3.168 \(\int (a+b \tanh ^2(c+d x))^4 \, dx\)

Optimal. Leaf size=110 \[ -\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tanh ^3(c+d x)}{3 d}-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^3 (4 a+b) \tanh ^5(c+d x)}{5 d}+x (a+b)^4-\frac {b^4 \tanh ^7(c+d x)}{7 d} \]

[Out]

(a+b)^4*x-b*(2*a+b)*(2*a^2+2*a*b+b^2)*tanh(d*x+c)/d-1/3*b^2*(6*a^2+4*a*b+b^2)*tanh(d*x+c)^3/d-1/5*b^3*(4*a+b)*
tanh(d*x+c)^5/d-1/7*b^4*tanh(d*x+c)^7/d

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Rubi [A]  time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3661, 390, 206} \[ -\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tanh ^3(c+d x)}{3 d}-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^3 (4 a+b) \tanh ^5(c+d x)}{5 d}+x (a+b)^4-\frac {b^4 \tanh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tanh[c + d*x]^2)^4,x]

[Out]

(a + b)^4*x - (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Tanh[c + d*x])/d - (b^2*(6*a^2 + 4*a*b + b^2)*Tanh[c + d*x]^3
)/(3*d) - (b^3*(4*a + b)*Tanh[c + d*x]^5)/(5*d) - (b^4*Tanh[c + d*x]^7)/(7*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 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 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \tanh ^2(c+d x)\right )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^4}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-b (2 a+b) \left (2 a^2+2 a b+b^2\right )-b^2 \left (6 a^2+4 a b+b^2\right ) x^2-b^3 (4 a+b) x^4-b^4 x^6+\frac {(a+b)^4}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tanh ^3(c+d x)}{3 d}-\frac {b^3 (4 a+b) \tanh ^5(c+d x)}{5 d}-\frac {b^4 \tanh ^7(c+d x)}{7 d}+\frac {(a+b)^4 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=(a+b)^4 x-\frac {b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \tanh ^3(c+d x)}{3 d}-\frac {b^3 (4 a+b) \tanh ^5(c+d x)}{5 d}-\frac {b^4 \tanh ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 128, normalized size = 1.16 \[ \frac {\tanh (c+d x) \left (\frac {105 (a+b)^4 \tanh ^{-1}\left (\sqrt {\tanh ^2(c+d x)}\right )}{\sqrt {\tanh ^2(c+d x)}}-b \left (35 b \left (6 a^2+4 a b+b^2\right ) \tanh ^2(c+d x)+105 \left (4 a^3+6 a^2 b+4 a b^2+b^3\right )+21 b^2 (4 a+b) \tanh ^4(c+d x)+15 b^3 \tanh ^6(c+d x)\right )\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tanh[c + d*x]*((105*(a + b)^4*ArcTanh[Sqrt[Tanh[c + d*x]^2]])/Sqrt[Tanh[c + d*x]^2] - b*(105*(4*a^3 + 6*a^2*b
+ 4*a*b^2 + b^3) + 35*b*(6*a^2 + 4*a*b + b^2)*Tanh[c + d*x]^2 + 21*b^2*(4*a + b)*Tanh[c + d*x]^4 + 15*b^3*Tan
h[c + d*x]^6)))/(105*d)

