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

Optimal. Leaf size=71 \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a x}{b^2}+\frac {\cosh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2746, 12, 2735, 2660, 618, 204} \[ -\frac {2 a^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^2 d \sqrt {a^2+b^2}}-\frac {a x}{b^2}+\frac {\cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x)/b^2) - (2*a^2*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^2*Sqrt[a^2 + b^2]*d) + Cosh[c + d
*x]/(b*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 204

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

Rule 618

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 2660

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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2746

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 74, normalized size = 1.04 \[ \frac {b \cosh (c+d x)-a \left (-\frac {2 a \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+c+d x\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.76, size = 331, normalized size = 4.66 \[ -\frac {2 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right ) - a^{2} b - b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a^{2} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \, {\left ({\left (a^{3} + a b^{2}\right )} d x - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a^3 + a*b^2)*d*x*cosh(d*x + c) - a^2*b - b^3 - (a^2*b + b^3)*cosh(d*x + c)^2 - (a^2*b + b^3)*sinh(d*x
+ c)^2 - 2*(a^2*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*co
sh(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)*d*x - (a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))/((a^2*b^2
+ b^4)*d*cosh(d*x + c) + (a^2*b^2 + b^4)*d*sinh(d*x + c))

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giac [A]  time = 0.29, size = 111, normalized size = 1.56 \[ \frac {\frac {2 \, a^{2} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {2 \, {\left (d x + c\right )} a}{b^{2}} + \frac {e^{\left (d x + c\right )}}{b} + \frac {e^{\left (-d x - c\right )}}{b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 132, normalized size = 1.86 \[ -\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{2}}+\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{2}}+\frac {2 a^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 119, normalized size = 1.68 \[ \frac {a^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2} d} - \frac {{\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{2 \, b d} + \frac {e^{\left (-d x - c\right )}}{2 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.37, size = 166, normalized size = 2.34 \[ \frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}+\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}-\frac {a\,x}{b^2}-\frac {a^2\,\ln \left (-\frac {2\,a^2\,{\mathrm {e}}^{c+d\,x}}{b^3}-\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,d\,\sqrt {a^2+b^2}}+\frac {a^2\,\ln \left (\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^3\,\sqrt {a^2+b^2}}-\frac {2\,a^2\,{\mathrm {e}}^{c+d\,x}}{b^3}\right )}{b^2\,d\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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