3.98 \(\int \cosh ^3(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=87 \[ \frac {b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac {b^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]

[Out]

1/2*b^2*(6*a+5*b)*arctan(sinh(d*x+c))/d+(a-2*b)*(a+b)^2*sinh(d*x+c)/d+1/3*(a+b)^3*sinh(d*x+c)^3/d-1/2*b^3*sech
(d*x+c)*tanh(d*x+c)/d

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Rubi [A]  time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 = {3676, 390, 385, 203} \[ \frac {b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac {b^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(6*a + 5*b)*ArcTan[Sinh[c + d*x]])/(2*d) + ((a - 2*b)*(a + b)^2*Sinh[c + d*x])/d + ((a + b)^3*Sinh[c + d*
x]^3)/(3*d) - (b^3*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

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])

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 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 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + 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]

Rubi steps

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

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Mathematica [C]  time = 6.89, size = 494, normalized size = 5.68 \[ \frac {\text {csch}^5(c+d x) \left (-256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b \sinh ^2(c+d x)\right )^3 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )+21 \left (a^3 \left (753 \sinh ^{10}(c+d x)+19579 \sinh ^8(c+d x)+89514 \sinh ^6(c+d x)+157878 \sinh ^4(c+d x)+124165 \sinh ^2(c+d x)+36015\right )+3 a^2 b \left (753 \sinh ^8(c+d x)+18826 \sinh ^6(c+d x)+69728 \sinh ^4(c+d x)+88150 \sinh ^2(c+d x)+36015\right ) \sinh ^2(c+d x)+3 a b^2 \left (753 \sinh ^6(c+d x)+18073 \sinh ^4(c+d x)+50695 \sinh ^2(c+d x)+36015\right ) \sinh ^4(c+d x)+b^3 \left (753 \sinh ^4(c+d x)+17320 \sinh ^2(c+d x)+32415\right ) \sinh ^6(c+d x)\right )-\frac {315 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (a^3 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right ) \cosh ^6(c+d x)+3 a^2 b \left (\sinh ^3(c+d x)+\sinh (c+d x)\right )^2 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right )+3 a b^2 \sinh ^4(c+d x) \left (\sinh ^8(c+d x)+244 \sinh ^6(c+d x)+2118 \sinh ^4(c+d x)+4180 \sinh ^2(c+d x)+2401\right )+b^3 \sinh ^6(c+d x) \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2161\right )\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{30240 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Csch[c + d*x]^5*(-256*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8
*(a + a*Sinh[c + d*x]^2 + b*Sinh[c + d*x]^2)^3 - (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^3*Sinh[c + d*x]^6*(21
61 + 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 + 1875*Sinh[c +
d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a^2*b*(Sinh[c + d*x] + Sinh[c + d*x]^3)^2*(2401 + 1875*Si
nh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a*b^2*Sinh[c + d*x]^4*(2401 + 4180*Sinh[c + d*x]^2
+ 2118*Sinh[c + d*x]^4 + 244*Sinh[c + d*x]^6 + Sinh[c + d*x]^8)))/Sqrt[-Sinh[c + d*x]^2] + 21*(b^3*Sinh[c + d*
x]^6*(32415 + 17320*Sinh[c + d*x]^2 + 753*Sinh[c + d*x]^4) + 3*a*b^2*Sinh[c + d*x]^4*(36015 + 50695*Sinh[c + d
*x]^2 + 18073*Sinh[c + d*x]^4 + 753*Sinh[c + d*x]^6) + 3*a^2*b*Sinh[c + d*x]^2*(36015 + 88150*Sinh[c + d*x]^2
+ 69728*Sinh[c + d*x]^4 + 18826*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*Sinh[c + d*x]^8 + 753*Sinh[c + d*x]^10))))/(30240*d)

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fricas [B]  time = 0.45, size = 1840, normalized size = 21.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^10 + 10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh
(d*x + c)^9 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^10 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d
*x + c)^8 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3 + 45*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d
*x + c)^8 + 8*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cos
h(d*x + c))*sinh(d*x + c)^7 + 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^6 + 2*(105*(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3 + 14*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*
b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 4*(63*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^5 + 14*(11*a^3 - 3
*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^3 + 3*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*sinh(d*x
+ c)^5 - 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^4 + 2*(105*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh
(d*x + c)^6 + 35*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^4 - 5*a^3 + 3*a^2*b + 21*a*b^2 + 25*b^3
+ 15*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(15*(a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*cosh(d*x + c)^7 + 7*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 - 3*a^2*b - 21*a*b
^2 - 25*b^3)*cosh(d*x + c)^3 - (5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - a^3 - 3*
a^2*b - 3*a*b^2 - b^3 - (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^2 + (45*(a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*cosh(d*x + c)^8 + 28*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^6 + 30*(5*a^3 - 3*a^2*b - 21*
a*b^2 - 25*b^3)*cosh(d*x + c)^4 - 11*a^3 + 3*a^2*b + 39*a*b^2 + 25*b^3 - 12*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b
^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 24*((6*a*b^2 + 5*b^3)*cosh(d*x + c)^7 + 7*(6*a*b^2 + 5*b^3)*cosh(d*x +
c)*sinh(d*x + c)^6 + (6*a*b^2 + 5*b^3)*sinh(d*x + c)^7 + 2*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + (12*a*b^2 + 10*
b^3 + 21*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 5*(7*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 2*(6*a*
b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + (6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + (35*(6*a*b^2 + 5*b^3)*cosh(d
*x + c)^4 + 6*a*b^2 + 5*b^3 + 20*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + (21*(6*a*b^2 + 5*b^3)*co
sh(d*x + c)^5 + 20*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 3*(6*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + (7
*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 10*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 3*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^
2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^9
+ 4*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^7 + 6*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x
+ c)^5 - 4*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^3 - (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cos
h(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*co
sh(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))*sin
h(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*c
osh(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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giac [B]  time = 0.66, size = 279, normalized size = 3.21 \[ \frac {24 \, {\left (6 \, a b^{2} e^{c} + 5 \, b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} - {\left (9 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 9 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 45 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 27 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} + {\left (a^{3} e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a^{2} b e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a b^{2} e^{\left (3 \, d x + 30 \, c\right )} + b^{3} e^{\left (3 \, d x + 30 \, c\right )} + 9 \, a^{3} e^{\left (d x + 28 \, c\right )} - 9 \, a^{2} b e^{\left (d x + 28 \, c\right )} - 45 \, a b^{2} e^{\left (d x + 28 \, c\right )} - 27 \, b^{3} e^{\left (d x + 28 \, c\right )}\right )} e^{\left (-27 \, c\right )} - \frac {24 \, {\left (b^{3} e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(6*a*b^2*e^c + 5*b^3*e^c)*arctan(e^(d*x + c))*e^(-c) - (9*a^3*e^(2*d*x + 2*c) - 9*a^2*b*e^(2*d*x + 2*
c) - 45*a*b^2*e^(2*d*x + 2*c) - 27*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-3*d*x - 3*c) + (a^
3*e^(3*d*x + 30*c) + 3*a^2*b*e^(3*d*x + 30*c) + 3*a*b^2*e^(3*d*x + 30*c) + b^3*e^(3*d*x + 30*c) + 9*a^3*e^(d*x
+ 28*c) - 9*a^2*b*e^(d*x + 28*c) - 45*a*b^2*e^(d*x + 28*c) - 27*b^3*e^(d*x + 28*c))*e^(-27*c) - 24*(b^3*e^(3*
d*x + 3*c) - b^3*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^2)/d

