3.57 \(\int (a+b \tanh (c+d x))^5 \, dx\)

Optimal. Leaf size=142 \[ -\frac {b \left (3 a^2+b^2\right ) (a+b \tanh (c+d x))^2}{2 d}-\frac {4 a b^2 \left (a^2+b^2\right ) \tanh (c+d x)}{d}+\frac {b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\cosh (c+d x))}{d}+a x \left (a^4+10 a^2 b^2+5 b^4\right )-\frac {b (a+b \tanh (c+d x))^4}{4 d}-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d} \]

[Out]

a*(a^4+10*a^2*b^2+5*b^4)*x+b*(5*a^4+10*a^2*b^2+b^4)*ln(cosh(d*x+c))/d-4*a*b^2*(a^2+b^2)*tanh(d*x+c)/d-1/2*b*(3
*a^2+b^2)*(a+b*tanh(d*x+c))^2/d-2/3*a*b*(a+b*tanh(d*x+c))^3/d-1/4*b*(a+b*tanh(d*x+c))^4/d

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Rubi [A]  time = 0.21, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3482, 3528, 3525, 3475} \[ -\frac {b \left (3 a^2+b^2\right ) (a+b \tanh (c+d x))^2}{2 d}-\frac {4 a b^2 \left (a^2+b^2\right ) \tanh (c+d x)}{d}+\frac {b \left (10 a^2 b^2+5 a^4+b^4\right ) \log (\cosh (c+d x))}{d}+a x \left (10 a^2 b^2+a^4+5 b^4\right )-\frac {b (a+b \tanh (c+d x))^4}{4 d}-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tanh[c + d*x])^5,x]

[Out]

a*(a^4 + 10*a^2*b^2 + 5*b^4)*x + (b*(5*a^4 + 10*a^2*b^2 + b^4)*Log[Cosh[c + d*x]])/d - (4*a*b^2*(a^2 + b^2)*Ta
nh[c + d*x])/d - (b*(3*a^2 + b^2)*(a + b*Tanh[c + d*x])^2)/(2*d) - (2*a*b*(a + b*Tanh[c + d*x])^3)/(3*d) - (b*
(a + b*Tanh[c + d*x])^4)/(4*d)

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int (a+b \tanh (c+d x))^5 \, dx &=-\frac {b (a+b \tanh (c+d x))^4}{4 d}+\int (a+b \tanh (c+d x))^3 \left (a^2+b^2+2 a b \tanh (c+d x)\right ) \, dx\\ &=-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d}-\frac {b (a+b \tanh (c+d x))^4}{4 d}+\int (a+b \tanh (c+d x))^2 \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \tanh (c+d x)\right ) \, dx\\ &=-\frac {b \left (3 a^2+b^2\right ) (a+b \tanh (c+d x))^2}{2 d}-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d}-\frac {b (a+b \tanh (c+d x))^4}{4 d}+\int (a+b \tanh (c+d x)) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \tanh (c+d x)\right ) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x-\frac {4 a b^2 \left (a^2+b^2\right ) \tanh (c+d x)}{d}-\frac {b \left (3 a^2+b^2\right ) (a+b \tanh (c+d x))^2}{2 d}-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d}-\frac {b (a+b \tanh (c+d x))^4}{4 d}+\left (b \left (5 a^4+10 a^2 b^2+b^4\right )\right ) \int \tanh (c+d x) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x+\frac {b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\cosh (c+d x))}{d}-\frac {4 a b^2 \left (a^2+b^2\right ) \tanh (c+d x)}{d}-\frac {b \left (3 a^2+b^2\right ) (a+b \tanh (c+d x))^2}{2 d}-\frac {2 a b (a+b \tanh (c+d x))^3}{3 d}-\frac {b (a+b \tanh (c+d x))^4}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 114, normalized size = 0.80 \[ -\frac {60 a b^2 \left (2 a^2+b^2\right ) \tanh (c+d x)+6 b^3 \left (10 a^2+b^2\right ) \tanh ^2(c+d x)+20 a b^4 \tanh ^3(c+d x)-6 (a-b)^5 \log (\tanh (c+d x)+1)+6 (a+b)^5 \log (1-\tanh (c+d x))+3 b^5 \tanh ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tanh[c + d*x])^5,x]

