3.312 \(\int \text {sech}^5(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=103 \[ \frac {3 (a-b) \left ((a+b)^2+4 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(a-b)^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}+\frac {3 (a-b)^2 (a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b^3 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3190, 390, 1157, 385, 203} \[ \frac {3 (a-b) \left ((a+b)^2+4 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(a-b)^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}+\frac {3 (a-b)^2 (a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b^3 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^5*(a + b*Sinh[c + d*x]^2)^3,x]

[Out]

(3*(a - b)*(4*b^2 + (a + b)^2)*ArcTan[Sinh[c + d*x]])/(8*d) + (b^3*Sinh[c + d*x])/d + (3*(a - b)^2*(a + 3*b)*S
ech[c + d*x]*Tanh[c + d*x])/(8*d) + ((a - b)^3*Sech[c + d*x]^3*Tanh[c + d*x])/(4*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 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 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \text {sech}^5(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b^3+\frac {a^3-b^3+3 b \left (a^2-b^2\right ) x^2+3 (a-b) b^2 x^4}{\left (1+x^2\right )^3}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {b^3 \sinh (c+d x)}{d}+\frac {\operatorname {Subst}\left (\int \frac {a^3-b^3+3 b \left (a^2-b^2\right ) x^2+3 (a-b) b^2 x^4}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {b^3 \sinh (c+d x)}{d}+\frac {(a-b)^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-3 (a-b) (a+b)^2-12 (a-b) b^2 x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 d}\\ &=\frac {b^3 \sinh (c+d x)}{d}+\frac {3 (a-b)^2 (a+3 b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {(a-b)^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}+\frac {\left (3 (a-b) \left (4 b^2+(a+b)^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{8 d}\\ &=\frac {3 (a-b) \left (4 b^2+(a+b)^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {b^3 \sinh (c+d x)}{d}+\frac {3 (a-b)^2 (a+3 b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {(a-b)^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}\\ \end {align*}

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Mathematica [C]  time = 10.07, size = 472, normalized size = 4.58 \[ -\frac {\text {csch}^5(c+d x) \left (256 \sinh ^8(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )+384 \sinh ^8(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \left (7 a+5 b \sinh ^2(c+d x)\right ) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )-21 \left (a^3 \left (8226 \sinh ^4(c+d x)+140965 \sinh ^2(c+d x)+252105\right )+9 a^2 b \left (2131 \sinh ^4(c+d x)+41615 \sinh ^2(c+d x)+72030\right ) \sinh ^2(c+d x)+15 a b^2 \left (1128 \sinh ^4(c+d x)+21529 \sinh ^2(c+d x)+36015\right ) \sinh ^4(c+d x)+b^3 \left (4887 \sinh ^4(c+d x)+90805 \sinh ^2(c+d x)+149460\right ) \sinh ^6(c+d x)\right )+\frac {315 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (a^3 \left (-62 \sinh ^6(c+d x)+2187 \sinh ^4(c+d x)+15000 \sinh ^2(c+d x)+16807\right )+9 a^2 b \left (3 \sinh ^6(c+d x)+640 \sinh ^4(c+d x)+4375 \sinh ^2(c+d x)+4802\right ) \sinh ^2(c+d x)+3 a b^2 \left (8 \sinh ^6(c+d x)+1701 \sinh ^4(c+d x)+11178 \sinh ^2(c+d x)+12005\right ) \sinh ^4(c+d x)+b^3 \left (7 \sinh ^6(c+d x)+1458 \sinh ^4(c+d x)+9375 \sinh ^2(c+d x)+9964\right ) \sinh ^6(c+d x)\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{60480 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[c + d*x]^5*(a + b*Sinh[c + d*x]^2)^3,x]

[Out]