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maple [B]  time = 0.38, size = 206, normalized size = 2.37 \[ \frac {2 a^{3} \sinh \left (d x +c \right )}{3 d}+\frac {a^{3} \sinh \left (d x +c \right ) \left (\cosh ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} b \left (\sinh ^{3}\left (d x +c \right )\right )}{d}+\frac {a \,b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{d}-\frac {3 a \,b^{2} \sinh \left (d x +c \right )}{d}+\frac {6 a \,b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b^{3} \left (\sinh ^{5}\left (d x +c \right )\right )}{3 d \cosh \left (d x +c \right )^{2}}-\frac {5 b^{3} \left (\sinh ^{3}\left (d x +c \right )\right )}{3 d \cosh \left (d x +c \right )^{2}}-\frac {5 b^{3} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{2}}+\frac {5 b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}+\frac {5 b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a^3*sinh(d*x+c)/d+1/3/d*a^3*sinh(d*x+c)*cosh(d*x+c)^2+a^2*b*sinh(d*x+c)^3/d+1/d*a*b^2*sinh(d*x+c)^3-3/d*a*
b^2*sinh(d*x+c)+6/d*a*b^2*arctan(exp(d*x+c))+1/3/d*b^3*sinh(d*x+c)^5/cosh(d*x+c)^2-5/3/d*b^3*sinh(d*x+c)^3/cos
h(d*x+c)^2-5/d*b^3*sinh(d*x+c)/cosh(d*x+c)^2+5/2/d*b^3*sech(d*x+c)*tanh(d*x+c)+5/d*b^3*arctan(exp(d*x+c))

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maxima [B]  time = 0.43, size = 284, normalized size = 3.26 \[ \frac {a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3}}{8 \, d} - \frac {1}{8} \, a b^{2} {\left (\frac {{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac {1}{24} \, b^{3} {\left (\frac {27 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac {120 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {25 \, e^{\left (-2 \, d x - 2 \, c\right )} + 77 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac {1}{24} \, a^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3/d - 1/8*a*b^2*((15*e^(-2*d*x - 2*c) - 1)*e^(3*d*x + 3*c)/d - (15*e^(-
d*x - c) - e^(-3*d*x - 3*c))/d + 48*arctan(e^(-d*x - c))/d) + 1/24*b^3*((27*e^(-d*x - c) - e^(-3*d*x - 3*c))/d
- 120*arctan(e^(-d*x - c))/d - (25*e^(-2*d*x - 2*c) + 77*e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) - 1)/(d*(e^(-3
*d*x - 3*c) + 2*e^(-5*d*x - 5*c) + e^(-7*d*x - 7*c)))) + 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)

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mupad [B]  time = 0.36, size = 232, normalized size = 2.67 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (5\,b^3\,\sqrt {d^2}+6\,a\,b^2\,\sqrt {d^2}\right )}{d\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}\right )\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}{\sqrt {d^2}}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}-\frac {b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{-c-d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}+\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((exp(d*x)*exp(c)*(5*b^3*(d^2)^(1/2) + 6*a*b^2*(d^2)^(1/2)))/(d*(60*a*b^5 + 25*b^6 + 36*a^2*b^4)^(1/2)))*
(60*a*b^5 + 25*b^6 + 36*a^2*b^4)^(1/2))/(d^2)^(1/2) - (exp(- 3*c - 3*d*x)*(a + b)^3)/(24*d) + (exp(3*c + 3*d*x
)*(a + b)^3)/(24*d) + (3*exp(c + d*x)*(a + b)^2*(a - 3*b))/(8*d) - (b^3*exp(c + d*x))/(d*(exp(2*c + 2*d*x) + 1
)) - (3*exp(- c - d*x)*(a + b)^2*(a - 3*b))/(8*d) + (2*b^3*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*
d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \cosh ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**3*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*cosh(c + d*x)**3, x)

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