3.313 \(\int \text{sech}^6(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=74 \[ \frac{\left (a^3-b^3\right ) \tanh (c+d x)}{d}+\frac{(a-b)^3 \tanh ^5(c+d x)}{5 d}-\frac{(a-b)^2 (2 a+b) \tanh ^3(c+d x)}{3 d}+b^3 x \]

[Out]

b^3*x + ((a^3 - b^3)*Tanh[c + d*x])/d - ((a - b)^2*(2*a + b)*Tanh[c + d*x]^3)/(3*d) + ((a - b)^3*Tanh[c + d*x]
^5)/(5*d)

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Rubi [A]  time = 0.0798852, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3191, 390, 206} \[ \frac{\left (a^3-b^3\right ) \tanh (c+d x)}{d}+\frac{(a-b)^3 \tanh ^5(c+d x)}{5 d}-\frac{(a-b)^2 (2 a+b) \tanh ^3(c+d x)}{3 d}+b^3 x \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^3,x]

[Out]

b^3*x + ((a^3 - b^3)*Tanh[c + d*x])/d - ((a - b)^2*(2*a + b)*Tanh[c + d*x]^3)/(3*d) + ((a - b)^3*Tanh[c + d*x]
^5)/(5*d)

Rule 3191

Int[cos[(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/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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])

Rubi steps

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

Mathematica [A]  time = 0.731567, size = 86, normalized size = 1.16 \[ \frac{(a-b) \tanh (c+d x) \left (\left (4 a^2+7 a b-11 b^2\right ) \text{sech}^2(c+d x)+8 a^2+3 (a-b)^2 \text{sech}^4(c+d x)+14 a b+23 b^2\right )+15 b^3 (c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^3,x]

[Out]

(15*b^3*(c + d*x) + (a - b)*(8*a^2 + 14*a*b + 23*b^2 + (4*a^2 + 7*a*b - 11*b^2)*Sech[c + d*x]^2 + 3*(a - b)^2*
Sech[c + d*x]^4)*Tanh[c + d*x])/(15*d)

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Maple [B]  time = 0.058, size = 199, normalized size = 2.7 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) \tanh \left ( dx+c \right ) +3\,{a}^{2}b \left ( -1/4\,{\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+1/4\, \left ({\frac{8}{15}}+1/5\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) \tanh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( -1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-3/8\,{\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+3/8\, \left ({\frac{8}{15}}+1/5\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) \tanh \left ( dx+c \right ) \right ) +{b}^{3} \left ( dx+c-\tanh \left ( dx+c \right ) -{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3}}-{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{5}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)+3*a^2*b*(-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(
8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c))+3*a*b^2*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x
+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c))+b^3*(d*x+c-tanh(d*x+c)-1/3*tanh
(d*x+c)^3-1/5*tanh(d*x+c)^5))

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Maxima [B]  time = 1.24156, size = 1112, normalized size = 15.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/15*b^3*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) + 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) + 45*e^(-8*d*x -
8*c) + 23)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d
*x - 10*c) + 1))) + 16/15*a^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x
- 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4
*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) +
10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 4/5*a^2*b*(5*e^(
-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10
*d*x - 10*c) + 1)) - 5*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5
*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 15*e^(-6*d*x - 6*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c
) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*
d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 6/5*a*b^2*(10*e^(-4*d*x -
4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10
*c) + 1)) + 5*e^(-8*d*x - 8*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*
x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5
*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)))

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Fricas [B]  time = 1.56797, size = 1287, normalized size = 17.39 \begin{align*} \frac{{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} +{\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \,{\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3} + 2 \,{\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \,{\left ({\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 16 \, a^{3} - 24 \, a^{2} b + 18 \, a b^{2} - 10 \, b^{3} + 3 \,{\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/15*((15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c)^5 + 5*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*
a*b^2 + 23*b^3)*cosh(d*x + c)*sinh(d*x + c)^4 + (8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^3)*sinh(d*x + c)^5 + 5*(15*b
^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c)^3 + 5*(8*a^3 + 6*a^2*b - 9*a*b^2 - 5*b^3 + 2*(8*a^3
+ 6*a^2*b + 9*a*b^2 - 23*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 5*(2*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2
+ 23*b^3)*cosh(d*x + c)^3 + 3*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c))*sinh(d*x + c)^
2 + 10*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c) + 5*((8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^
3)*cosh(d*x + c)^4 + 16*a^3 - 24*a^2*b + 18*a*b^2 - 10*b^3 + 3*(8*a^3 + 6*a^2*b - 9*a*b^2 - 5*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cos
h(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(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(sech(d*x+c)**6*(a+b*sinh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.38027, size = 288, normalized size = 3.89 \begin{align*} \frac{{\left (d x + c\right )} b^{3}}{d} - \frac{2 \,{\left (45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 30 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

(d*x + c)*b^3/d - 2/15*(45*a*b^2*e^(8*d*x + 8*c) - 45*b^3*e^(8*d*x + 8*c) + 90*a^2*b*e^(6*d*x + 6*c) - 90*b^3*
e^(6*d*x + 6*c) + 80*a^3*e^(4*d*x + 4*c) - 30*a^2*b*e^(4*d*x + 4*c) + 90*a*b^2*e^(4*d*x + 4*c) - 140*b^3*e^(4*
d*x + 4*c) + 40*a^3*e^(2*d*x + 2*c) + 30*a^2*b*e^(2*d*x + 2*c) - 70*b^3*e^(2*d*x + 2*c) + 8*a^3 + 6*a^2*b + 9*
a*b^2 - 23*b^3)/(d*(e^(2*d*x + 2*c) + 1)^5)