-1/60480*(Csch[c + d*x]^5*(256*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*S
inh[c + d*x]^8*(a + b*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*Sinh[c + d*x]^2)^2*(7*a + 5*b*Sinh[c + d*x]^2) - 21*(15*a*b^2*Sinh[c + d*x]^4*
(36015 + 21529*Sinh[c + d*x]^2 + 1128*Sinh[c + d*x]^4) + 9*a^2*b*Sinh[c + d*x]^2*(72030 + 41615*Sinh[c + d*x]^
2 + 2131*Sinh[c + d*x]^4) + b^3*Sinh[c + d*x]^6*(149460 + 90805*Sinh[c + d*x]^2 + 4887*Sinh[c + d*x]^4) + a^3*
(252105 + 140965*Sinh[c + d*x]^2 + 8226*Sinh[c + d*x]^4)) + (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(a^3*(16807 +
15000*Sinh[c + d*x]^2 + 2187*Sinh[c + d*x]^4 - 62*Sinh[c + d*x]^6) + 9*a^2*b*Sinh[c + d*x]^2*(4802 + 4375*Sin
h[c + d*x]^2 + 640*Sinh[c + d*x]^4 + 3*Sinh[c + d*x]^6) + b^3*Sinh[c + d*x]^6*(9964 + 9375*Sinh[c + d*x]^2 + 1
458*Sinh[c + d*x]^4 + 7*Sinh[c + d*x]^6) + 3*a*b^2*Sinh[c + d*x]^4*(12005 + 11178*Sinh[c + d*x]^2 + 1701*Sinh[
c + d*x]^4 + 8*Sinh[c + d*x]^6)))/Sqrt[-Sinh[c + d*x]^2]))/d

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fricas [B]  time = 1.03, size = 2245, normalized size = 21.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/4*(2*b^3*cosh(d*x + c)^10 + 20*b^3*cosh(d*x + c)*sinh(d*x + c)^9 + 2*b^3*sinh(d*x + c)^10 + 3*(a^3 + a^2*b -
5*a*b^2 + 5*b^3)*cosh(d*x + c)^8 + 3*(30*b^3*cosh(d*x + c)^2 + a^3 + a^2*b - 5*a*b^2 + 5*b^3)*sinh(d*x + c)^8
+ 24*(10*b^3*cosh(d*x + c)^3 + (a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + (11*a^3 - 21*
a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + (420*b^3*cosh(d*x + c)^4 + 11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3 + 84
*(a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 6*(84*b^3*cosh(d*x + c)^5 + 28*(a^3 + a^2*
b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + (11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 -
(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + (420*b^3*cosh(d*x + c)^6 + 210*(a^3 + a^2*b - 5*a*b^2
+ 5*b^3)*cosh(d*x + c)^4 - 11*a^3 + 21*a^2*b - 9*a*b^2 - 5*b^3 + 15*(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh
(d*x + c)^2)*sinh(d*x + c)^4 + 4*(60*b^3*cosh(d*x + c)^7 + 42*(a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^5
+ 5*(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^3 - (11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c
))*sinh(d*x + c)^3 - 2*b^3 - 3*(a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^2 + 3*(30*b^3*cosh(d*x + c)^8 + 2
8*(a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 5*(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^4 -
a^3 - a^2*b + 5*a*b^2 - 5*b^3 - 2*(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*(
(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^9 + 9*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)*sinh(d*x + c
)^8 + (a^3 + a^2*b + 3*a*b^2 - 5*b^3)*sinh(d*x + c)^9 + 4*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^7 + 4*
(a^3 + a^2*b + 3*a*b^2 - 5*b^3 + 9*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 28*(3*(a
^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^6
+ 6*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^5 + 6*(21*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^4 +
a^3 + a^2*b + 3*a*b^2 - 5*b^3 + 14*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 2*(63*(
a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^5 + 70*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^3 + 15*(a^3
+ a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 4*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^3
+ 4*(21*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^6 + 35*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^4 +
a^3 + a^2*b + 3*a*b^2 - 5*b^3 + 15*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 12*(3*(
a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^7 + 7*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^5 + 5*(a^3 +
a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + (a
^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c) + (9*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^8 + 28*(a^3 + a
^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^6 + 30*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^4 + a^3 + a^2*b + 3
*a*b^2 - 5*b^3 + 12*(a^3 + a^2*b + 3*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sin
h(d*x + c)) + 2*(10*b^3*cosh(d*x + c)^9 + 12*(a^3 + a^2*b - 5*a*b^2 + 5*b^3)*cosh(d*x + c)^7 + 3*(11*a^3 - 21*
a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^5 - 2*(11*a^3 - 21*a^2*b + 9*a*b^2 + 5*b^3)*cosh(d*x + c)^3 - 3*(a^3 +
a^2*b - 5*a*b^2 + 5*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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giac [B]  time = 0.23, size = 301, normalized size = 2.92 \[ \frac {8 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (a^{3} + a^{2} b + 3 \, a b^{2} - 5 \, b^{3}\right )} + \frac {4 \, {\left (3 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 15 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 9 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 20 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 12 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 36 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 28 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/16*(8*b^3*(e^(d*x + c) - e^(-d*x - c)) + 3*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(a^3 + a^
2*b + 3*a*b^2 - 5*b^3) + 4*(3*a^3*(e^(d*x + c) - e^(-d*x - c))^3 + 3*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3 - 15
*a*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 9*b^3*(e^(d*x + c) - e^(-d*x - c))^3 + 20*a^3*(e^(d*x + c) - e^(-d*x -
c)) - 12*a^2*b*(e^(d*x + c) - e^(-d*x - c)) - 36*a*b^2*(e^(d*x + c) - e^(-d*x - c)) + 28*b^3*(e^(d*x + c) - e
^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2)/d

