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

Optimal. Leaf size=93 \[ \frac{3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{a^3 \sinh (c+d x)}{d}+\frac{3 b^2 (4 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}+\frac{b^3 \tanh (c+d x) \text{sech}^3(c+d x)}{4 d} \]

[Out]

(3*b*(8*a^2 + 4*a*b + b^2)*ArcTan[Sinh[c + d*x]])/(8*d) + (a^3*Sinh[c + d*x])/d + (3*b^2*(4*a + b)*Sech[c + d*
x]*Tanh[c + d*x])/(8*d) + (b^3*Sech[c + d*x]^3*Tanh[c + d*x])/(4*d)

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Rubi [A]  time = 0.102681, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4147, 390, 1157, 385, 203} \[ \frac{3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{a^3 \sinh (c+d x)}{d}+\frac{3 b^2 (4 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}+\frac{b^3 \tanh (c+d x) \text{sech}^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(8*a^2 + 4*a*b + b^2)*ArcTan[Sinh[c + d*x]])/(8*d) + (a^3*Sinh[c + d*x])/d + (3*b^2*(4*a + b)*Sech[c + d*
x]*Tanh[c + d*x])/(8*d) + (b^3*Sech[c + d*x]^3*Tanh[c + d*x])/(4*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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 (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 )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^3+\frac{b \left (3 a^2+3 a b+b^2\right )+3 a b (2 a+b) x^2+3 a^2 b x^4}{\left (1+x^2\right )^3}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^3 \sinh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{b \left (3 a^2+3 a b+b^2\right )+3 a b (2 a+b) x^2+3 a^2 b x^4}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^3 \sinh (c+d x)}{d}+\frac{b^3 \text{sech}^3(c+d x) \tanh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-3 b (2 a+b)^2-12 a^2 b x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 d}\\ &=\frac{a^3 \sinh (c+d x)}{d}+\frac{3 b^2 (4 a+b) \text{sech}(c+d x) \tanh (c+d x)}{8 d}+\frac{b^3 \text{sech}^3(c+d x) \tanh (c+d x)}{4 d}+\frac{\left (3 b \left (8 a^2+4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{8 d}\\ &=\frac{3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac{a^3 \sinh (c+d x)}{d}+\frac{3 b^2 (4 a+b) \text{sech}(c+d x) \tanh (c+d x)}{8 d}+\frac{b^3 \text{sech}^3(c+d x) \tanh (c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 7.90187, size = 575, normalized size = 6.18 \[ -\frac{\cosh (c+d x) \coth ^5(c+d x) \left (a+b \text{sech}^2(c+d x)\right )^3 \left (256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{11}{2}\right \},-\sinh ^2(c+d x)\right )+384 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2 \left (a \left (5 \sinh ^2(c+d x)+7\right )+7 b\right ) \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{11}{2}\right \},-\sinh ^2(c+d x)\right )-21 \left (3 a^2 b \left (5640 \sinh ^8(c+d x)+120431 \sinh ^6(c+d x)+437991 \sinh ^4(c+d x)+573145 \sinh ^2(c+d x)+252105\right )+a^3 \left (4887 \sinh ^{10}(c+d x)+107725 \sinh ^8(c+d x)+491574 \sinh ^6(c+d x)+922986 \sinh ^4(c+d x)+789235 \sinh ^2(c+d x)+252105\right )+3 a b^2 \left (6393 \sinh ^6(c+d x)+133071 \sinh ^4(c+d x)+357055 \sinh ^2(c+d x)+252105\right )+b^3 \left (8226 \sinh ^4(c+d x)+140965 \sinh ^2(c+d x)+252105\right )\right )+\frac{315 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right ) \left (3 a^2 b \left (8 \sinh ^{10}(c+d x)+1719 \sinh ^8(c+d x)+14956 \sinh ^6(c+d x)+40442 \sinh ^4(c+d x)+43812 \sinh ^2(c+d x)+16807\right )+a^3 \left (7 \sinh ^8(c+d x)+1468 \sinh ^6(c+d x)+11562 \sinh ^4(c+d x)+24604 \sinh ^2(c+d x)+16807\right ) \cosh ^4(c+d x)+3 a b^2 \left (9 \sinh ^8(c+d x)+1858 \sinh ^6(c+d x)+15312 \sinh ^4(c+d x)+29406 \sinh ^2(c+d x)+16807\right )+b^3 \left (-62 \sinh ^6(c+d x)+2187 \sinh ^4(c+d x)+15000 \sinh ^2(c+d x)+16807\right )\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{7560 d (a \cosh (2 c+2 d x)+a+2 b)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Cosh[c + d*x]*Coth[c + d*x]^5*(a + b*Sech[c + d*x]^2)^3*(256*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1,
1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8*(a + b + a*Sinh[c + d*x]^2)^3 + 384*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)^2*(7*b + a*(7 + 5*Sin
h[c + d*x]^2)) + (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^3*(16807 + 15000*Sinh[c + d*x]^2 + 2187*Sinh[c + d*x]
^4 - 62*Sinh[c + d*x]^6) + a^3*Cosh[c + d*x]^4*(16807 + 24604*Sinh[c + d*x]^2 + 11562*Sinh[c + d*x]^4 + 1468*S
inh[c + d*x]^6 + 7*Sinh[c + d*x]^8) + 3*a*b^2*(16807 + 29406*Sinh[c + d*x]^2 + 15312*Sinh[c + d*x]^4 + 1858*Si
nh[c + d*x]^6 + 9*Sinh[c + d*x]^8) + 3*a^2*b*(16807 + 43812*Sinh[c + d*x]^2 + 40442*Sinh[c + d*x]^4 + 14956*Si
nh[c + d*x]^6 + 1719*Sinh[c + d*x]^8 + 8*Sinh[c + d*x]^10)))/Sqrt[-Sinh[c + d*x]^2] - 21*(b^3*(252105 + 140965
*Sinh[c + d*x]^2 + 8226*Sinh[c + d*x]^4) + 3*a*b^2*(252105 + 357055*Sinh[c + d*x]^2 + 133071*Sinh[c + d*x]^4 +
6393*Sinh[c + d*x]^6) + 3*a^2*b*(252105 + 573145*Sinh[c + d*x]^2 + 437991*Sinh[c + d*x]^4 + 120431*Sinh[c + d
*x]^6 + 5640*Sinh[c + d*x]^8) + a^3*(252105 + 789235*Sinh[c + d*x]^2 + 922986*Sinh[c + d*x]^4 + 491574*Sinh[c
+ d*x]^6 + 107725*Sinh[c + d*x]^8 + 4887*Sinh[c + d*x]^10))))/(7560*d*(a + 2*b + a*Cosh[2*c + 2*d*x])^3)

