3.2.2 \(\int \frac {1}{(a+b \sinh (c+d x))^2} \, dx\) [102]

Optimal. Leaf size=79 \[ -\frac {2 a \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {b \cosh (c+d x)}{\left (a^2+b^2\right ) d (a+b \sinh (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2743, 12, 2739, 632, 210} \begin {gather*} -\frac {2 a \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 {b \cosh (c+d x)}{d \left (a^2+b^2\right ) (a+b \sinh (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^(-2),x]

[Out]

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

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 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 85, normalized size = 1.08 \begin {gather*} -\frac {\frac {2 a \text {ArcTan}\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}}+\frac {b \cosh (c+d x)}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.13, size = 118, normalized size = 1.49

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}+b^{2}\right )}-\frac {b}{a^{2}+b^{2}}\right )}{a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(118\)
default	\(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}+b^{2}\right )}-\frac {b}{a^{2}+b^{2}}\right )}{a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(118\)
risch	\(\frac {2 a \,{\mathrm e}^{d x +c}-2 b}{d \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}+\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2*(-b^2/a/(a^2+b^2)*tanh(1/2*d*x+1/2*c)-b/(a^2+b^2))/(a*tanh(1/2*d*x+1/2*c)^2-2*b*tanh(1/2*d*x+1/2*c)-a)
+2*a/(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 = 0.50, size = 138, normalized size = 1.75 \begin {gather*} \frac {a \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 + b^{3} + 2 \, {\left (a^{3} + a b^{2}\right )} e^{\left (-d x - c\right )} - {\left (a^{2} b + b^{3}\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(1/(a+b*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

a*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 + b^3 + 2*(a^3 + a*b^2)*e^(-d*x - c) - (a^2*b + b^3)*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. 423 vs. \(2 (76) = 152\).
time = 0.37, size = 423, normalized size = 5.35 \begin {gather*} -\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) - a b + 2 \, {\left (a b \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right )\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} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \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 ) - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d + 2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

-(2*a^2*b + 2*b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) - a*b + 2*(a*b*cosh(d*x +
c) + a^2)*sinh(d*x + c))*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*b + 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) - (a^4*b + 2*
a^2*b^3 + b^5)*d + 2*((a^4*b + 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sinh {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))**2,x)

[Out]

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

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Giac [A]
time = 0.43, size = 119, normalized size = 1.51 \begin {gather*} \frac {\frac {a \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 (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.88, size = 200, normalized size = 2.53 \begin {gather*} \frac {a\,\ln \left (\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2+b^2\right )}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {a\,\ln \left (-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2+b^2\right )}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,b^2}{d\,\left (a^2\,b+b^3\right )}-\frac {2\,a\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sinh(c + d*x))^2,x)

[Out]

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

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