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maple [B]  time = 0.14, size = 376, normalized size = 3.65 \[ \frac {a^{3} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}+\frac {3 a^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}+\frac {3 a^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d}-\frac {a^{2} b \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{4}}+\frac {a^{2} b \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}+\frac {3 a^{2} b \,\mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}+\frac {3 a^{2} b \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d}-\frac {3 a \,b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{4}}-\frac {3 a \,b^{2} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{4}}+\frac {3 a \,b^{2} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}+\frac {9 a \,b^{2} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}+\frac {9 a \,b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d}+\frac {b^{3} \left (\sinh ^{5}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{4}}+\frac {5 b^{3} \left (\sinh ^{3}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{4}}+\frac {5 b^{3} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{4}}-\frac {5 b^{3} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}-\frac {15 b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}-\frac {15 b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^5*(a+b*sinh(d*x+c)^2)^3,x)

[Out]

1/4/d*a^3*tanh(d*x+c)*sech(d*x+c)^3+3/8/d*a^3*sech(d*x+c)*tanh(d*x+c)+3/4/d*a^3*arctan(exp(d*x+c))-1/d*a^2*b*s
inh(d*x+c)/cosh(d*x+c)^4+1/4/d*a^2*b*tanh(d*x+c)*sech(d*x+c)^3+3/8/d*a^2*b*sech(d*x+c)*tanh(d*x+c)+3/4/d*a^2*b
*arctan(exp(d*x+c))-3/d*a*b^2*sinh(d*x+c)^3/cosh(d*x+c)^4-3/d*a*b^2*sinh(d*x+c)/cosh(d*x+c)^4+3/4/d*a*b^2*tanh
(d*x+c)*sech(d*x+c)^3+9/8/d*a*b^2*sech(d*x+c)*tanh(d*x+c)+9/4/d*a*b^2*arctan(exp(d*x+c))+1/d*b^3*sinh(d*x+c)^5
/cosh(d*x+c)^4+5/d*b^3*sinh(d*x+c)^3/cosh(d*x+c)^4+5/d*b^3*sinh(d*x+c)/cosh(d*x+c)^4-5/4/d*b^3*tanh(d*x+c)*sec
h(d*x+c)^3-15/8/d*b^3*sech(d*x+c)*tanh(d*x+c)-15/4/d*b^3*arctan(exp(d*x+c))

