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(d*x+c)/d-1/3*(a-b)^2*(2*a+b)*tanh(d*x+c)^3/d+1/5*(a-b)^3*tanh(d*x+c)^5/d

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Rubi [A]  time = 0.08, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.77, 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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fricas [B]  time = 0.58, size = 530, normalized size = 7.16 \[ \frac {{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3} + 2 \, {\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \, {\left ({\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 16 \, a^{3} - 24 \, a^{2} b + 18 \, a b^{2} - 10 \, b^{3} + 3 \, {\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.25, size = 213, normalized size = 2.88 \[ \frac {15 \, {\left (d x + c\right )} b^{3} - \frac {2 \, {\left (45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 30 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]

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]

1/15*(15*(d*x + c)*b^3 - 2*(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)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [B]  time = 0.12, size = 199, normalized size = 2.69 \[ \frac {a^{3} \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )+3 a \,b^{2} \left (-\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )+b^{3} \left (d x +c -\tanh \left (d x +c \right )-\frac {\left (\tanh ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\tanh ^{5}\left (d x +c \right )\right )}{5}\right )}{d} \]

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 = 0.39, size = 824, normalized size = 11.14 \[ \text {result too large to display} \]

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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mupad [B]  time = 0.84, size = 563, normalized size = 7.61 \[ b^3\,x-\frac {\frac {2\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{15\,d}-\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {6\,\left (a\,b^2-b^3\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {6\,\left (a\,b^2-b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*x - ((2*(9*a*b^2 - 12*a^2*b + 8*a^3 - 5*b^3))/(15*d) - (12*exp(2*c + 2*d*x)*(a*b^2 - a^2*b))/(5*d) + (6*ex
p(4*c + 4*d*x)*(a*b^2 - b^3))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) + ((6*(a
*b^2 - a^2*b))/(5*d) - (6*exp(2*c + 2*d*x)*(a*b^2 - b^3))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) +
((6*(a*b^2 - a^2*b))/(5*d) + (18*exp(4*c + 4*d*x)*(a*b^2 - a^2*b))/(5*d) - (2*exp(2*c + 2*d*x)*(9*a*b^2 - 12*
a^2*b + 8*a^3 - 5*b^3))/(5*d) - (6*exp(6*c + 6*d*x)*(a*b^2 - b^3))/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*
d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((6*(a*b^2 - b^3))/(5*d) - (24*exp(2*c + 2*d*x)*(a*b^2 - a
^2*b))/(5*d) - (24*exp(6*c + 6*d*x)*(a*b^2 - a^2*b))/(5*d) + (4*exp(4*c + 4*d*x)*(9*a*b^2 - 12*a^2*b + 8*a^3 -
5*b^3))/(5*d) + (6*exp(8*c + 8*d*x)*(a*b^2 - b^3))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(
6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - (6*(a*b^2 - b^3))/(5*d*(exp(2*c + 2*d*x) + 1))

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

[Out]

Timed out

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