3.29 \(\int \frac{\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ -\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b^3 d \sqrt{a-b}}+\frac{x \left (8 a^2+4 a b+3 b^2\right )}{8 b^3}-\frac{(4 a+3 b) \sinh (c+d x) \cosh (c+d x)}{8 b^2 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x)}{4 b d} \]

[Out]

((8*a^2 + 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a - b]*b^3*
d) - ((4*a + 3*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*b^2*d) + (Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*b*d)

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Rubi [A]  time = 0.235035, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3187, 470, 578, 522, 206, 208} \[ -\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b^3 d \sqrt{a-b}}+\frac{x \left (8 a^2+4 a b+3 b^2\right )}{8 b^3}-\frac{(4 a+3 b) \sinh (c+d x) \cosh (c+d x)}{8 b^2 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^6/(a + b*Sinh[c + d*x]^2),x]

[Out]

((8*a^2 + 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a - b]*b^3*
d) - ((4*a + 3*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*b^2*d) + (Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*b*d)

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 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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 ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3 \left (a-(a-b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x) \sinh ^3(c+d x)}{4 b d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(a+3 b) x^2\right )}{\left (1-x^2\right )^2 \left (a+(-a+b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{4 b d}\\ &=-\frac{(4 a+3 b) \cosh (c+d x) \sinh (c+d x)}{8 b^2 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x)}{4 b d}-\frac{\operatorname{Subst}\left (\int \frac{-a (4 a+3 b)+\left (-4 a^2-a b-3 b^2\right ) x^2}{\left (1-x^2\right ) \left (a+(-a+b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 b^2 d}\\ &=-\frac{(4 a+3 b) \cosh (c+d x) \sinh (c+d x)}{8 b^2 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x)}{4 b d}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\tanh (c+d x)\right )}{b^3 d}+\frac{\left (8 a^2+4 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 b^3 d}\\ &=\frac{\left (8 a^2+4 a b+3 b^2\right ) x}{8 b^3}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b} b^3 d}-\frac{(4 a+3 b) \cosh (c+d x) \sinh (c+d x)}{8 b^2 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x)}{4 b d}\\ \end{align*}

Mathematica [A]  time = 0.475127, size = 97, normalized size = 0.8 \[ \frac{4 \left (8 a^2+4 a b+3 b^2\right ) (c+d x)-\frac{32 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b}}-8 b (a+b) \sinh (2 (c+d x))+b^2 \sinh (4 (c+d x))}{32 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^6/(a + b*Sinh[c + d*x]^2),x]

[Out]

(4*(8*a^2 + 4*a*b + 3*b^2)*(c + d*x) - (32*a^(5/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/Sqrt[a - b] -
8*b*(a + b)*Sinh[2*(c + d*x)] + b^2*Sinh[4*(c + d*x)])/(32*b^3*d)

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Maple [B]  time = 0.08, size = 670, normalized size = 5.5 \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)^6/(a+b*sinh(d*x+c)^2),x)

[Out]