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Maple [A]  time = 0.048, size = 125, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}\sinh \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{3\,a{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+3\,{\frac{a{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{3} \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}\tanh \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{b}^{3}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^3*sinh(d*x+c)+6/d*a^2*b*arctan(exp(d*x+c))+3/2/d*a*b^2*sech(d*x+c)*tanh(d*x+c)+3/d*a*b^2*arctan(exp(d*x+
c))+1/4*b^3*sech(d*x+c)^3*tanh(d*x+c)/d+3/8*b^3*sech(d*x+c)*tanh(d*x+c)/d+3/4/d*b^3*arctan(exp(d*x+c))

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Maxima [B]  time = 1.68732, size = 298, normalized size = 3.2 \begin{align*} -\frac{1}{4} \, b^{3}{\left (\frac{3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{3 \, e^{\left (-d x - c\right )} + 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} - 3 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - 3 \, a b^{2}{\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{6 \, a^{2} b \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{a^{3} \sinh \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^3*(3*arctan(e^(-d*x - c))/d - (3*e^(-d*x - c) + 11*e^(-3*d*x - 3*c) - 11*e^(-5*d*x - 5*c) - 3*e^(-7*d*x
- 7*c))/(d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) - 3*a*b^2*
(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))) -
6*a^2*b*arctan(e^(-d*x - c))/d + a^3*sinh(d*x + c)/d

