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

Optimal. Leaf size=92 \[ \frac {3 \cosh (c+d x)}{2 d (a+b)^2}-\frac {\cosh (c+d x)}{2 d (a+b) \left (a-b \text {sech}^2(c+d x)+b\right )}-\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}} \]

[Out]

3/2*cosh(d*x+c)/(a+b)^2/d-1/2*cosh(d*x+c)/(a+b)/d/(a+b-b*sech(d*x+c)^2)-3/2*arctanh(sech(d*x+c)*b^(1/2)/(a+b)^
(1/2))*b^(1/2)/(a+b)^(5/2)/d

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Rubi [A]  time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3664, 290, 325, 208} \[ \frac {3 \cosh (c+d x)}{2 d (a+b)^2}-\frac {\cosh (c+d x)}{2 d (a+b) \left (a-b \text {sech}^2(c+d x)+b\right )}-\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[b]*ArcTanh[(Sqrt[b]*Sech[c + d*x])/Sqrt[a + b]])/(2*(a + b)^(5/2)*d) + (3*Cosh[c + d*x])/(2*(a + b)^2
*d) - Cosh[c + d*x]/(2*(a + b)*d*(a + b - b*Sech[c + d*x]^2))

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]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sinh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b-b x^2\right )^2} \, dx,x,\text {sech}(c+d x)\right )}{d}\\ &=-\frac {\cosh (c+d x)}{2 (a+b) d \left (a+b-b \text {sech}^2(c+d x)\right )}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b-b x^2\right )} \, dx,x,\text {sech}(c+d x)\right )}{2 (a+b) d}\\ &=\frac {3 \cosh (c+d x)}{2 (a+b)^2 d}-\frac {\cosh (c+d x)}{2 (a+b) d \left (a+b-b \text {sech}^2(c+d x)\right )}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\text {sech}(c+d x)\right )}{2 (a+b)^2 d}\\ &=-\frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}+\frac {3 \cosh (c+d x)}{2 (a+b)^2 d}-\frac {\cosh (c+d x)}{2 (a+b) d \left (a+b-b \text {sech}^2(c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 0.77, size = 133, normalized size = 1.45 \[ \frac {\frac {2 \cosh (c+d x) \left (1-\frac {b}{(a+b) \cosh (2 (c+d x))+a-b}\right )}{(a+b)^2}-\frac {3 i \sqrt {b} \left (\tan ^{-1}\left (\frac {-\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a+b}}{\sqrt {b}}\right )+\tan ^{-1}\left (\frac {\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )-i \sqrt {a+b}}{\sqrt {b}}\right )\right )}{(a+b)^{5/2}}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)*Sqrt[b]*(ArcTan[((-I)*Sqrt[a + b] - Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[b]] + ArcTan[((-I)*Sqrt[a + b] +
Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[b]]))/(a + b)^(5/2) + (2*Cosh[c + d*x]*(1 - b/(a - b + (a + b)*Cosh[2*(c + d*x
)])))/(a + b)^2)/(2*d)