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fricas [B]  time = 0.43, size = 1176, normalized size = 10.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*c
osh(d*x + c)^7 + 7*(420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 +
b^4)*d*x)*cosh(d*x + c)*sinh(d*x + c)^6 - 4*(105*a^3*b + 210*a^2*b^2 + 161*a*b^3 + 44*b^4)*sinh(d*x + c)^7 +
7*(420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d
*x + c)^5 - 28*(75*a^3*b + 120*a^2*b^2 + 71*a*b^3 + 14*b^4 + 3*(105*a^3*b + 210*a^2*b^2 + 161*a*b^3 + 44*b^4)*
cosh(d*x + c)^2)*sinh(d*x + c)^5 + 35*((420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6
*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c)^3 + (420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4
*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c))*sinh(d*x + c)^4 + 21*(420*a^3*b + 840*a^2*b^2 + 644*a*
b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c)^3 - 28*(5*(105*a^3*b + 210*
a^2*b^2 + 161*a*b^3 + 44*b^4)*cosh(d*x + c)^4 + 135*a^3*b + 180*a^2*b^2 + 123*a*b^3 + 42*b^4 + 10*(75*a^3*b +
120*a^2*b^2 + 71*a*b^3 + 14*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 7*(3*(420*a^3*b + 840*a^2*b^2 + 644*a*b^3
+ 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c)^5 + 10*(420*a^3*b + 840*a^2*b^2
+ 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c)^3 + 9*(420*a^3*b +
840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*x)*cosh(d*x + c))*sinh(
d*x + c)^2 + 35*(420*a^3*b + 840*a^2*b^2 + 644*a*b^3 + 176*b^4 + 105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^
4)*d*x)*cosh(d*x + c) - 28*((105*a^3*b + 210*a^2*b^2 + 161*a*b^3 + 44*b^4)*cosh(d*x + c)^6 + 5*(75*a^3*b + 120
*a^2*b^2 + 71*a*b^3 + 14*b^4)*cosh(d*x + c)^4 + 75*a^3*b + 90*a^2*b^2 + 75*a*b^3 + 9*(45*a^3*b + 60*a^2*b^2 +
41*a*b^3 + 14*b^4)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + 7*
d*cosh(d*x + c)^5 + 35*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^4 + 21*d*cosh(d*x + c)^3 + 7*(3*d*c
osh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 9*d*cosh(d*x + c))*sinh(d*x + c)^2 + 35*d*cosh(d*x + c))

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giac [B]  time = 0.22, size = 447, normalized size = 4.06 \[ \frac {105 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (d x + c\right )} + \frac {8 \, {\left (105 \, a^{3} b e^{\left (12 \, d x + 12 \, c\right )} + 315 \, a^{2} b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 315 \, a b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 105 \, b^{4} e^{\left (12 \, d x + 12 \, c\right )} + 630 \, a^{3} b e^{\left (10 \, d x + 10 \, c\right )} + 1575 \, a^{2} b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 1260 \, a b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 315 \, b^{4} e^{\left (10 \, d x + 10 \, c\right )} + 1575 \, a^{3} b e^{\left (8 \, d x + 8 \, c\right )} + 3360 \, a^{2} b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2555 \, a b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 770 \, b^{4} e^{\left (8 \, d x + 8 \, c\right )} + 2100 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} + 3990 \, a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 3080 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 770 \, b^{4} e^{\left (6 \, d x + 6 \, c\right )} + 1575 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} + 2835 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2121 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 609 \, b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 630 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} + 1155 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 812 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 203 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 105 \, a^{3} b + 210 \, a^{2} b^{2} + 161 \, a b^{3} + 44 \, b^{4}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{7}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(d*x + c) + 8*(105*a^3*b*e^(12*d*x + 12*c) + 315*a^2*b^
2*e^(12*d*x + 12*c) + 315*a*b^3*e^(12*d*x + 12*c) + 105*b^4*e^(12*d*x + 12*c) + 630*a^3*b*e^(10*d*x + 10*c) +
1575*a^2*b^2*e^(10*d*x + 10*c) + 1260*a*b^3*e^(10*d*x + 10*c) + 315*b^4*e^(10*d*x + 10*c) + 1575*a^3*b*e^(8*d*
x + 8*c) + 3360*a^2*b^2*e^(8*d*x + 8*c) + 2555*a*b^3*e^(8*d*x + 8*c) + 770*b^4*e^(8*d*x + 8*c) + 2100*a^3*b*e^
(6*d*x + 6*c) + 3990*a^2*b^2*e^(6*d*x + 6*c) + 3080*a*b^3*e^(6*d*x + 6*c) + 770*b^4*e^(6*d*x + 6*c) + 1575*a^3
*b*e^(4*d*x + 4*c) + 2835*a^2*b^2*e^(4*d*x + 4*c) + 2121*a*b^3*e^(4*d*x + 4*c) + 609*b^4*e^(4*d*x + 4*c) + 630
*a^3*b*e^(2*d*x + 2*c) + 1155*a^2*b^2*e^(2*d*x + 2*c) + 812*a*b^3*e^(2*d*x + 2*c) + 203*b^4*e^(2*d*x + 2*c) +
105*a^3*b + 210*a^2*b^2 + 161*a*b^3 + 44*b^4)/(e^(2*d*x + 2*c) + 1)^7)/d

