Integrand size = 23, antiderivative size = 83 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {(a+b)^2 \tanh (c+d x)}{d}-\frac {(a+b)^2 \tanh ^3(c+d x)}{3 d}-\frac {b (2 a+b) \tanh ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^7(c+d x)}{7 d} \] Output:
(a+b)^2*x-(a+b)^2*tanh(d*x+c)/d-1/3*(a+b)^2*tanh(d*x+c)^3/d-1/5*b*(2*a+b)* tanh(d*x+c)^5/d-1/7*b^2*tanh(d*x+c)^7/d
Leaf count is larger than twice the leaf count of optimal. \(190\) vs. \(2(83)=166\).
Time = 0.06 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.29 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \text {arctanh}(\tanh (c+d x))}{d}+\frac {2 a b \text {arctanh}(\tanh (c+d x))}{d}+\frac {b^2 \text {arctanh}(\tanh (c+d x))}{d}-\frac {a^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d}-\frac {a^2 \tanh ^3(c+d x)}{3 d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}-\frac {2 a b \tanh ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^7(c+d x)}{7 d} \] Input:
Integrate[Tanh[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(a^2*ArcTanh[Tanh[c + d*x]])/d + (2*a*b*ArcTanh[Tanh[c + d*x]])/d + (b^2*A rcTanh[Tanh[c + d*x]])/d - (a^2*Tanh[c + d*x])/d - (2*a*b*Tanh[c + d*x])/d - (b^2*Tanh[c + d*x])/d - (a^2*Tanh[c + d*x]^3)/(3*d) - (2*a*b*Tanh[c + d *x]^3)/(3*d) - (b^2*Tanh[c + d*x]^3)/(3*d) - (2*a*b*Tanh[c + d*x]^5)/(5*d) - (b^2*Tanh[c + d*x]^5)/(5*d) - (b^2*Tanh[c + d*x]^7)/(7*d)
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4153, 364, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (i c+i d x)^4 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {\int \left (-b^2 \tanh ^6(c+d x)-b (2 a+b) \tanh ^4(c+d x)-(a+b)^2 \tanh ^2(c+d x)-(a+b)^2+\frac {a^2+2 b a+b^2}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b)^2 \text {arctanh}(\tanh (c+d x))-\frac {1}{5} b (2 a+b) \tanh ^5(c+d x)-\frac {1}{3} (a+b)^2 \tanh ^3(c+d x)-(a+b)^2 \tanh (c+d x)-\frac {1}{7} b^2 \tanh ^7(c+d x)}{d}\) |
Input:
Int[Tanh[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
((a + b)^2*ArcTanh[Tanh[c + d*x]] - (a + b)^2*Tanh[c + d*x] - ((a + b)^2*T anh[c + d*x]^3)/3 - (b*(2*a + b)*Tanh[c + d*x]^5)/5 - (b^2*Tanh[c + d*x]^7 )/7)/d
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.63
| method | result | size |
| parallelrisch | \(-\frac {15 b^{2} \tanh \left (d x +c \right )^{7}+42 a b \tanh \left (d x +c \right )^{5}+21 b^{2} \tanh \left (d x +c \right )^{5}+35 a^{2} \tanh \left (d x +c \right )^{3}+70 a b \tanh \left (d x +c \right )^{3}+35 b^{2} \tanh \left (d x +c \right )^{3}-105 a^{2} d x -210 a b d x -105 b^{2} d x +105 a^{2} \tanh \left (d x +c \right )+210 a b \tanh \left (d x +c \right )+105 b^{2} \tanh \left (d x +c \right )}{105 d}\) | \(135\) |
| derivativedivides | \(\frac {-2 a b \tanh \left (d x +c \right )-\frac {2 a b \tanh \left (d x +c \right )^{5}}{5}-\frac {2 a b \tanh \left (d x +c \right )^{3}}{3}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{5}}{5}-\frac {a^{2} \tanh \left (d x +c \right )^{3}}{3}-\frac {b^{2} \tanh \left (d x +c \right )^{3}}{3}-a^{2} \tanh \left (d x +c \right )-b^{2} \tanh \left (d x +c \right )-\frac {b^{2} \tanh \left (d x +c \right )^{7}}{7}}{d}\) | \(158\) |
| default | \(\frac {-2 a b \tanh \left (d x +c \right )-\frac {2 a b \tanh \left (d x +c \right )^{5}}{5}-\frac {2 a b \tanh \left (d x +c \right )^{3}}{3}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{5}}{5}-\frac {a^{2} \tanh \left (d x +c \right )^{3}}{3}-\frac {b^{2} \tanh \left (d x +c \right )^{3}}{3}-a^{2} \tanh \left (d x +c \right )-b^{2} \tanh \left (d x +c \right )-\frac {b^{2} \tanh \left (d x +c \right )^{7}}{7}}{d}\) | \(158\) |
| parts | \(\frac {a^{2} \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\tanh \left (d x +c \right )^{7}}{7}-\frac {\tanh \left (d x +c \right )^{5}}{5}-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \left (-\frac {\tanh \left (d x +c \right )^{5}}{5}-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}\) | \(176\) |