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fricas [B]  time = 0.61, size = 2252, normalized size = 24.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a + b)*cosh(d*x + c)^6 + 12*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(a + b)*sinh(d*x + c)^6 + 6*(a
- b)*cosh(d*x + c)^4 + 6*(5*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^4 + 8*(5*(a + b)*cosh(d*x + c)^3 +
3*(a - b)*cosh(d*x + c))*sinh(d*x + c)^3 + 6*(a - b)*cosh(d*x + c)^2 + 6*(5*(a + b)*cosh(d*x + c)^4 + 6*(a - b
)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 3*((a + b)*cosh(d*x + c)^5 + 5*(a + b)*cosh(d*x + c)*sinh(d*x + c
)^4 + (a + b)*sinh(d*x + c)^5 + 2*(a - b)*cosh(d*x + c)^3 + 2*(5*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c
)^3 + 2*(5*(a + b)*cosh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c))*sinh(d*x + c)^2 + (a + b)*cosh(d*x + c) + (5*(a
+ b)*cosh(d*x + c)^4 + 6*(a - b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c))*sqrt(b/(a + b))*log(((a + b)*cosh(d*x
+ c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a + 3*b)*cosh(d*x + c)^2 + 2*
(3*(a + b)*cosh(d*x + c)^2 + a + 3*b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a + 3*b)*cosh(d*x + c))*
sinh(d*x + c) - 4*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3
+ (a + b)*cosh(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c))*sqrt(b/(a + b)) + a + b)/((a + b
)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c
)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x +
c))*sinh(d*x + c) + a + b)) + 12*((a + b)*cosh(d*x + c)^5 + 2*(a - b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))
*sinh(d*x + c) + 2*a + 2*b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^5 + 5*(a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*d*cosh(d*x + c)*sinh(d*x + c)^4 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)^5 + 2*(a^3 + a^2*b - a*
b^2 - b^3)*d*cosh(d*x + c)^3 + 2*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2 + (a^3 + a^2*b - a*b^2 -
b^3)*d)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c) + 2*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b
^3)*d*cosh(d*x + c)^3 + 3*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c))*sinh(d*x + c)^2 + (5*(a^3 + 3*a^2*b + 3
*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 6*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^2 + (a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*d)*sinh(d*x + c)), 1/2*((a + b)*cosh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh
(d*x + c)^6 + 3*(a - b)*cosh(d*x + c)^4 + 3*(5*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^4 + 4*(5*(a + b)
*cosh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*
x + c)^4 + 6*(a - b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 - 3*((a + b)*cosh(d*x + c)^5 + 5*(a + b)*cosh(d*
x + c)*sinh(d*x + c)^4 + (a + b)*sinh(d*x + c)^5 + 2*(a - b)*cosh(d*x + c)^3 + 2*(5*(a + b)*cosh(d*x + c)^2 +
a - b)*sinh(d*x + c)^3 + 2*(5*(a + b)*cosh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c))*sinh(d*x + c)^2 + (a + b)*cos
h(d*x + c) + (5*(a + b)*cosh(d*x + c)^4 + 6*(a - b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c))*sqrt(-b/(a + b))*a
rctan(1/2*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3 + (a -
3*b)*cosh(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 + a - 3*b)*sinh(d*x + c))*sqrt(-b/(a + b))/b) + 3*((a + b)*cos
h(d*x + c)^5 + 5*(a + b)*cosh(d*x + c)*sinh(d*x + c)^4 + (a + b)*sinh(d*x + c)^5 + 2*(a - b)*cosh(d*x + c)^3 +
2*(5*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^3 + 2*(5*(a + b)*cosh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c
))*sinh(d*x + c)^2 + (a + b)*cosh(d*x + c) + (5*(a + b)*cosh(d*x + c)^4 + 6*(a - b)*cosh(d*x + c)^2 + a + b)*s
inh(d*x + c))*sqrt(-b/(a + b))*arctan(1/2*((a + b)*cosh(d*x + c) + (a + b)*sinh(d*x + c))*sqrt(-b/(a + b))/b)
+ 6*((a + b)*cosh(d*x + c)^5 + 2*(a - b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)/((a^3
+ 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^5 + 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c
)^4 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)^5 + 2*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^3 + 2*
(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2 + (a^3 + a^2*b - a*b^2 - b^3)*d)*sinh(d*x + c)^3 + (a^3 +
3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c) + 2*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^3 + 3*(a^3 +
a^2*b - a*b^2 - b^3)*d*cosh(d*x + c))*sinh(d*x + c)^2 + (5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 +
6*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)*sinh(d*x + c))]

