3.5.43 \(\int \frac {\text {csch}(c+d x) \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [443]

Optimal. Leaf size=113 \[ -\frac {\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{3/2} d}+\frac {\text {sech}(c+d x)}{a d}-\frac {b \text {sech}(c+d x) (b+a \sinh (c+d x))}{a \left (a^2+b^2\right ) d} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2977, 2702, 327, 213, 2775, 12, 2739, 632, 210} \begin {gather*} -\frac {b \text {sech}(c+d x) (a \sinh (c+d x)+b)}{a d \left (a^2+b^2\right )}+\frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{3/2}}+\frac {\text {sech}(c+d x)}{a d}-\frac {\tanh ^{-1}(\cosh (c+d x))}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cosh[c + d*x]]/(a*d)) + (2*b^3*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a*(a^2 + b^2)^(3
/2)*d) + Sech[c + d*x]/(a*d) - (b*Sech[c + d*x]*(b + a*Sinh[c + d*x]))/(a*(a^2 + b^2)*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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

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

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 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2977

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 171, normalized size = 1.51 \begin {gather*} -\frac {-2 b^3 \text {ArcTan}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )+a^2 \sqrt {-a^2-b^2} \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+b^2 \sqrt {-a^2-b^2} \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sqrt {-a^2-b^2} \text {sech}(c+d x)-a b \sqrt {-a^2-b^2} \tanh (c+d x)}{a \left (-a^2-b^2\right )^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((-2*b^3*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]] + a^2*Sqrt[-a^2 - b^2]*Log[Tanh[(c + d*x)/2]] + b
^2*Sqrt[-a^2 - b^2]*Log[Tanh[(c + d*x)/2]] + a^2*Sqrt[-a^2 - b^2]*Sech[c + d*x] - a*b*Sqrt[-a^2 - b^2]*Tanh[c
+ d*x])/(a*(-a^2 - b^2)^(3/2)*d))

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Maple [A]
time = 1.64, size = 106, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b^{3} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(106\)
default	\(\frac {-\frac {2 \left (b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b^{3} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(106\)
risch	\(\frac {2 a \,{\mathrm e}^{d x +c}+2 b}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}+\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d a}-\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d a}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/(a^2+b^2)*(b*tanh(1/2*d*x+1/2*c)-a)/(tanh(1/2*d*x+1/2*c)^2+1)+1/a*ln(tanh(1/2*d*x+1/2*c))-2/a*b^3/(a^2
+b^2)^(3/2)*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 [A]
time = 0.48, size = 168, normalized size = 1.49 \begin {gather*} -\frac {b^{3} \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 )}{{\left (a^{3} + a b^{2}\right )} \sqrt {a^{2} + b^{2}} d} + \frac {2 \, {\left (a e^{\left (-d x - c\right )} - b\right )}}{{\left (a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (110) = 220\).
time = 0.50, size = 581, normalized size = 5.14 \begin {gather*} \frac {2 \, a^{3} b + 2 \, a b^{3} + {\left (b^{3} \cosh \left (d x + c\right )^{2} + 2 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} \sinh \left (d x + c\right )^{2} + b^{3}\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 (a^{4} + a^{2} b^{2}\right )} \cosh \left (d x + c\right ) - {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) + 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \sinh \left (d x + c\right )^{2} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*b + 2*a*b^3 + (b^3*cosh(d*x + c)^2 + 2*b^3*cosh(d*x + c)*sinh(d*x + c) + b^3*sinh(d*x + c)^2 + b^3)*sqr
t(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)) + 2*(a^4 + a^2*b^2)*cosh
(d*x + c) - (a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*cosh(
d*x + c)*sinh(d*x + c) + (a^4 + 2*a^2*b^2 + b^4)*sinh(d*x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + (a^
4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*cosh(d*x + c)*sinh(d
*x + c) + (a^4 + 2*a^2*b^2 + b^4)*sinh(d*x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(a^4 + a^2*b^2)*
sinh(d*x + c))/((a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c)*sinh
(d*x + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d*sinh(d*x + c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {csch}{\left (c + d x \right )} \operatorname {sech}^{2}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 146, normalized size = 1.29 \begin {gather*} -\frac {\frac {b^{3} \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 )}{{\left (a^{3} + a b^{2}\right )} \sqrt {a^{2} + b^{2}}} + \frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a} - \frac {2 \, {\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.82, size = 668, normalized size = 5.91 \begin {gather*} \frac {\frac {2\,b}{d\,\left (a^2+b^2\right )}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}-1\right )}{a\,d}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1\right )}{a\,d}-\frac {b^3\,\ln \left (\frac {32\,\left (-4\,{\mathrm {e}}^{c+d\,x}\,a^3+2\,a^2\,b-5\,{\mathrm {e}}^{c+d\,x}\,a\,b^2+2\,b^3\right )}{b^2\,{\left (a^2+b^2\right )}^2}-\frac {128\,a^{10}\,{\mathrm {e}}^{c+d\,x}-64\,a^9\,b-96\,a\,b^9+64\,b^7\,\sqrt {{\left (a^2+b^2\right )}^3}-384\,a^3\,b^7-512\,a^5\,b^5-288\,a^7\,b^3+288\,a^2\,b^8\,{\mathrm {e}}^{c+d\,x}+960\,a^4\,b^6\,{\mathrm {e}}^{c+d\,x}+1152\,a^6\,b^4\,{\mathrm {e}}^{c+d\,x}+608\,a^8\,b^2\,{\mathrm {e}}^{c+d\,x}-64\,a\,b^6\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+32\,a^3\,b^4\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}}{b^2\,{\left ({\left (a^2+b^2\right )}^3\right )}^{3/2}\,\left (a^2+b^2\right )}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^7+3\,d\,a^5\,b^2+3\,d\,a^3\,b^4+d\,a\,b^6}+\frac {b^3\,\ln \left (\frac {32\,\left (-4\,{\mathrm {e}}^{c+d\,x}\,a^3+2\,a^2\,b-5\,{\mathrm {e}}^{c+d\,x}\,a\,b^2+2\,b^3\right )}{b^2\,{\left (a^2+b^2\right )}^2}-\frac {96\,a\,b^9+64\,a^9\,b-128\,a^{10}\,{\mathrm {e}}^{c+d\,x}+64\,b^7\,\sqrt {{\left (a^2+b^2\right )}^3}+384\,a^3\,b^7+512\,a^5\,b^5+288\,a^7\,b^3-288\,a^2\,b^8\,{\mathrm {e}}^{c+d\,x}-960\,a^4\,b^6\,{\mathrm {e}}^{c+d\,x}-1152\,a^6\,b^4\,{\mathrm {e}}^{c+d\,x}-608\,a^8\,b^2\,{\mathrm {e}}^{c+d\,x}-64\,a\,b^6\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+32\,a^3\,b^4\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}}{b^2\,{\left ({\left (a^2+b^2\right )}^3\right )}^{3/2}\,\left (a^2+b^2\right )}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^7+3\,d\,a^5\,b^2+3\,d\,a^3\,b^4+d\,a\,b^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^2*sinh(c + d*x)*(a + b*sinh(c + d*x))),x)

