3.42 \(\int \frac{\sinh ^4(c+d x)}{(a+b \sinh ^2(c+d x))^2} \, dx\)
Optimal. Leaf size=102 \[ -\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 b^2 d (a-b)^{3/2}}-\frac{a \tanh (c+d x)}{2 b d (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{x}{b^2} \]
[Out]
x/b^2 - (Sqrt[a]*(2*a - 3*b)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(2*(a - b)^(3/2)*b^2*d) - (a*Tanh[c
+ d*x])/(2*(a - b)*b*d*(a - (a - b)*Tanh[c + d*x]^2))
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Rubi [A] time = 0.169097, antiderivative size = 102, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{3187, 470, 522, 206, 208} \[ -\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 b^2 d (a-b)^{3/2}}-\frac{a \tanh (c+d x)}{2 b d (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{x}{b^2} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^2)^2,x]
[Out]
x/b^2 - (Sqrt[a]*(2*a - 3*b)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(2*(a - b)^(3/2)*b^2*d) - (a*Tanh[c
+ d*x])/(2*(a - b)*b*d*(a - (a - b)*Tanh[c + d*x]^2))
Rule 3187
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Rule 470
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
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]
Rubi steps
\begin{align*} \int \frac{\sinh ^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right ) \left (a-(a-b) x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{a \tanh (c+d x)}{2 (a-b) b d \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+(a-2 b) x^2}{\left (1-x^2\right ) \left (a+(-a+b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 (a-b) b d}\\ &=-\frac{a \tanh (c+d x)}{2 (a-b) b d \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{b^2 d}-\frac{(a (2 a-3 b)) \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\tanh (c+d x)\right )}{2 (a-b) b^2 d}\\ &=\frac{x}{b^2}-\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 (a-b)^{3/2} b^2 d}-\frac{a \tanh (c+d x)}{2 (a-b) b d \left (a-(a-b) \tanh ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.832878, size = 99, normalized size = 0.97 \[ -\frac{\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{(a-b)^{3/2}}+\frac{a b \sinh (2 (c+d x))}{(a-b) (2 a+b \cosh (2 (c+d x))-b)}-2 (c+d x)}{2 b^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^2)^2,x]
[Out]
-(-2*(c + d*x) + (Sqrt[a]*(2*a - 3*b)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(a - b)^(3/2) + (a*b*Sinh[
2*(c + d*x)])/((a - b)*(2*a - b + b*Cosh[2*(c + d*x)])))/(2*b^2*d)
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Maple [B] time = 0.051, size = 798, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2)^2,x)
[Out]
1/d/b^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/b^2*ln(tanh(1/2*d*x+1/2*c)-1)-1/d*a/b/(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*a/b/(tanh(1/2*d*x+1/2*c)^4*a-2*t
anh(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/d*a^2/b^2/(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/d*a^2/b/(a-b)/(-b
*(a-b))^(1/2)/((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/d*a^2/b^2/(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/d*a^2/b/(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))+3/2/d*a/b/(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))-3/2/d*a/(a-b)/(-b*(a-b))^(1/2)/((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))-3/2/d*a/b/(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))-3/2/d*a/(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))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.55959, size = 4199, normalized size = 41.17 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
[1/4*(4*(a*b - b^2)*d*x*cosh(d*x + c)^4 + 16*(a*b - b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a*b - b^2)*d*x