[Out]

-1/12*(6*(a + b)^5*Log[1 - Tanh[c + d*x]] - 6*(a - b)^5*Log[1 + Tanh[c + d*x]] + 60*a*b^2*(2*a^2 + b^2)*Tanh[c
+ d*x] + 6*b^3*(10*a^2 + b^2)*Tanh[c + d*x]^2 + 20*a*b^4*Tanh[c + d*x]^3 + 3*b^5*Tanh[c + d*x]^4)/d

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fricas [B]  time = 0.52, size = 2739, normalized size = 19.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(d*x+c))^5,x, algorithm="fricas")

[Out]

1/3*(3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^8 + 24*(a^5 - 5*a^4*b + 10*
a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a
^2*b^3 + 5*a*b^4 - b^5)*d*x*sinh(d*x + c)^8 + 12*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*
a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^6 + 12*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + 7*(a^
5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^2 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*
a^2*b^3 + 5*a*b^4 - b^5)*d*x)*sinh(d*x + c)^6 + 24*(7*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5
)*d*x*cosh(d*x + c)^3 + 3*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 +
5*a*b^4 - b^5)*d*x)*cosh(d*x + c))*sinh(d*x + c)^5 + 60*a^3*b^2 + 40*a*b^4 + 6*(30*a^3*b^2 + 20*a^2*b^3 + 20*a
*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^4 + 6*(35*(a^5 -
5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^4 + 30*a^3*b^2 + 20*a^2*b^3 + 20*a*b^4 +
2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x + 30*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4
+ b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 24*
(7*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^5 + 10*(5*a^3*b^2 + 5*a^2*b^3 +
5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^3 + (30*a^3*b^2
+ 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x +
c))*sinh(d*x + c)^3 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x + 4*(45*a^3*b^2 + 15*a^2
*b^3 + 25*a*b^4 + 3*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^2 + 4
*(21*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^6 + 45*a^3*b^2 + 15*a^2*b^3 +
25*a*b^4 + 3*b^5 + 45*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a
*b^4 - b^5)*d*x)*cosh(d*x + c)^4 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x + 9*(30*a^3
*b^2 + 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d
*x + c)^2)*sinh(d*x + c)^2 + 3*((5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^8 + 8*(5*a^4*b + 10*a^2*b^3 + b^5)*
cosh(d*x + c)*sinh(d*x + c)^7 + (5*a^4*b + 10*a^2*b^3 + b^5)*sinh(d*x + c)^8 + 4*(5*a^4*b + 10*a^2*b^3 + b^5)*
cosh(d*x + c)^6 + 4*(5*a^4*b + 10*a^2*b^3 + b^5 + 7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^2)*sinh(d*x + c
)^6 + 8*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^3 + 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d
*x + c)^5 + 5*a^4*b + 10*a^2*b^3 + b^5 + 6*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 2*(15*a^4*b + 30*a^2
*b^3 + 3*b^5 + 35*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 30*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)
^2)*sinh(d*x + c)^4 + 8*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^5 + 10*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh
(d*x + c)^3 + 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(5*a^4*b + 10*a^2*b^3 + b^5)*c
osh(d*x + c)^2 + 4*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^6 + 5*a^4*b + 10*a^2*b^3 + b^5 + 15*(5*a^4*b
+ 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 9*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8*((5*
a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^7 + 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^5 + 3*(5*a^4*b + 10*a
^2*b^3 + b^5)*cosh(d*x + c)^3 + (5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)
/(cosh(d*x + c) - sinh(d*x + c))) + 8*(3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*
x + c)^7 + 9*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5
)*d*x)*cosh(d*x + c)^5 + 3*(30*a^3*b^2 + 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^
2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^3 + (45*a^3*b^2 + 15*a^2*b^3 + 25*a*b^4 + 3*b^5 + 3*(a^5 - 5*a^4*b +
10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x
+ c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6
+ 8*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4
+ 30*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x +
c))*sinh(d*x + c)^3 + 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 + 15*d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)
^2 + d)*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*
sinh(d*x + c) + d)