[Out]

((2*b)/(d*(a^2 + b^2)) + (2*a*exp(c + d*x))/(d*(a^2 + b^2)))/(exp(2*c + 2*d*x) + 1) + log(exp(c + d*x) - 1)/(a
*d) - log(exp(c + d*x) + 1)/(a*d) - (b^3*log((32*(2*a^2*b - 4*a^3*exp(c + d*x) + 2*b^3 - 5*a*b^2*exp(c + d*x))
)/(b^2*(a^2 + b^2)^2) - (128*a^10*exp(c + d*x) - 64*a^9*b - 96*a*b^9 + 64*b^7*((a^2 + b^2)^3)^(1/2) - 384*a^3*
b^7 - 512*a^5*b^5 - 288*a^7*b^3 + 288*a^2*b^8*exp(c + d*x) + 960*a^4*b^6*exp(c + d*x) + 1152*a^6*b^4*exp(c + d
*x) + 608*a^8*b^2*exp(c + d*x) - 64*a*b^6*exp(c + d*x)*((a^2 + b^2)^3)^(1/2) + 32*a^3*b^4*exp(c + d*x)*((a^2 +
b^2)^3)^(1/2))/(b^2*((a^2 + b^2)^3)^(3/2)*(a^2 + b^2)))*((a^2 + b^2)^3)^(1/2))/(a^7*d + 3*a^3*b^4*d + 3*a^5*b
^2*d + a*b^6*d) + (b^3*log((32*(2*a^2*b - 4*a^3*exp(c + d*x) + 2*b^3 - 5*a*b^2*exp(c + d*x)))/(b^2*(a^2 + b^2)
^2) - (96*a*b^9 + 64*a^9*b - 128*a^10*exp(c + d*x) + 64*b^7*((a^2 + b^2)^3)^(1/2) + 384*a^3*b^7 + 512*a^5*b^5
+ 288*a^7*b^3 - 288*a^2*b^8*exp(c + d*x) - 960*a^4*b^6*exp(c + d*x) - 1152*a^6*b^4*exp(c + d*x) - 608*a^8*b^2*
exp(c + d*x) - 64*a*b^6*exp(c + d*x)*((a^2 + b^2)^3)^(1/2) + 32*a^3*b^4*exp(c + d*x)*((a^2 + b^2)^3)^(1/2))/(b
^2*((a^2 + b^2)^3)^(3/2)*(a^2 + b^2)))*((a^2 + b^2)^3)^(1/2))/(a^7*d + 3*a^3*b^4*d + 3*a^5*b^2*d + a*b^6*d)

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