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

Optimal. Leaf size=84 \[ \frac {\sinh (c+d x) \cosh (c+d x)}{2 d (a-b) \left (a+b \sinh ^2(c+d x)\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} d (a-b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3173, 12, 3181, 208} \[ \frac {\sinh (c+d x) \cosh (c+d x)}{2 d (a-b) \left (a+b \sinh ^2(c+d x)\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} d (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 208

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

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 81, normalized size = 0.96 \[ \frac {\frac {\sinh (2 (c+d x))}{(a-b) (2 a+b \cosh (2 (c+d x))-b)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{3/2}}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.53, size = 1523, normalized size = 18.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*a^2*b - 4*a*b^2 + 4*(2*a^3 - 3*a^2*b + a*b^2)*cosh(d*x + c)^2 + 8*(2*a^3 - 3*a^2*b + a*b^2)*cosh(d*x
+ c)*sinh(d*x + c) + 4*(2*a^3 - 3*a^2*b + a*b^2)*sinh(d*x + c)^2 + (b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*
sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b
^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 - a*
b)*log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2)*cosh
(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b^2)*sinh(d*x + c)^2 + 8*a^2 - 8*a*b + b^2 + 4*(b^2*cosh(d*x
+ c)^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) +
b*sinh(d*x + c)^2 + 2*a - b)*sqrt(a^2 - a*b))/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh
(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x
+ c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)))/((a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^4 + 4*(a
^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*sinh(d*x + c)^4
+ 2*(2*a^4*b - 5*a^3*b^2 + 4*a^2*b^3 - a*b^4)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*
x + c)^2 + (2*a^4*b - 5*a^3*b^2 + 4*a^2*b^3 - a*b^4)*d)*sinh(d*x + c)^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*d + 4*
((a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^3 + (2*a^4*b - 5*a^3*b^2 + 4*a^2*b^3 - a*b^4)*d*cosh(d*x + c))*
sinh(d*x + c)), -1/2*(2*a^2*b - 2*a*b^2 + 2*(2*a^3 - 3*a^2*b + a*b^2)*cosh(d*x + c)^2 + 4*(2*a^3 - 3*a^2*b + a
*b^2)*cosh(d*x + c)*sinh(d*x + c) + 2*(2*a^3 - 3*a^2*b + a*b^2)*sinh(d*x + c)^2 - (b^2*cosh(d*x + c)^4 + 4*b^2
*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c
)^2 + 2*a*b - b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c)
)*sqrt(-a^2 + a*b)*arctan(-1/2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a
- b)*sqrt(-a^2 + a*b)/(a^2 - a*b)))/((a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b^2 - 2*a^2*b^3
+ a*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*sinh(d*x + c)^4 + 2*(2*a^4*b - 5*a^
3*b^2 + 4*a^2*b^3 - a*b^4)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 - 2*a^2*b^3 + a*b^4)*d*cosh(d*x + c)^2 + (2*a^4*b
- 5*a^3*b^2 + 4*a^2*b^3 - a*b^4)*d)*sinh(d*x + c)^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*d + 4*((a^3*b^2 - 2*a^2*b
^3 + a*b^4)*d*cosh(d*x + c)^3 + (2*a^4*b - 5*a^3*b^2 + 4*a^2*b^3 - a*b^4)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [A]  time = 2.12, size = 135, normalized size = 1.61 \[ -\frac {\frac {\arctan \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} {\left (a - b\right )}} + \frac {2 \, {\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}}{{\left (a b - b^{2}\right )} {\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*(a - b)) + 2*(2*a*e^(2*d*x
+ 2*c) - b*e^(2*d*x + 2*c) + b)/((a*b - b^2)*(b*e^(4*d*x + 4*c) + 4*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) +
b)))/d

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maple [B]  time = 0.07, size = 428, normalized size = 5.10 \[ \frac {\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right ) \left (a -b \right )}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right ) \left (a -b \right )}+\frac {\arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 d \left (a -b \right ) \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}+\frac {\arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right ) b}{2 d \left (a -b \right ) \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 d \left (a -b \right ) \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right ) b}{2 d \left (a -b \right ) \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/(a-b)*tanh(1/2*d*x+1/2*c)^
3+1/d/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/(a-b)*tanh(1/2*d*x+1/2*c
)+1/2/d/(a-b)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)
^(1/2))+1/2/d/(a-b)/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b
*(a-b))^(1/2)-a+2*b)*a)^(1/2))*b-1/2/d/(a-b)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c
)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/2/d/(a-b)/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arct
anh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))*b

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^2}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**2)**2,x)

[Out]

Timed out

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