-1/4/d/b/(tanh(1/2*d*x+1/2*c)+1)^4+1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)^3+1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^2*a+1/8
/d/b/(tanh(1/2*d*x+1/2*c)+1)^2-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)*a-3/8/d/b/(tanh(1/2*d*x+1/2*c)+1)+1/d/b^3*ln(
tanh(1/2*d*x+1/2*c)+1)*a^2+1/2/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)+1)+3/8/d/b*ln(tanh(1/2*d*x+1/2*c)+1)+1/4/d/b/(ta
nh(1/2*d*x+1/2*c)-1)^4+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^3-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)^2*a-1/8/d/b/(tanh(1
/2*d*x+1/2*c)-1)^2-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)*a-3/8/d/b/(tanh(1/2*d*x+1/2*c)-1)-1/d/b^3*ln(tanh(1/2*d*x
+1/2*c)-1)*a^2-1/2/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)-1)-3/8/d/b*ln(tanh(1/2*d*x+1/2*c)-1)-1/d*a^3/b^3/((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^3/b^2/(-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^3/b^3/((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^3/b^2/(-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)^6/(a+b*sinh(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.05375, size = 4313, normalized size = 35.64 \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)^6/(a+b*sinh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/64*(b^2*cosh(d*x + c)^8 + 8*b^2*cosh(d*x + c)*sinh(d*x + c)^7 + b^2*sinh(d*x + c)^8 + 8*(8*a^2 + 4*a*b + 3*
b^2)*d*x*cosh(d*x + c)^4 - 8*(a*b + b^2)*cosh(d*x + c)^6 + 4*(7*b^2*cosh(d*x + c)^2 - 2*a*b - 2*b^2)*sinh(d*x
+ c)^6 + 8*(7*b^2*cosh(d*x + c)^3 - 6*(a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*b^2*cosh(d*x + c)^4 +
4*(8*a^2 + 4*a*b + 3*b^2)*d*x - 60*(a*b + b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*b^2*cosh(d*x + c)^5 +
4*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c) - 20*(a*b + b^2)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a*b + b^2)*
cosh(d*x + c)^2 + 4*(7*b^2*cosh(d*x + c)^6 + 12*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^2 - 30*(a*b + b^2)*c
osh(d*x + c)^4 + 2*a*b + 2*b^2)*sinh(d*x + c)^2 + 32*(a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)^3*sinh(d*x + c
) + 6*a^2*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4)*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*cos
h(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)*cos
h(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*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)) - b^2 +
8*(b^2*cosh(d*x + c)^7 + 4*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^3 - 6*(a*b + b^2)*cosh(d*x + c)^5 + 2*(a*
b + b^2)*cosh(d*x + c))*sinh(d*x + c))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^3*
d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*d*sinh(d*x + c)^4), 1/64*(b^2*
cosh(d*x + c)^8 + 8*b^2*cosh(d*x + c)*sinh(d*x + c)^7 + b^2*sinh(d*x + c)^8 + 8*(8*a^2 + 4*a*b + 3*b^2)*d*x*co
sh(d*x + c)^4 - 8*(a*b + b^2)*cosh(d*x + c)^6 + 4*(7*b^2*cosh(d*x + c)^2 - 2*a*b - 2*b^2)*sinh(d*x + c)^6 + 8*
(7*b^2*cosh(d*x + c)^3 - 6*(a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*b^2*cosh(d*x + c)^4 + 4*(8*a^2 +
4*a*b + 3*b^2)*d*x - 60*(a*b + b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*b^2*cosh(d*x + c)^5 + 4*(8*a^2 +
4*a*b + 3*b^2)*d*x*cosh(d*x + c) - 20*(a*b + b^2)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a*b + b^2)*cosh(d*x +
c)^2 + 4*(7*b^2*cosh(d*x + c)^6 + 12*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^2 - 30*(a*b + b^2)*cosh(d*x + c
)^4 + 2*a*b + 2*b^2)*sinh(d*x + c)^2 - 64*(a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^2*c
osh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4)*sqrt(-a/(a - b))*a
rctan(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) - b^2 + 8*(b^2*cosh(d*x + c)^7 + 4*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^3 - 6*(a*b + b^2)*cosh(d*x +
c)^5 + 2*(a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)^3*sinh(d*x +
c) + 6*b^3*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*d*sinh(d*x + c)^4)
]

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

[Out]

Timed out

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Giac [B]  time = 1.20858, size = 296, normalized size = 2.45 \begin{align*} -\frac{a^{3} \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} b^{3} d} + \frac{{\left (8 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )}{\left (d x + c\right )}}{8 \, b^{3} d} - \frac{{\left (48 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 18 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, b^{3} d} + \frac{b d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^6/(a+b*sinh(d*x+c)^2),x, algorithm="giac")

[Out]

-a^3*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*b^3*d) + 1/8*(8*a^2 + 4*a*b
+ 3*b^2)*(d*x + c)/(b^3*d) - 1/64*(48*a^2*e^(4*d*x + 4*c) + 24*a*b*e^(4*d*x + 4*c) + 18*b^2*e^(4*d*x + 4*c) -
8*a*b*e^(2*d*x + 2*c) - 8*b^2*e^(2*d*x + 2*c) + b^2)*e^(-4*d*x - 4*c)/(b^3*d) + 1/64*(b*d*e^(4*d*x + 4*c) - 8*
a*d*e^(2*d*x + 2*c) - 8*b*d*e^(2*d*x + 2*c))/(b^2*d^2)