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maxima [B]  time = 0.49, size = 489, normalized size = 4.75 \[ \frac {1}{4} \, b^{3} {\left (\frac {15 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {2 \, e^{\left (-d x - c\right )}}{d} + \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 13 \, e^{\left (-4 \, d x - 4 \, c\right )} + 7 \, e^{\left (-6 \, d x - 6 \, c\right )} - 7 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2}{d {\left (e^{\left (-d x - c\right )} + 4 \, e^{\left (-3 \, d x - 3 \, c\right )} + 6 \, e^{\left (-5 \, d x - 5 \, c\right )} + 4 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} - \frac {3}{4} \, a b^{2} {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {5 \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, 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 )} - \frac {1}{4} \, a^{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 )} - \frac {3}{4} \, a^{2} b {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - 7 \, e^{\left (-3 \, d x - 3 \, c\right )} + 7 \, e^{\left (-5 \, d x - 5 \, c\right )} - 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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/4*b^3*(15*arctan(e^(-d*x - c))/d - 2*e^(-d*x - c)/d + (17*e^(-2*d*x - 2*c) + 13*e^(-4*d*x - 4*c) + 7*e^(-6*d
*x - 6*c) - 7*e^(-8*d*x - 8*c) + 2)/(d*(e^(-d*x - c) + 4*e^(-3*d*x - 3*c) + 6*e^(-5*d*x - 5*c) + 4*e^(-7*d*x -
7*c) + e^(-9*d*x - 9*c)))) - 3/4*a*b^2*(3*arctan(e^(-d*x - c))/d + (5*e^(-d*x - c) - 3*e^(-3*d*x - 3*c) + 3*e
^(-5*d*x - 5*c) - 5*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))) - 1/4*a^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))) - 3/4*a^2*b*(arctan(e^(-d*x - c))/d - (e^(-d*x - c) - 7*e^(-3*d*x - 3*c) + 7*e^(-5*d*x - 5*c) - 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)))

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mupad [B]  time = 0.19, size = 430, normalized size = 4.17 \[ \frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a^3\,\sqrt {d^2}-5\,b^3\,\sqrt {d^2}+3\,a\,b^2\,\sqrt {d^2}+a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {a^6+2\,a^5\,b+7\,a^4\,b^2-4\,a^3\,b^3-a^2\,b^4-30\,a\,b^5+25\,b^6}}\right )\,\sqrt {a^6+2\,a^5\,b+7\,a^4\,b^2-4\,a^3\,b^3-a^2\,b^4-30\,a\,b^5+25\,b^6}}{4\,\sqrt {d^2}}+\frac {b^3\,{\mathrm {e}}^{c+d\,x}}{2\,d}-\frac {b^3\,{\mathrm {e}}^{-c-d\,x}}{2\,d}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,\left (a^3+a^2\,b-5\,a\,b^2+3\,b^3\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (a^3-15\,a^2\,b+27\,a\,b^2-13\,b^3\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {6\,{\mathrm {e}}^{c+d\,x}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {4\,{\mathrm {e}}^{c+d\,x}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^2)^3/cosh(c + d*x)^5,x)

[Out]

(3*atan((exp(d*x)*exp(c)*(a^3*(d^2)^(1/2) - 5*b^3*(d^2)^(1/2) + 3*a*b^2*(d^2)^(1/2) + a^2*b*(d^2)^(1/2)))/(d*(
2*a^5*b - 30*a*b^5 + a^6 + 25*b^6 - a^2*b^4 - 4*a^3*b^3 + 7*a^4*b^2)^(1/2)))*(2*a^5*b - 30*a*b^5 + a^6 + 25*b^
6 - a^2*b^4 - 4*a^3*b^3 + 7*a^4*b^2)^(1/2))/(4*(d^2)^(1/2)) + (b^3*exp(c + d*x))/(2*d) - (b^3*exp(- c - d*x))/
(2*d) + (3*exp(c + d*x)*(a^2*b - 5*a*b^2 + a^3 + 3*b^3))/(4*d*(exp(2*c + 2*d*x) + 1)) + (exp(c + d*x)*(27*a*b^
2 - 15*a^2*b + a^3 - 13*b^3))/(2*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (6*exp(c + d*x)*(3*a*b^2 - 3
*a^2*b + a^3 - b^3))/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (4*exp(c + d*x)*(3
*a*b^2 - 3*a^2*b + a^3 - b^3))/(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))

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

[Out]

Timed out

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