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maple [B]  time = 0.02, size = 344, normalized size = 3.13 \[ \frac {\ln \left (1+\tanh \left (d x +c \right )\right ) a^{4}}{2 d}+\frac {2 \ln \left (1+\tanh \left (d x +c \right )\right ) a^{3} b}{d}+\frac {3 \ln \left (1+\tanh \left (d x +c \right )\right ) a^{2} b^{2}}{d}+\frac {2 \ln \left (1+\tanh \left (d x +c \right )\right ) a \,b^{3}}{d}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) b^{4}}{2 d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) a^{4}}{2 d}-\frac {2 \ln \left (\tanh \left (d x +c \right )-1\right ) a^{3} b}{d}-\frac {3 \ln \left (\tanh \left (d x +c \right )-1\right ) a^{2} b^{2}}{d}-\frac {2 \ln \left (\tanh \left (d x +c \right )-1\right ) a \,b^{3}}{d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) b^{4}}{2 d}-\frac {\left (\tanh ^{5}\left (d x +c \right )\right ) b^{4}}{5 d}-\frac {b^{4} \tanh \left (d x +c \right )}{d}-\frac {b^{4} \left (\tanh ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{4} \left (\tanh ^{7}\left (d x +c \right )\right )}{7 d}-\frac {2 \left (\tanh ^{3}\left (d x +c \right )\right ) a^{2} b^{2}}{d}-\frac {4 \left (\tanh ^{3}\left (d x +c \right )\right ) a \,b^{3}}{3 d}-\frac {4 \left (\tanh ^{5}\left (d x +c \right )\right ) a \,b^{3}}{5 d}-\frac {4 a \,b^{3} \tanh \left (d x +c \right )}{d}-\frac {4 a^{3} b \tanh \left (d x +c \right )}{d}-\frac {6 a^{2} b^{2} \tanh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tanh(d*x+c)^2)^4,x)

[Out]

1/2/d*ln(1+tanh(d*x+c))*a^4+2/d*ln(1+tanh(d*x+c))*a^3*b+3/d*ln(1+tanh(d*x+c))*a^2*b^2+2/d*ln(1+tanh(d*x+c))*a*
b^3+1/2/d*ln(1+tanh(d*x+c))*b^4-1/2/d*ln(tanh(d*x+c)-1)*a^4-2/d*ln(tanh(d*x+c)-1)*a^3*b-3/d*ln(tanh(d*x+c)-1)*
a^2*b^2-2/d*ln(tanh(d*x+c)-1)*a*b^3-1/2/d*ln(tanh(d*x+c)-1)*b^4-1/5/d*tanh(d*x+c)^5*b^4-1/d*b^4*tanh(d*x+c)-1/
3/d*tanh(d*x+c)^3*b^4-1/7*b^4*tanh(d*x+c)^7/d-2/d*tanh(d*x+c)^3*a^2*b^2-4/3/d*tanh(d*x+c)^3*a*b^3-4/5/d*tanh(d
*x+c)^5*a*b^3-4/d*a*b^3*tanh(d*x+c)-4/d*a^3*b*tanh(d*x+c)-6/d*a^2*b^2*tanh(d*x+c)