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giac [A]  time = 0.19, size = 224, normalized size = 1.58 \[ \frac {3 \, {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (d x + c\right )} + 3 \, {\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {4 \, {\left (15 \, a^{3} b^{2} + 10 \, a b^{4} + 3 \, {\left (5 \, a^{3} b^{2} + 5 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 3 \, {\left (15 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 10 \, a b^{4} + b^{5}\right )} e^{\left (4 \, d x + 4 \, c\right )} + {\left (45 \, a^{3} b^{2} + 15 \, a^{2} b^{3} + 25 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(d*x+c))^5,x, algorithm="giac")

[Out]

1/3*(3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(d*x + c) + 3*(5*a^4*b + 10*a^2*b^3 + b^5)*lo
g(e^(2*d*x + 2*c) + 1) + 4*(15*a^3*b^2 + 10*a*b^4 + 3*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5)*e^(6*d*x + 6*c)
+ 3*(15*a^3*b^2 + 10*a^2*b^3 + 10*a*b^4 + b^5)*e^(4*d*x + 4*c) + (45*a^3*b^2 + 15*a^2*b^3 + 25*a*b^4 + 3*b^5)*
e^(2*d*x + 2*c))/(e^(2*d*x + 2*c) + 1)^4)/d

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maple [B]  time = 0.01, size = 322, normalized size = 2.27 \[ -\frac {\left (\tanh ^{2}\left (d x +c \right )\right ) b^{5}}{2 d}-\frac {a^{5} \ln \left (\tanh \left (d x +c \right )-1\right )}{2 d}-\frac {5 \ln \left (\tanh \left (d x +c \right )-1\right ) a^{4} b}{2 d}-\frac {5 \ln \left (\tanh \left (d x +c \right )-1\right ) a^{3} b^{2}}{d}-\frac {5 \ln \left (\tanh \left (d x +c \right )-1\right ) a^{2} b^{3}}{d}-\frac {5 \ln \left (\tanh \left (d x +c \right )-1\right ) a \,b^{4}}{2 d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) b^{5}}{2 d}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) a^{5}}{2 d}-\frac {5 \ln \left (1+\tanh \left (d x +c \right )\right ) a^{4} b}{2 d}+\frac {5 \ln \left (1+\tanh \left (d x +c \right )\right ) a^{3} b^{2}}{d}-\frac {5 \ln \left (1+\tanh \left (d x +c \right )\right ) a^{2} b^{3}}{d}+\frac {5 \ln \left (1+\tanh \left (d x +c \right )\right ) a \,b^{4}}{2 d}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) b^{5}}{2 d}-\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right ) a \,b^{4}}{3 d}-\frac {5 \left (\tanh ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{d}-\frac {5 a \,b^{4} \tanh \left (d x +c \right )}{d}-\frac {10 a^{3} b^{2} \tanh \left (d x +c \right )}{d}-\frac {b^{5} \left (\tanh ^{4}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tanh(d*x+c))^5,x)

[Out]

-1/2/d*tanh(d*x+c)^2*b^5-1/2/d*a^5*ln(tanh(d*x+c)-1)-5/2/d*ln(tanh(d*x+c)-1)*a^4*b-5/d*ln(tanh(d*x+c)-1)*a^3*b
^2-5/d*ln(tanh(d*x+c)-1)*a^2*b^3-5/2/d*ln(tanh(d*x+c)-1)*a*b^4-1/2/d*ln(tanh(d*x+c)-1)*b^5+1/2/d*ln(1+tanh(d*x
+c))*a^5-5/2/d*ln(1+tanh(d*x+c))*a^4*b+5/d*ln(1+tanh(d*x+c))*a^3*b^2-5/d*ln(1+tanh(d*x+c))*a^2*b^3+5/2/d*ln(1+
tanh(d*x+c))*a*b^4-1/2/d*ln(1+tanh(d*x+c))*b^5-5/3/d*tanh(d*x+c)^3*a*b^4-5/d*tanh(d*x+c)^2*a^2*b^3-5/d*a*b^4*t
anh(d*x+c)-10/d*a^3*b^2*tanh(d*x+c)-1/4/d*b^5*tanh(d*x+c)^4

