3.4.12 \(\int \frac {\text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [312]

Optimal. Leaf size=77 \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}+\frac {\text {sech}(c+d x) (b+a \sinh (c+d x))}{\left (a^2+b^2\right ) d} \]

[Out]

-2*b^2*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)*(b+a*sinh(d*x+c))/(a^2
+b^2)/d

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Rubi [A]
time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2775, 12, 2739, 632, 210} \begin {gather*} \frac {\text {sech}(c+d x) (a \sinh (c+d x)+b)}{d \left (a^2+b^2\right )}-\frac {2 b^2 \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 104, normalized size = 1.35 \begin {gather*} -\frac {2 b^2 \text {ArcTan}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )+b \sqrt {-a^2-b^2} \text {sech}(c+d x)+a \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]^2/(a + b*Sinh[c + d*x]),x]

[Out]

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

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Maple [A]
time = 1.25, size = 90, normalized size = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(90\)
default	\(\frac {-\frac {2 \left (-a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(90\)
risch	\(-\frac {2 \left (-b \,{\mathrm e}^{d x +c}+a \right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {b^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/(a^2+b^2)*(-a*tanh(1/2*d*x+1/2*c)-b)/(tanh(1/2*d*x+1/2*c)^2+1)+2*b^2/(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 = 0.47, size = 115, normalized size = 1.49 \begin {gather*} \frac {b^{2} \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 (b e^{\left (-d x - c\right )} + a\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)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

b^2*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*(b*e^(-d*x - c) + a)/((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. 353 vs. \(2 (74) = 148\).
time = 0.34, size = 353, normalized size = 4.58 \begin {gather*} -\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\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^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) - 2 \, {\left (a^{2} b + b^{3}\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)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(2*a^3 + 2*a*b^2 - (b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*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^2*b + b^3)*cosh(d*
x + c) - 2*(a^2*b + b^3)*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 {\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(sech(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

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

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Giac [A]
time = 0.60, size = 108, normalized size = 1.40 \begin {gather*} \frac {\frac {b^{2} \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 (b e^{\left (d x + c\right )} - a\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)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.38, size = 413, normalized size = 5.36 \begin {gather*} -\frac {\frac {2\,a}{d\,\left (a^2+b^2\right )}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {2\,\mathrm {atan}\left (\left (\frac {b^3\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}{2}+\frac {a^2\,b\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}{2}\right )\,\left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (\frac {2}{d\,\sqrt {b^4}\,{\left (a^2+b^2\right )}^2}+\frac {2\,a\,\left (a^3\,d\,\sqrt {b^4}+a\,b^2\,d\,\sqrt {b^4}\right )}{b^4\,\sqrt {-d^2\,{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}\right )-\frac {2\,a\,\left (b^3\,d\,\sqrt {b^4}+a^2\,b\,d\,\sqrt {b^4}\right )}{b^4\,\sqrt {-d^2\,{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}\right )\right )\,\sqrt {b^4}}{\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*a)/(d*(a^2 + b^2)) - (2*b*exp(c + d*x))/(d*(a^2 + b^2)))/(exp(2*c + 2*d*x) + 1) - (2*atan(((b^3*(- a^6*d
^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b^2*d^2)^(1/2))/2 + (a^2*b*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b
^2*d^2)^(1/2))/2)*(exp(d*x)*exp(c)*(2/(d*(b^4)^(1/2)*(a^2 + b^2)^2) + (2*a*(a^3*d*(b^4)^(1/2) + a*b^2*d*(b^4)^
(1/2)))/(b^4*(-d^2*(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b^2*d^2)^(1/2
))) - (2*a*(b^3*d*(b^4)^(1/2) + a^2*b*d*(b^4)^(1/2)))/(b^4*(-d^2*(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6*d^2 -
b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b^2*d^2)^(1/2))))*(b^4)^(1/2))/(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b
^2*d^2)^(1/2)

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