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maxima [B]  time = 0.35, size = 410, normalized size = 3.73 \[ \frac {1}{105} \, b^{4} {\left (105 \, x + \frac {105 \, c}{d} - \frac {8 \, {\left (203 \, e^{\left (-2 \, d x - 2 \, c\right )} + 609 \, e^{\left (-4 \, d x - 4 \, c\right )} + 770 \, e^{\left (-6 \, d x - 6 \, c\right )} + 770 \, e^{\left (-8 \, d x - 8 \, c\right )} + 315 \, e^{\left (-10 \, d x - 10 \, c\right )} + 105 \, e^{\left (-12 \, d x - 12 \, c\right )} + 44\right )}}{d {\left (7 \, e^{\left (-2 \, d x - 2 \, c\right )} + 21 \, e^{\left (-4 \, d x - 4 \, c\right )} + 35 \, e^{\left (-6 \, d x - 6 \, c\right )} + 35 \, e^{\left (-8 \, d x - 8 \, c\right )} + 21 \, e^{\left (-10 \, d x - 10 \, c\right )} + 7 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )} + 1\right )}}\right )} + \frac {4}{15} \, a b^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 2 \, a^{2} b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + 4 \, a^{3} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{4} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*b^4*(105*x + 105*c/d - 8*(203*e^(-2*d*x - 2*c) + 609*e^(-4*d*x - 4*c) + 770*e^(-6*d*x - 6*c) + 770*e^(-8
*d*x - 8*c) + 315*e^(-10*d*x - 10*c) + 105*e^(-12*d*x - 12*c) + 44)/(d*(7*e^(-2*d*x - 2*c) + 21*e^(-4*d*x - 4*
c) + 35*e^(-6*d*x - 6*c) + 35*e^(-8*d*x - 8*c) + 21*e^(-10*d*x - 10*c) + 7*e^(-12*d*x - 12*c) + e^(-14*d*x - 1
4*c) + 1))) + 4/15*a*b^3*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) + 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c)
+ 45*e^(-8*d*x - 8*c) + 23)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x -
8*c) + e^(-10*d*x - 10*c) + 1))) + 2*a^2*b^2*(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))) + 4*a^3*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*
c) + 1))) + a^4*x

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mupad [B]  time = 0.20, size = 133, normalized size = 1.21 \[ x\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{3\,d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^5\,\left (b^4+4\,a\,b^3\right )}{5\,d}-\frac {b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^7}{7\,d}-\frac {b\,\mathrm {tanh}\left (c+d\,x\right )\,\left (4\,a^3+6\,a^2\,b+4\,a\,b^2+b^3\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^2)^4,x)

[Out]

x*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2) - (tanh(c + d*x)^3*(4*a*b^3 + b^4 + 6*a^2*b^2))/(3*d) - (tanh(c
+ d*x)^5*(4*a*b^3 + b^4))/(5*d) - (b^4*tanh(c + d*x)^7)/(7*d) - (b*tanh(c + d*x)*(4*a*b^2 + 6*a^2*b + 4*a^3 +
b^3))/d

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sympy [A]  time = 1.28, size = 209, normalized size = 1.90 \[ \begin {cases} a^{4} x + 4 a^{3} b x - \frac {4 a^{3} b \tanh {\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x - \frac {2 a^{2} b^{2} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac {6 a^{2} b^{2} \tanh {\left (c + d x \right )}}{d} + 4 a b^{3} x - \frac {4 a b^{3} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {4 a b^{3} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a b^{3} \tanh {\left (c + d x \right )}}{d} + b^{4} x - \frac {b^{4} \tanh ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{4} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{4} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{4} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(d*x+c)**2)**4,x)

[Out]

Piecewise((a**4*x + 4*a**3*b*x - 4*a**3*b*tanh(c + d*x)/d + 6*a**2*b**2*x - 2*a**2*b**2*tanh(c + d*x)**3/d - 6
*a**2*b**2*tanh(c + d*x)/d + 4*a*b**3*x - 4*a*b**3*tanh(c + d*x)**5/(5*d) - 4*a*b**3*tanh(c + d*x)**3/(3*d) -
4*a*b**3*tanh(c + d*x)/d + b**4*x - b**4*tanh(c + d*x)**7/(7*d) - b**4*tanh(c + d*x)**5/(5*d) - b**4*tanh(c +
d*x)**3/(3*d) - b**4*tanh(c + d*x)/d, Ne(d, 0)), (x*(a + b*tanh(c)**2)**4, True))

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