3.151 \(\int \frac{\tanh ^2(c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} d \sqrt{a+b}}+\frac{x}{a^2}-\frac{\tanh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.169109, antiderivative size = 85, 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 = {4141, 1975, 471, 522, 206, 208} \[ -\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} d \sqrt{a+b}}+\frac{x}{a^2}-\frac{\tanh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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{\tanh ^2(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{\tanh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 a d}\\ &=-\frac{\tanh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d}-\frac{(a+2 b) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 a^2 d}\\ &=\frac{x}{a^2}-\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} \sqrt{a+b} d}-\frac{\tanh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 4.48855, size = 326, normalized size = 3.84 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b)^2 \left (\frac{\left (a^2+8 a b+8 b^2\right ) \text{sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2 b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}+\frac{\left (-6 a^2 b+a^3-24 a b^2-16 b^3\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{a^2 b d (a+b)^{3/2} \sqrt{b (\cosh (c)-\sinh (c))^4}}+\frac{16 x}{a^2}-\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{b^{3/2} d (a+b)^{3/2}}+\frac{a \sinh (2 (c+d x))}{b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}\right )}{64 \left (a+b \text{sech}^2(c+d x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*((16*x)/a^2 - ((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sq
rt[a + b]])/(b^(3/2)*(a + b)^(3/2)*d) + ((a^3 - 6*a^2*b - 24*a*b^2 - 16*b^3)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - S
inh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c]
- Sinh[2*c]))/(a^2*b*(a + b)^(3/2)*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + ((a^2 + 8*a*b + 8*b^2)*Sech[2*c]*((a +
2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(a^2*b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)])) + (a*Sinh[2*(c + d*x)])/(b*
(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)]))))/(64*(a + b*Sech[c + d*x]^2)^2)

________________________________________________________________________________________

Maple [B]  time = 0.081, size = 417, normalized size = 4.9 \begin{align*}{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}-{\frac{1}{da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}-{\frac{1}{4\,da}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{a+b}}}}+{\frac{1}{4\,da}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ){\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2\,d{a}^{2}}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2\,d{a}^{2}}\sqrt{b}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/a/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)
^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/2*c)^3-1/d/a/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)
^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/2*c)-1/4/d/a/b^(1/2)/(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))+1/4/d/a/b^(1/2)/(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^(1/2)/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/a^2*b^(1/2)/(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/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1
)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.50604, size = 4431, normalized size = 52.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 16*(a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a^2*b
+ a*b^2)*d*x*sinh(d*x + c)^4 + 4*a^2*b + 4*a*b^2 + 4*(a^2*b + a*b^2)*d*x + 4*(a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2
*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 4*(6*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 + a^2*b + 3*a*b^2 + 2*b^
3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*sinh(d*x + c)^2 + ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d
*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2
+ 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)
^3 + (a^2 + 4*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*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))*si
nh(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)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d
*x + c) + a)) + 8*(2*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 + (a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b
^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^4*b + a^3*b^2)*d*cosh(d*x +
c)*sinh(d*x + c)^3 + (a^4*b + a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c)^
2 + 2*(3*(a^4*b + a^3*b^2)*d*cosh(d*x + c)^2 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + a
^3*b^2)*d + 4*((a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x
+ c)), 1/2*(2*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(
a^2*b + a*b^2)*d*x*sinh(d*x + c)^4 + 2*a^2*b + 2*a*b^2 + 2*(a^2*b + a*b^2)*d*x + 2*(a^2*b + 3*a*b^2 + 2*b^3 +
2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 2*(6*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 + a^2*b + 3*a*b^2
+ 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*sinh(d*x + c)^2 - ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*
cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3
*(a^2 + 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*
x + c)^3 + (a^2 + 4*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*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)) + 4*(2*(a^2*b +
a*b^2)*d*x*cosh(d*x + c)^3 + (a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c))*sinh(
d*x + c))/((a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4*b
+ a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^4*b + a^3*b^2)*d*
cosh(d*x + c)^2 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + a^3*b^2)*d + 4*((a^4*b + a^3*b
^2)*d*cosh(d*x + c)^3 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(c + d*x)**2/(a + b*sech(c + d*x)**2)**2, x)

________________________________________________________________________________________

Giac [A]  time = 1.92285, size = 198, normalized size = 2.33 \begin{align*} -\frac{\frac{{\left (a e^{\left (2 \, c\right )} + 2 \, b e^{\left (2 \, c\right )}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right ) e^{\left (-2 \, c\right )}}{\sqrt{-a b - b^{2}} a^{2}} - \frac{2 \, d x}{a^{2}} - \frac{2 \,{\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*e^(2*c) + 2*b*e^(2*c))*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))*e^(-2*c)/(sqrt(-a*b
- b^2)*a^2) - 2*d*x/a^2 - 2*(a*e^(2*d*x + 2*c) + 2*b*e^(2*d*x + 2*c) + a)/((a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x
+ 2*c) + 4*b*e^(2*d*x + 2*c) + a)*a^2))/d