3.66 \(\int \cosh ^3(c+d x) (a+b \text{sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=81 \[ \frac{a^2 (a+3 b) \sinh (c+d x)}{d}+\frac{a^3 \sinh ^3(c+d x)}{3 d}+\frac{b^2 (6 a+b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^3 \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

[Out]

(b^2*(6*a + b)*ArcTan[Sinh[c + d*x]])/(2*d) + (a^2*(a + 3*b)*Sinh[c + d*x])/d + (a^3*Sinh[c + d*x]^3)/(3*d) +
(b^3*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

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Rubi [A]  time = 0.0931799, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4147, 390, 385, 203} \[ \frac{a^2 (a+3 b) \sinh (c+d x)}{d}+\frac{a^3 \sinh ^3(c+d x)}{3 d}+\frac{b^2 (6 a+b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^3 \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(6*a + b)*ArcTan[Sinh[c + d*x]])/(2*d) + (a^2*(a + 3*b)*Sinh[c + d*x])/d + (a^3*Sinh[c + d*x]^3)/(3*d) +
(b^3*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 6.83592, size = 483, normalized size = 5.96 \[ \frac{\coth ^3(c+d x) \text{csch}^2(c+d x) (a \cosh (c+d x)+b \text{sech}(c+d x))^3 \left (-256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{11}{2}\right \},-\sinh ^2(c+d x)\right )+21 \left (3 a^2 b \left (753 \sinh ^8(c+d x)+19786 \sinh ^6(c+d x)+69728 \sinh ^4(c+d x)+88150 \sinh ^2(c+d x)+36015\right )+a^3 \left (753 \sinh ^{10}(c+d x)+19579 \sinh ^8(c+d x)+89514 \sinh ^6(c+d x)+157878 \sinh ^4(c+d x)+124165 \sinh ^2(c+d x)+36015\right )+3 a b^2 \left (753 \sinh ^6(c+d x)+17593 \sinh ^4(c+d x)+52135 \sinh ^2(c+d x)+36015\right )+b^3 \left (1473 \sinh ^4(c+d x)+16120 \sinh ^2(c+d x)+36015\right )\right )-\frac{315 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right ) \left (3 a^2 b \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right ) \cosh ^4(c+d x)+a^3 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right ) \cosh ^6(c+d x)+3 a b^2 \left (\sinh ^8(c+d x)+148 \sinh ^6(c+d x)+2118 \sinh ^4(c+d x)+4276 \sinh ^2(c+d x)+2401\right )+b^3 \left (-47 \sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right )\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{3780 d (a \cosh (2 c+2 d x)+a+2 b)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Coth[c + d*x]^3*Csch[c + d*x]^2*(a*Cosh[c + d*x] + b*Sech[c + d*x])^3*(-256*HypergeometricPFQ[{3/2, 2, 2, 2,
2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8*(a + b + a*Sinh[c + d*x]^2)^3 - (315*ArcTanh[Sqrt[-Sinh
[c + d*x]^2]]*(b^3*(2401 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 - 47*Sinh[c + d*x]^6) + 3*a^2*b*Cosh[c +
d*x]^4*(2401 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + a^3*Cosh[c + d*x]^6*(2401 + 18
75*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a*b^2*(2401 + 4276*Sinh[c + d*x]^2 + 2118*Sinh
[c + d*x]^4 + 148*Sinh[c + d*x]^6 + Sinh[c + d*x]^8)))/Sqrt[-Sinh[c + d*x]^2] + 21*(b^3*(36015 + 16120*Sinh[c
+ d*x]^2 + 1473*Sinh[c + d*x]^4) + 3*a*b^2*(36015 + 52135*Sinh[c + d*x]^2 + 17593*Sinh[c + d*x]^4 + 753*Sinh[c
+ d*x]^6) + 3*a^2*b*(36015 + 88150*Sinh[c + d*x]^2 + 69728*Sinh[c + d*x]^4 + 19786*Sinh[c + d*x]^6 + 753*Sinh
[c + d*x]^8) + a^3*(36015 + 124165*Sinh[c + d*x]^2 + 157878*Sinh[c + d*x]^4 + 89514*Sinh[c + d*x]^6 + 19579*Si
nh[c + d*x]^8 + 753*Sinh[c + d*x]^10))))/(3780*d*(a + 2*b + a*Cosh[2*c + 2*d*x])^3)

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Maple [A]  time = 0.045, size = 103, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{3}\sinh \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+3\,{\frac{{a}^{2}b\sinh \left ( dx+c \right ) }{d}}+6\,{\frac{a{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{3}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*a^3*sinh(d*x+c)+1/3/d*a^3*sinh(d*x+c)*cosh(d*x+c)^2+3*a^2*b*sinh(d*x+c)/d+6/d*a*b^2*arctan(exp(d*x+c))+1
/2*b^3*sech(d*x+c)*tanh(d*x+c)/d+1/d*b^3*arctan(exp(d*x+c))

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Maxima [B]  time = 1.7009, size = 242, normalized size = 2.99 \begin{align*} -b^{3}{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{1}{24} \, a^{3}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} - \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac{3}{2} \, a^{2} b{\left (\frac{e^{\left (d x + c\right )}}{d} - \frac{e^{\left (-d x - c\right )}}{d}\right )} - \frac{6 \, a b^{2} \arctan \left (e^{\left (-d x - c\right )}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3*(arctan(e^(-d*x - c))/d - (e^(-d*x - c) - e^(-3*d*x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1
))) + 1/24*a^3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3*c)/d) + 3/2*a^2*b*(e^(d
*x + c)/d - e^(-d*x - c)/d) - 6*a*b^2*arctan(e^(-d*x - c))/d