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giac [B]  time = 0.41, size = 924, normalized size = 10.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*(5*a^2*b - 10*a*b^2 + b^3 - (a^2 - 10*a*b + 5*b^2)*sqrt(-a*b))*sqrt(a^2 - b^2 + 2*sqrt(-a*b)*(a + b))*a
bs(a*e^(2*c) + b*e^(2*c))*arctan(e^(d*x)/sqrt((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*e^(2*c) - b^3*e^(2*c) + sqr
t((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*e^(2*c) - b^3*e^(2*c))^2 - (a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(
4*c) + b^3*e^(4*c))*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)))/(a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e
^(4*c))))*e^(-2*c)/(a^7 - 11*a^6*b - 39*a^5*b^2 - 27*a^4*b^3 + 27*a^3*b^4 + 39*a^2*b^5 + 11*a*b^6 - b^7 + 2*(3
*a^6 + 2*a^5*b - 19*a^4*b^2 - 36*a^3*b^3 - 19*a^2*b^4 + 2*a*b^5 + 3*b^6)*sqrt(-a*b)) + 3*(5*a^2*b - 10*a*b^2 +
b^3 + (a^2 - 10*a*b + 5*b^2)*sqrt(-a*b))*sqrt(a^2 - b^2 - 2*sqrt(-a*b)*(a + b))*abs(a*e^(2*c) + b*e^(2*c))*ar
ctan(e^(d*x)/sqrt((a^3*e^(2*c) + a^2*b*e^(2*c) - a*b^2*e^(2*c) - b^3*e^(2*c) - sqrt((a^3*e^(2*c) + a^2*b*e^(2*
c) - a*b^2*e^(2*c) - b^3*e^(2*c))^2 - (a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c))*(a^3 + 3
*a^2*b + 3*a*b^2 + b^3)))/(a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c))))*e^(-2*c)/(a^7 - 11
*a^6*b - 39*a^5*b^2 - 27*a^4*b^3 + 27*a^3*b^4 + 39*a^2*b^5 + 11*a*b^6 - b^7 - 2*(3*a^6 + 2*a^5*b - 19*a^4*b^2
- 36*a^3*b^3 - 19*a^2*b^4 + 2*a*b^5 + 3*b^6)*sqrt(-a*b)) + e^(d*x + 10*c)/(a^2*e^(9*c) + 2*a*b*e^(9*c) + b^2*e
^(9*c)) + (a*e^(4*d*x + 4*c) - b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 4*b*e^(2*d*x + 2*c) + a + b)/((a^2*e^
c + 2*a*b*e^c + b^2*e^c)*(a*e^(5*d*x + 4*c) + b*e^(5*d*x + 4*c) + 2*a*e^(3*d*x + 2*c) - 2*b*e^(3*d*x + 2*c) +
a*e^(d*x) + b*e^(d*x))))/d

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maple [B]  time = 0.32, size = 167, normalized size = 1.82 \[ \frac {-\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b \left (\frac {-\frac {\left (a +2 b \right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {1}{2}}{\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a}-\frac {3 \arctanh \left (\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (a e^{\left (6 \, c\right )} + b e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \, {\left (a e^{\left (4 \, c\right )} - b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \, {\left (a e^{\left (2 \, c\right )} - b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + a + b}{2 \, {\left ({\left (a^{3} d e^{\left (5 \, c\right )} + 3 \, a^{2} b d e^{\left (5 \, c\right )} + 3 \, a b^{2} d e^{\left (5 \, c\right )} + b^{3} d e^{\left (5 \, c\right )}\right )} e^{\left (5 \, d x\right )} + 2 \, {\left (a^{3} d e^{\left (3 \, c\right )} + a^{2} b d e^{\left (3 \, c\right )} - a b^{2} d e^{\left (3 \, c\right )} - b^{3} d e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + {\left (a^{3} d e^{c} + 3 \, a^{2} b d e^{c} + 3 \, a b^{2} d e^{c} + b^{3} d e^{c}\right )} e^{\left (d x\right )}\right )}} + \frac {1}{2} \, \int \frac {6 \, {\left (b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + {\left (a^{3} e^{\left (4 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, c\right )} + b^{3} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \, {\left (a^{3} e^{\left (2 \, c\right )} + a^{2} b e^{\left (2 \, c\right )} - a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*e^(6*c) + b*e^(6*c))*e^(6*d*x) + 3*(a*e^(4*c) - b*e^(4*c))*e^(4*d*x) + 3*(a*e^(2*c) - b*e^(2*c))*e^(2*
d*x) + a + b)/((a^3*d*e^(5*c) + 3*a^2*b*d*e^(5*c) + 3*a*b^2*d*e^(5*c) + b^3*d*e^(5*c))*e^(5*d*x) + 2*(a^3*d*e^
(3*c) + a^2*b*d*e^(3*c) - a*b^2*d*e^(3*c) - b^3*d*e^(3*c))*e^(3*d*x) + (a^3*d*e^c + 3*a^2*b*d*e^c + 3*a*b^2*d*
e^c + b^3*d*e^c)*e^(d*x)) + 1/2*integrate(6*(b*e^(3*d*x + 3*c) - b*e^(d*x + c))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3
+ (a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c))*e^(4*d*x) + 2*(a^3*e^(2*c) + a^2*b*e^(2*c)
- a*b^2*e^(2*c) - b^3*e^(2*c))*e^(2*d*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (c+d\,x\right )}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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