)}{3 b d}\\ \end{align*}

Mathematica [A]  time = 0.391655, size = 123, normalized size = 0.87 \[ \frac{3 b \left (4 a^2+b^2\right ) \cosh (c+d x)-3 a \left (2 \left (2 a^2+b^2\right ) (c+d x)+8 a \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )+b^2 \sinh (2 (c+d x))\right )+b^3 \cosh (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(4*a^2 + b^2)*Cosh[c + d*x] + b^3*Cosh[3*(c + d*x)] - 3*a*(2*(2*a^2 + b^2)*(c + d*x) + 8*a*Sqrt[-a^2 - b^
2]*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]] + b^2*Sinh[2*(c + d*x)]))/(12*b^4*d)

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Maple [B]  time = 0.037, size = 398, normalized size = 2.8 \begin{align*}{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{{a}^{2}}{d{b}^{3}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{a}{2\,d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{a}^{2}}{d{b}^{3}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{a}{2\,d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{2}\sqrt{{a}^{2}+{b}^{2}}}{d{b}^{4}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d/b/(tanh(1/2*d*x+1/2*c)+1)^3-1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^2*a+1/d/
b^3/(tanh(1/2*d*x+1/2*c)+1)*a^2-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)*a+1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/d*a^3/b^
4*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/3/d/b/(tanh(1/2*d*x+1/2*c)-1)^3-1/2/d/b^2/
(tanh(1/2*d*x+1/2*c)-1)^2*a-1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^2-1/d/b^3/(tanh(1/2*d*x+1/2*c)-1)*a^2-1/2/d/b^2/(t
anh(1/2*d*x+1/2*c)-1)*a-1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)+1/d*a^3/b^4*ln(tanh(1/2*d*x+1/2*c)-1)+1/2/d*a/b^2*ln(t
anh(1/2*d*x+1/2*c)-1)+2/d*a^2*(a^2+b^2)^(1/2)/b^4*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^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(cosh(d*x+c)^2*sinh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13802, size = 1882, normalized size = 13.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^3*cosh(d*x + c)^6 + b^3*sinh(d*x + c)^6 - 3*a*b^2*cosh(d*x + c)^5 - 12*(2*a^3 + a*b^2)*d*x*cosh(d*x +
c)^3 + 3*(2*b^3*cosh(d*x + c) - a*b^2)*sinh(d*x + c)^5 + 3*(4*a^2*b + b^3)*cosh(d*x + c)^4 + 3*(5*b^3*cosh(d*x
+ c)^2 - 5*a*b^2*cosh(d*x + c) + 4*a^2*b + b^3)*sinh(d*x + c)^4 + 3*a*b^2*cosh(d*x + c) + 2*(10*b^3*cosh(d*x
+ c)^3 - 15*a*b^2*cosh(d*x + c)^2 - 6*(2*a^3 + a*b^2)*d*x + 6*(4*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c)^3 +
b^3 + 3*(4*a^2*b + b^3)*cosh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 - 10*a*b^2*cosh(d*x + c)^3 - 12*(2*a^3 + a
*b^2)*d*x*cosh(d*x + c) + 4*a^2*b + b^3 + 6*(4*a^2*b + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 24*(a^2*cosh(d*
x + c)^3 + 3*a^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*cosh(d*x + c)*sinh(d*x + c)^2 + a^2*sinh(d*x + c)^3)*sq
rt(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*cosh(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)) + 3*(2*b^3*cosh(d*x + c
)^5 - 5*a*b^2*cosh(d*x + c)^4 - 12*(2*a^3 + a*b^2)*d*x*cosh(d*x + c)^2 + 4*(4*a^2*b + b^3)*cosh(d*x + c)^3 + a
*b^2 + 2*(4*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))/(b^4*d*cosh(d*x + c)^3 + 3*b^4*d*cosh(d*x + c)^2*sinh(d
*x + c) + 3*b^4*d*cosh(d*x + c)*sinh(d*x + c)^2 + b^4*d*sinh(d*x + c)^3)

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

[Out]

Timed out

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Giac [A]  time = 1.16917, size = 309, normalized size = 2.19 \begin{align*} -\frac{{\left (2 \, a^{3} + a b^{2}\right )}{\left (d x + c\right )}}{2 \, b^{4} d} + \frac{{\left (3 \, a b^{2} e^{\left (d x + c\right )} + b^{3} + 3 \,{\left (4 \, a^{2} b + b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, b^{4} d} + \frac{{\left (a^{4} + a^{2} b^{2}\right )} \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^{4} d} + \frac{b^{2} d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 3 \, a b d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, a^{2} d^{2} e^{\left (d x + c\right )} + 3 \, b^{2} d^{2} e^{\left (d x + c\right )}}{24 \, b^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3 + a*b^2)*(d*x + c)/(b^4*d) + 1/24*(3*a*b^2*e^(d*x + c) + b^3 + 3*(4*a^2*b + b^3)*e^(2*d*x + 2*c))*
e^(-3*d*x - 3*c)/(b^4*d) + (a^4 + a^2*b^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^4*d) + 1/24*(b^2*d^2*e^(3*d*x + 3*c) - 3*a*b*d^2*e^(2*d*x +
2*c) + 12*a^2*d^2*e^(d*x + c) + 3*b^2*d^2*e^(d*x + c))/(b^3*d^3)