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Fricas [B]  time = 2.28865, size = 3555, normalized size = 43.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*cosh(d*x + c)^10 + 10*a^3*cosh(d*x + c)*sinh(d*x + c)^9 + a^3*sinh(d*x + c)^10 + (11*a^3 + 36*a^2*b)
*cosh(d*x + c)^8 + (45*a^3*cosh(d*x + c)^2 + 11*a^3 + 36*a^2*b)*sinh(d*x + c)^8 + 8*(15*a^3*cosh(d*x + c)^3 +
(11*a^3 + 36*a^2*b)*cosh(d*x + c))*sinh(d*x + c)^7 + 2*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c)^6 + 2*(105*a^
3*cosh(d*x + c)^4 + 5*a^3 + 18*a^2*b + 12*b^3 + 14*(11*a^3 + 36*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 4*(6
3*a^3*cosh(d*x + c)^5 + 14*(11*a^3 + 36*a^2*b)*cosh(d*x + c)^3 + 3*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c))*
sinh(d*x + c)^5 - 2*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c)^4 + 2*(105*a^3*cosh(d*x + c)^6 + 35*(11*a^3 + 36
*a^2*b)*cosh(d*x + c)^4 - 5*a^3 - 18*a^2*b - 12*b^3 + 15*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c)^2)*sinh(d*x
+ c)^4 + 8*(15*a^3*cosh(d*x + c)^7 + 7*(11*a^3 + 36*a^2*b)*cosh(d*x + c)^5 + 5*(5*a^3 + 18*a^2*b + 12*b^3)*co
sh(d*x + c)^3 - (5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - a^3 - (11*a^3 + 36*a^2*b)*cosh(d*
x + c)^2 + (45*a^3*cosh(d*x + c)^8 + 28*(11*a^3 + 36*a^2*b)*cosh(d*x + c)^6 + 30*(5*a^3 + 18*a^2*b + 12*b^3)*c
osh(d*x + c)^4 - 11*a^3 - 36*a^2*b - 12*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 24*((6*
a*b^2 + b^3)*cosh(d*x + c)^7 + 7*(6*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^6 + (6*a*b^2 + b^3)*sinh(d*x + c)
^7 + 2*(6*a*b^2 + b^3)*cosh(d*x + c)^5 + (12*a*b^2 + 2*b^3 + 21*(6*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)
^5 + 5*(7*(6*a*b^2 + b^3)*cosh(d*x + c)^3 + 2*(6*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + (6*a*b^2 + b^3)
*cosh(d*x + c)^3 + (35*(6*a*b^2 + b^3)*cosh(d*x + c)^4 + 6*a*b^2 + b^3 + 20*(6*a*b^2 + b^3)*cosh(d*x + c)^2)*s
inh(d*x + c)^3 + (21*(6*a*b^2 + b^3)*cosh(d*x + c)^5 + 20*(6*a*b^2 + b^3)*cosh(d*x + c)^3 + 3*(6*a*b^2 + b^3)*
cosh(d*x + c))*sinh(d*x + c)^2 + (7*(6*a*b^2 + b^3)*cosh(d*x + c)^6 + 10*(6*a*b^2 + b^3)*cosh(d*x + c)^4 + 3*(
6*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(5*a^3*cosh(d*x + c)^
9 + 4*(11*a^3 + 36*a^2*b)*cosh(d*x + c)^7 + 6*(5*a^3 + 18*a^2*b + 12*b^3)*cosh(d*x + c)^5 - 4*(5*a^3 + 18*a^2*
b + 12*b^3)*cosh(d*x + c)^3 - (11*a^3 + 36*a^2*b)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(
d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 2*d*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c
)^5 + 5*(7*d*cosh(d*x + c)^3 + 2*d*cosh(d*x + c))*sinh(d*x + c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4
+ 20*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + (21*d*cosh(d*x + c)^5 + 20*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)
)*sinh(d*x + c)^2 + (7*d*cosh(d*x + c)^6 + 10*d*cosh(d*x + c)^4 + 3*d*cosh(d*x + c)^2)*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(cosh(d*x+c)**3*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.19224, size = 240, normalized size = 2.96 \begin{align*} \frac{b^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )} d} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (6 \, a b^{2} + b^{3}\right )}}{4 \, d} + \frac{a^{3} d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 12 \, a^{3} d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 36 \, a^{2} b d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{24 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*(e^(d*x + c) - e^(-d*x - c))/(((e^(d*x + c) - e^(-d*x - c))^2 + 4)*d) + 1/4*(pi + 2*arctan(1/2*(e^(2*d*x +
2*c) - 1)*e^(-d*x - c)))*(6*a*b^2 + b^3)/d + 1/24*(a^3*d^2*(e^(d*x + c) - e^(-d*x - c))^3 + 12*a^3*d^2*(e^(d*
x + c) - e^(-d*x - c)) + 36*a^2*b*d^2*(e^(d*x + c) - e^(-d*x - c)))/d^3