3.25 \(\int \frac {\sinh ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=117 \[ -\frac {\sqrt {b} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^3 d}-\frac {(5 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac {x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}+\frac {\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]

[Out]

1/8*(3*a^2+12*a*b+8*b^2)*x/a^3-1/8*(5*a+4*b)*cosh(d*x+c)*sinh(d*x+c)/a^2/d+1/4*cosh(d*x+c)^3*sinh(d*x+c)/a/d-(
a+b)^(3/2)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))*b^(1/2)/a^3/d

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Rubi [A]  time = 0.20, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 470, 527, 522, 206, 208} \[ \frac {x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}-\frac {\sqrt {b} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^3 d}-\frac {(5 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac {\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]

[Out]

((3*a^2 + 12*a*b + 8*b^2)*x)/(8*a^3) - (Sqrt[b]*(a + b)^(3/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a
^3*d) - ((5*a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*a^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*a*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 208

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

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 4132

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

Rubi steps

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

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Mathematica [B]  time = 2.48, size = 294, normalized size = 2.51 \[ -\frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\sqrt {b} \left (3 a^3+34 a^2 b+64 a b^2+32 b^3\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )-\sqrt {b (\cosh (c)-\sinh (c))^4} \left (\sqrt {b} \sqrt {a+b} \left (a^2 \sinh (4 (c+d x))-2 a^2 c+12 a^2 d x-8 a (a+b) \sinh (2 (c+d x))+48 a b d x+32 b^2 d x\right )+a^2 (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )\right )\right )}{64 a^3 \sqrt {b} d \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]

[Out]

-1/64*((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(Sqrt[b]*(3*a^3 + 34*a^2*b + 64*a*b^2 + 32*b^3)*ArcTanh
[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c]
- Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]) - Sqrt[b*(Cosh[c] - Sinh[c])^4]*(a^2*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tanh
[c + d*x])/Sqrt[a + b]] + Sqrt[b]*Sqrt[a + b]*(-2*a^2*c + 12*a^2*d*x + 48*a*b*d*x + 32*b^2*d*x - 8*a*(a + b)*S
inh[2*(c + d*x)] + a^2*Sinh[4*(c + d*x)]))))/(a^3*Sqrt[b]*Sqrt[a + b]*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c
] - Sinh[c])^4])

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fricas [B]  time = 0.59, size = 1681, normalized size = 14.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/64*(a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 + 12*a*b + 8
*b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x
+ c)^6 + 8*(7*a^2*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4
+ 4*(3*a^2 + 12*a*b + 8*b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5
+ 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*
b)*cosh(d*x + c)^2 + 4*(7*a^2*cosh(d*x + c)^6 + 12*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*
b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*sinh(d*x + c)^2 + 32*((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)^3*
sinh(d*x + c) + 6*(a + b)*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*
sinh(d*x + c)^4)*sqrt(a*b + b^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x
+ c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*
a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*cosh(d*x + c)^2 + 2*a
*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x
+ c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*si
nh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - a^2 + 8*(a^2*cosh(d*x +
c)^7 + 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x
+ c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^3*d*cosh(d*x + c)^2
*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x + c)^4), 1/64*(a^2*cosh(d*x + c)^8 +
8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^4 -
8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7*a^2*cosh(d*x
+ c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 + 4*(3*a^2 + 12*a*b + 8*b^2)
*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 + 4*(3*a^2 + 12*a*b + 8*b^2)
*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cosh(d*x + c)^2 + 4*(7*a^
2*cosh(d*x + c)^6 + 12*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*b)*cosh(d*x + c)^4 + 2*a^2 +
2*a*b)*sinh(d*x + c)^2 - 64*((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a + b)*co
sh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4)*sqrt(-a*b -
b^2)*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-a*b
- b^2)/(a*b + b^2)) - a^2 + 8*(a^2*cosh(d*x + c)^7 + 4*(3*a^2 + 12*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 - 6*(a^2
+ a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x
+ c)^3*sinh(d*x + c) + 6*a^3*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*
d*sinh(d*x + c)^4)]

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giac [B]  time = 2.88, size = 220, normalized size = 1.88 \[ \frac {\frac {8 \, {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {a e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b e^{\left (2 \, d x + 2 \, c\right )}}{a^{2}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 72 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{a^{3}} - \frac {64 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*(8*(3*a^2 + 12*a*b + 8*b^2)*(d*x + c)/a^3 + (a*e^(4*d*x + 4*c) - 8*a*e^(2*d*x + 2*c) - 8*b*e^(2*d*x + 2*c
))/a^2 - (18*a^2*e^(4*d*x + 4*c) + 72*a*b*e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) - 8
*a*b*e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c)/a^3 - 64*(a^2*b + 2*a*b^2 + b^3)*arctan(1/2*(a*e^(2*d*x + 2*c) +
a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^3))/d

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maple [B]  time = 0.41, size = 708, normalized size = 6.05 \[ \frac {1}{4 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{8 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d a}-\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b}{2 d \,a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}}{d \,a^{3}}-\frac {1}{4 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{8 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d a}+\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b}{2 d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}}{d \,a^{3}}+\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{d \,a^{2} \sqrt {a +b}}+\frac {b^{\frac {5}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{3} \sqrt {a +b}}-\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{d \,a^{2} \sqrt {a +b}}-\frac {b^{\frac {5}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{3} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x)