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Fricas [B]  time = 2.67461, size = 4986, normalized size = 53.61 \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)*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/4*(2*a^3*cosh(d*x + c)^10 + 20*a^3*cosh(d*x + c)*sinh(d*x + c)^9 + 2*a^3*sinh(d*x + c)^10 + 3*(2*a^3 + 4*a*b
^2 + b^3)*cosh(d*x + c)^8 + 3*(30*a^3*cosh(d*x + c)^2 + 2*a^3 + 4*a*b^2 + b^3)*sinh(d*x + c)^8 + 24*(10*a^3*co
sh(d*x + c)^3 + (2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x
+ c)^6 + (420*a^3*cosh(d*x + c)^4 + 4*a^3 + 12*a*b^2 + 11*b^3 + 84*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*si
nh(d*x + c)^6 + 6*(84*a^3*cosh(d*x + c)^5 + 28*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (4*a^3 + 12*a*b^2 + 1
1*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 - (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^4 + (420*a^3*cosh(d*x + c)^6
+ 210*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^4 - 4*a^3 - 12*a*b^2 - 11*b^3 + 15*(4*a^3 + 12*a*b^2 + 11*b^3)*co
sh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(60*a^3*cosh(d*x + c)^7 + 42*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^5 + 5*(4
*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - 2*a^3
- 3*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^2 + 3*(30*a^3*cosh(d*x + c)^8 + 28*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x
+ c)^6 + 5*(4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^4 - 2*a^3 - 4*a*b^2 - b^3 - 2*(4*a^3 + 12*a*b^2 + 11*b^3
)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*((8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^9 + 9*(8*a^2*b + 4*a*b^2 + b^3
)*cosh(d*x + c)*sinh(d*x + c)^8 + (8*a^2*b + 4*a*b^2 + b^3)*sinh(d*x + c)^9 + 4*(8*a^2*b + 4*a*b^2 + b^3)*cosh
(d*x + c)^7 + 4*(8*a^2*b + 4*a*b^2 + b^3 + 9*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 28*(
3*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^6 + 6*(8*
a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^5 + 6*(21*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 +
b^3 + 14*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 2*(63*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*
x + c)^5 + 70*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + 15*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x
+ c)^4 + 4*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + 4*(21*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^6 + 35*(
8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 + b^3 + 15*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c
)^2)*sinh(d*x + c)^3 + 12*(3*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^7 + 7*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x
+ c)^5 + 5*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^
2 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c) + (9*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^8 + 28*(8*a^2*b + 4*a
*b^2 + b^3)*cosh(d*x + c)^6 + 30*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 + b^3 + 12*(8*a
^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(10*a^3*cosh(d
*x + c)^9 + 12*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^7 + 3*(4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^5 - 2*(4*
a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - 3*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x
+ c)^9 + 9*d*cosh(d*x + c)*sinh(d*x + c)^8 + d*sinh(d*x + c)^9 + 4*d*cosh(d*x + c)^7 + 4*(9*d*cosh(d*x + c)^2
+ d)*sinh(d*x + c)^7 + 28*(3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^6 + 6*d*cosh(d*x + c)^5 + 6*(
21*d*cosh(d*x + c)^4 + 14*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^5 + 2*(63*d*cosh(d*x + c)^5 + 70*d*cosh(d*x + c
)^3 + 15*d*cosh(d*x + c))*sinh(d*x + c)^4 + 4*d*cosh(d*x + c)^3 + 4*(21*d*cosh(d*x + c)^6 + 35*d*cosh(d*x + c)
^4 + 15*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 12*(3*d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)^5 + 5*d*cosh(d*x
+ c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (9*d*cosh(d*x + c)^8 + 28*d*cosh(d*x + c)^6 + 30
*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + d)*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)*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.18348, size = 274, normalized size = 2.95 \begin{align*} \frac{a^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{2 \, d} + \frac{3 \,{\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 (8 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )}}{16 \, d} + \frac{12 \, a b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, b^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 48 \, a b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 20 \, b^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{4 \,{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*(e^(d*x + c) - e^(-d*x - c))/d + 3/16*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(8*a^2*b
+ 4*a*b^2 + b^3)/d + 1/4*(12*a*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 3*b^3*(e^(d*x + c) - e^(-d*x - c))^3 + 48
*a*b^2*(e^(d*x + c) - e^(-d*x - c)) + 20*b^3*(e^(d*x + c) - e^(-d*x - c)))/(((e^(d*x + c) - e^(-d*x - c))^2 +
4)^2*d)