*sinh(d*x + c)^4 + 4*(a*b - b^2)*d*x + 4*(2*(2*a^2 - 3*a*b + b^2)*d*x + 2*a^2 - a*b)*cosh(d*x + c)^2 + 4*(6*(a
*b - b^2)*d*x*cosh(d*x + c)^2 + 2*(2*a^2 - 3*a*b + b^2)*d*x + 2*a^2 - a*b)*sinh(d*x + c)^2 + ((2*a*b - 3*b^2)*
cosh(d*x + c)^4 + 4*(2*a*b - 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a*b - 3*b^2)*sinh(d*x + c)^4 + 2*(4*a^2
- 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 2*(3*(2*a*b - 3*b^2)*cosh(d*x + c)^2 + 4*a^2 - 8*a*b + 3*b^2)*sinh(d*x + c
)^2 + 2*a*b - 3*b^2 + 4*((2*a*b - 3*b^2)*cosh(d*x + c)^3 + (4*a^2 - 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c
))*sqrt(a/(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*((a*b - b^2)*cosh(d*x + c)^2 + 2*(a*b
- b^2)*cosh(d*x + c)*sinh(d*x + c) + (a*b - b^2)*sinh(d*x + c)^2 + 2*a^2 - 3*a*b + b^2)*sqrt(a/(a - b)))/(b*co
sh(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*c
osh(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)
) + 4*a*b + 8*(2*(a*b - b^2)*d*x*cosh(d*x + c)^3 + (2*(2*a^2 - 3*a*b + b^2)*d*x + 2*a^2 - a*b)*cosh(d*x + c))*
sinh(d*x + c))/((a*b^3 - b^4)*d*cosh(d*x + c)^4 + 4*(a*b^3 - b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b^3 - b
^4)*d*sinh(d*x + c)^4 + 2*(2*a^2*b^2 - 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + 2*(3*(a*b^3 - b^4)*d*cosh(d*x + c)^2
+ (2*a^2*b^2 - 3*a*b^3 + b^4)*d)*sinh(d*x + c)^2 + (a*b^3 - b^4)*d + 4*((a*b^3 - b^4)*d*cosh(d*x + c)^3 + (2*
a^2*b^2 - 3*a*b^3 + b^4)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a*b - b^2)*d*x*cosh(d*x + c)^4 + 8*(a*b - b^
2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a*b - b^2)*d*x*sinh(d*x + c)^4 + 2*(a*b - b^2)*d*x + 2*(2*(2*a^2 - 3
*a*b + b^2)*d*x + 2*a^2 - a*b)*cosh(d*x + c)^2 + 2*(6*(a*b - b^2)*d*x*cosh(d*x + c)^2 + 2*(2*a^2 - 3*a*b + b^2
)*d*x + 2*a^2 - a*b)*sinh(d*x + c)^2 - ((2*a*b - 3*b^2)*cosh(d*x + c)^4 + 4*(2*a*b - 3*b^2)*cosh(d*x + c)*sinh
(d*x + c)^3 + (2*a*b - 3*b^2)*sinh(d*x + c)^4 + 2*(4*a^2 - 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 2*(3*(2*a*b - 3*b^
2)*cosh(d*x + c)^2 + 4*a^2 - 8*a*b + 3*b^2)*sinh(d*x + c)^2 + 2*a*b - 3*b^2 + 4*((2*a*b - 3*b^2)*cosh(d*x + c)
^3 + (4*a^2 - 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-a/(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/(a - b))/a) + 2*a*b + 4*(2*(a*b - b^2)*d*
x*cosh(d*x + c)^3 + (2*(2*a^2 - 3*a*b + b^2)*d*x + 2*a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/((a*b^3 - b^4)*d
*cosh(d*x + c)^4 + 4*(a*b^3 - b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b^3 - b^4)*d*sinh(d*x + c)^4 + 2*(2*a^
2*b^2 - 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + 2*(3*(a*b^3 - b^4)*d*cosh(d*x + c)^2 + (2*a^2*b^2 - 3*a*b^3 + b^4)*
d)*sinh(d*x + c)^2 + (a*b^3 - b^4)*d + 4*((a*b^3 - b^4)*d*cosh(d*x + c)^3 + (2*a^2*b^2 - 3*a*b^3 + b^4)*d*cosh
(d*x + c))*sinh(d*x + c))]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)**4/(a+b*sinh(d*x+c)**2)**2,x)
[Out]
Timed out
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Giac [A] time = 1.28282, size = 228, normalized size = 2.24 \begin{align*} -\frac{{\left (2 \, a^{2} - 3 \, a b\right )} \arctan \left (\frac{b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt{-a^{2} + a b}}\right )}{2 \,{\left (a b^{2} d - b^{3} d\right )} \sqrt{-a^{2} + a b}} + \frac{2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (2 \, d x + 2 \, c\right )} + a b}{{\left (a b^{2} d - b^{3} d\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 )}} + \frac{d x + c}{b^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")
[Out]
-1/2*(2*a^2 - 3*a*b)*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/((a*b^2*d - b^3*d)*sqrt(-a^2 +
a*b)) + (2*a^2*e^(2*d*x + 2*c) - a*b*e^(2*d*x + 2*c) + a*b)/((a*b^2*d - b^3*d)*(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*x + c)/(b^2*d)