[Out]

1/4/d/a/(tanh(1/2*d*x+1/2*c)-1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)^3-1/8/d/a/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/d/a^
2/(tanh(1/2*d*x+1/2*c)-1)^2*b-3/8/d/a/(tanh(1/2*d*x+1/2*c)-1)-1/2/d/a^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/a*ln(t
anh(1/2*d*x+1/2*c)-1)-3/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)*b-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)-1)*b^2-1/4/d/a/(tan
h(1/2*d*x+1/2*c)+1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)^3+1/8/d/a/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/a^2/(tanh(1/2*
d*x+1/2*c)+1)^2*b-3/8/d/a/(tanh(1/2*d*x+1/2*c)+1)-1/2/d/a^2/(tanh(1/2*d*x+1/2*c)+1)*b+3/8/d/a*ln(tanh(1/2*d*x+
1/2*c)+1)+3/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)*b+1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2+1/2/d/a*b^(1/2)/(a+b)^(1
/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))+1/d/a^2*b^(3/2)/(a+b)^(1/2
)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))+1/2/d/a^3*b^(5/2)/(a+b)^(1/2
)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))-1/2/d/a*b^(1/2)/(a+b)^(1/2)*
ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))-1/d/a^2*b^(3/2)/(a+b)^(1/2)*ln
((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))-1/2/d/a^3*b^(5/2)/(a+b)^(1/2)*ln
((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))

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maxima [B]  time = 0.46, size = 526, normalized size = 4.50 \[ \frac {3 \, b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, \sqrt {{\left (a + b\right )} b} a d} + \frac {3 \, {\left (d x + c\right )}}{8 \, a d} - \frac {{\left (8 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, a^{2} d} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d} + \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, a^{2} d} - \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, a^{2} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} + \frac {{\left (a b + 2 \, b^{2}\right )} {\left (d x + c\right )}}{2 \, a^{3} d} + \frac {8 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, a^{2} d} + \frac {{\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, \sqrt {{\left (a + b\right )} b} a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

3/16*b*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b
)))/(sqrt((a + b)*b)*a*d) + 3/8*(d*x + c)/(a*d) - 1/64*(8*b*e^(-2*d*x - 2*c) - a)*e^(4*d*x + 4*c)/(a^2*d) - 1/
8*e^(2*d*x + 2*c)/(a*d) + 1/8*e^(-2*d*x - 2*c)/(a*d) + 1/4*b*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*
c) + a)/(a^2*d) - 1/4*b*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a^2*d) - 1/8*(a*b + 2*b^2)
*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqr
t((a + b)*b)*a^2*d) + 1/8*(a*b + 2*b^2)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x -
2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^2*d) + 1/2*(a*b + 2*b^2)*(d*x + c)/(a^3*d) + 1/64*(8*b
*e^(-2*d*x - 2*c) - a*e^(-4*d*x - 4*c))/(a^2*d) + 1/16*(a^2*b + 8*a*b^2 + 8*b^3)*log((a*e^(-2*d*x - 2*c) + a +
2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^3*d)

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mupad [B]  time = 2.43, size = 328, normalized size = 2.80 \[ \frac {x\,\left (3\,a^2+12\,a\,b+8\,b^2\right )}{8\,a^3}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,a\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,a\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+b\right )}{8\,a^2\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{8\,a^2\,d}+\frac {\sqrt {b}\,\ln \left (\frac {4\,b\,{\left (a+b\right )}^3\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^8}-\frac {8\,b^{3/2}\,{\left (a+b\right )}^{7/2}\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^8}\right )\,{\left (a+b\right )}^{3/2}}{2\,a^3\,d}-\frac {\sqrt {b}\,\ln \left (\frac {8\,b^{3/2}\,{\left (a+b\right )}^{7/2}\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^8}+\frac {4\,b\,{\left (a+b\right )}^3\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^8}\right )\,{\left (a+b\right )}^{3/2}}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^4/(a + b/cosh(c + d*x)^2),x)

[Out]

(x*(12*a*b + 3*a^2 + 8*b^2))/(8*a^3) - exp(- 4*c - 4*d*x)/(64*a*d) + exp(4*c + 4*d*x)/(64*a*d) + (exp(- 2*c -
2*d*x)*(a + b))/(8*a^2*d) - (exp(2*c + 2*d*x)*(a + b))/(8*a^2*d) + (b^(1/2)*log((4*b*(a + b)^3*(2*a*b + a^2 +
a^2*exp(2*c + 2*d*x) + 8*b^2*exp(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/a^8 - (8*b^(3/2)*(a + b)^(7/2)*(a + 2
*a*exp(2*c + 2*d*x) + 4*b*exp(2*c + 2*d*x)))/a^8)*(a + b)^(3/2))/(2*a^3*d) - (b^(1/2)*log((8*b^(3/2)*(a + b)^(
7/2)*(a + 2*a*exp(2*c + 2*d*x) + 4*b*exp(2*c + 2*d*x)))/a^8 + (4*b*(a + b)^3*(2*a*b + a^2 + a^2*exp(2*c + 2*d*
x) + 8*b^2*exp(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/a^8)*(a + b)^(3/2))/(2*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{4}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(sinh(c + d*x)**4/(a + b*sech(c + d*x)**2), x)

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