3.4.56 \(\int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [356]

Optimal. Leaf size=78 \[ \frac {2 a b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {\text {sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2945, 12, 2739, 632, 210} \begin {gather*} \frac {2 a b \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {\text {sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(3/2)*d) - (Sech[c + d*x]*(a - b*Sinh[
c + d*x]))/((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 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 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 2945

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

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 104, normalized size = 1.33 \begin {gather*} -\frac {-2 a b \text {ArcTan}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-a \sqrt {-a^2-b^2} \text {sech}(c+d x)+b \sqrt {-a^2-b^2} \tanh (c+d x)}{\left (-a^2-b^2\right )^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.19, size = 101, normalized size = 1.29

method	result	size
derivativedivides	\(\frac {-\frac {4 a b \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(101\)
default	\(\frac {-\frac {4 a b \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(101\)
risch	\(-\frac {2 \left (a \,{\mathrm e}^{d x +c}+b \right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {b a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {b a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 117, normalized size = 1.50 \begin {gather*} -\frac {a b \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^{2} + b^{2}\right )}^{\frac {3}{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (75) = 150\).
time = 0.37, size = 350, normalized size = 4.49 \begin {gather*} -\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b \sinh \left (d x + c\right )^{2} + a b\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^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )^{2} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b + 2*b^3 - (a*b*cosh(d*x + c)^2 + 2*a*b*cosh(d*x + c)*sinh(d*x + c) + a*b*sinh(d*x + c)^2 + a*b)*sqrt
(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^3 + a*b^2)*cosh(d*
x + c) + 2*(a^3 + a*b^2)*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)
*d*cosh(d*x + c)*sinh(d*x + c) + (a^4 + 2*a^2*b^2 + b^4)*d*sinh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d)

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

[Out]

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

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Giac [A]
time = 0.46, size = 106, normalized size = 1.36 \begin {gather*} -\frac {\frac {a b \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^{2} + b^{2}\right )}^{\frac {3}{2}}} + \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(sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

-(a*b*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^2
+ b^2)^(3/2) + 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 = 0.60, size = 170, normalized size = 2.18 \begin {gather*} \frac {a\,b\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{a^2+b^2}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {a\,b\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{a^2+b^2}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b*log((2*a*exp(c + d*x))/(a^2 + b^2) + (2*a*(b - a*exp(c + d*x)))/(a^2 + b^2)^(3/2)))/(d*(a^2 + b^2)^(3/2))
- (a*b*log((2*a*exp(c + d*x))/(a^2 + b^2) - (2*a*(b - a*exp(c + d*x)))/(a^2 + b^2)^(3/2)))/(d*(a^2 + b^2)^(3/
2)) - ((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)

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