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maxima [B]  time = 0.45, size = 310, normalized size = 2.18 \[ \frac {5}{3} \, a b^{4} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + b^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 10 \, a^{2} b^{3} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + 10 \, a^{3} b^{2} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{5} x + \frac {5 \, a^{4} b \log \left (\cosh \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(d*x+c))^5,x, algorithm="maxima")

[Out]

5/3*a*b^4*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x
- 4*c) + e^(-6*d*x - 6*c) + 1))) + b^5*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*
x - 4*c) + e^(-6*d*x - 6*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))) + 10*a^2*b^3*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^
(-4*d*x - 4*c) + 1))) + 10*a^3*b^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + a^5*x + 5*a^4*b*log(cosh(d*x + c
))/d

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mupad [B]  time = 1.16, size = 153, normalized size = 1.08 \[ x\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )-\frac {5\,\mathrm {tanh}\left (c+d\,x\right )\,\left (2\,a^3\,b^2+a\,b^4\right )}{d}-\frac {b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^4}{4\,d}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )\,\left (5\,a^4\,b+10\,a^2\,b^3+b^5\right )}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (10\,a^2\,b^3+b^5\right )}{2\,d}-\frac {5\,a\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^3}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x))^5,x)

[Out]

x*(5*a*b^4 + 5*a^4*b + a^5 + b^5 + 10*a^2*b^3 + 10*a^3*b^2) - (5*tanh(c + d*x)*(a*b^4 + 2*a^3*b^2))/d - (b^5*t
anh(c + d*x)^4)/(4*d) - (log(tanh(c + d*x) + 1)*(5*a^4*b + b^5 + 10*a^2*b^3))/d - (tanh(c + d*x)^2*(b^5 + 10*a
^2*b^3))/(2*d) - (5*a*b^4*tanh(c + d*x)^3)/(3*d)

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sympy [A]  time = 0.63, size = 211, normalized size = 1.49 \[ \begin {cases} a^{5} x + 5 a^{4} b x - \frac {5 a^{4} b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + 10 a^{3} b^{2} x - \frac {10 a^{3} b^{2} \tanh {\left (c + d x \right )}}{d} + 10 a^{2} b^{3} x - \frac {10 a^{2} b^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {5 a^{2} b^{3} \tanh ^{2}{\left (c + d x \right )}}{d} + 5 a b^{4} x - \frac {5 a b^{4} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {5 a b^{4} \tanh {\left (c + d x \right )}}{d} + b^{5} x - \frac {b^{5} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {b^{5} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{5} \tanh ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh {\relax (c )}\right )^{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(d*x+c))**5,x)

[Out]

Piecewise((a**5*x + 5*a**4*b*x - 5*a**4*b*log(tanh(c + d*x) + 1)/d + 10*a**3*b**2*x - 10*a**3*b**2*tanh(c + d*
x)/d + 10*a**2*b**3*x - 10*a**2*b**3*log(tanh(c + d*x) + 1)/d - 5*a**2*b**3*tanh(c + d*x)**2/d + 5*a*b**4*x -
5*a*b**4*tanh(c + d*x)**3/(3*d) - 5*a*b**4*tanh(c + d*x)/d + b**5*x - b**5*log(tanh(c + d*x) + 1)/d - b**5*tan
h(c + d*x)**4/(4*d) - b**5*tanh(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tanh(c))**5, True))

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