| risch | \(a^{2} x +2 a b x +b^{2} x +\frac {4 a^{2} {\mathrm e}^{12 d x +12 c}+12 a b \,{\mathrm e}^{12 d x +12 c}+8 b^{2} {\mathrm e}^{12 d x +12 c}+20 a^{2} {\mathrm e}^{10 d x +10 c}+48 a b \,{\mathrm e}^{10 d x +10 c}+24 b^{2} {\mathrm e}^{10 d x +10 c}+\frac {128 a^{2} {\mathrm e}^{8 d x +8 c}}{3}+\frac {292 a b \,{\mathrm e}^{8 d x +8 c}}{3}+\frac {176 b^{2} {\mathrm e}^{8 d x +8 c}}{3}+\frac {152 a^{2} {\mathrm e}^{6 d x +6 c}}{3}+\frac {352 a b \,{\mathrm e}^{6 d x +6 c}}{3}+\frac {176 b^{2} {\mathrm e}^{6 d x +6 c}}{3}+36 a^{2} {\mathrm e}^{4 d x +4 c}+\frac {404 a b \,{\mathrm e}^{4 d x +4 c}}{5}+\frac {232 b^{2} {\mathrm e}^{4 d x +4 c}}{5}+\frac {44 a^{2} {\mathrm e}^{2 d x +2 c}}{3}+\frac {464 a b \,{\mathrm e}^{2 d x +2 c}}{15}+\frac {232 b^{2} {\mathrm e}^{2 d x +2 c}}{15}+\frac {8 a^{2}}{3}+\frac {92 a b}{15}+\frac {352 b^{2}}{105}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{7}}\) | \(296\) |
Input:
int(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/105*(15*b^2*tanh(d*x+c)^7+42*a*b*tanh(d*x+c)^5+21*b^2*tanh(d*x+c)^5+35* a^2*tanh(d*x+c)^3+70*a*b*tanh(d*x+c)^3+35*b^2*tanh(d*x+c)^3-105*a^2*d*x-21 0*a*b*d*x-105*b^2*d*x+105*a^2*tanh(d*x+c)+210*a*b*tanh(d*x+c)+105*b^2*tanh (d*x+c))/d
Leaf count of result is larger than twice the leaf count of optimal. 796 vs. \(2 (77) = 154\).
Time = 0.09 (sec) , antiderivative size = 796, normalized size of antiderivative = 9.59 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
integrate(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/105*((105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d* x + c)^7 + 7*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*c osh(d*x + c)*sinh(d*x + c)^6 - 2*(70*a^2 + 161*a*b + 88*b^2)*sinh(d*x + c) ^7 + 7*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d* x + c)^5 - 14*(3*(70*a^2 + 161*a*b + 88*b^2)*cosh(d*x + c)^2 + 40*a^2 + 71 *a*b + 28*b^2)*sinh(d*x + c)^5 + 35*((105*(a^2 + 2*a*b + b^2)*d*x + 140*a^ 2 + 322*a*b + 176*b^2)*cosh(d*x + c)^3 + (105*(a^2 + 2*a*b + b^2)*d*x + 14 0*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 + 21*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^3 - 14*(5*( 70*a^2 + 161*a*b + 88*b^2)*cosh(d*x + c)^4 + 10*(40*a^2 + 71*a*b + 28*b^2) *cosh(d*x + c)^2 + 60*a^2 + 123*a*b + 84*b^2)*sinh(d*x + c)^3 + 7*(3*(105* (a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^5 + 1 0*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c )^3 + 9*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d *x + c))*sinh(d*x + c)^2 + 35*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322 *a*b + 176*b^2)*cosh(d*x + c) - 14*((70*a^2 + 161*a*b + 88*b^2)*cosh(d*x + c)^6 + 5*(40*a^2 + 71*a*b + 28*b^2)*cosh(d*x + c)^4 + 9*(20*a^2 + 41*a*b + 28*b^2)*cosh(d*x + c)^2 + 30*a^2 + 75*a*b)*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*cos h(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^4 + 21*d*cosh(d*x + c)^3 ...
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (70) = 140\).
Time = 0.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.99 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\begin {cases} a^{2} x - \frac {a^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \tanh {\left (c + d x \right )}}{d} + 2 a b x - \frac {2 a b \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 a b \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b \tanh {\left (c + d x \right )}}{d} + b^{2} x - \frac {b^{2} \tanh ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{2} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{2} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{2} \tanh ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tanh(d*x+c)**4*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Piecewise((a**2*x - a**2*tanh(c + d*x)**3/(3*d) - a**2*tanh(c + d*x)/d + 2 *a*b*x - 2*a*b*tanh(c + d*x)**5/(5*d) - 2*a*b*tanh(c + d*x)**3/(3*d) - 2*a *b*tanh(c + d*x)/d + b**2*x - b**2*tanh(c + d*x)**7/(7*d) - b**2*tanh(c + d*x)**5/(5*d) - b**2*tanh(c + d*x)**3/(3*d) - b**2*tanh(c + d*x)/d, Ne(d, 0)), (x*(a + b*tanh(c)**2)**2*tanh(c)**4, True))
Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (77) = 154\).
Time = 0.05 (sec) , antiderivative size = 369, normalized size of antiderivative = 4.45 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{105} \, b^{2} {\left (105 \, x + \frac {105 \, c}{d} - \frac {8 \, {\left (203 \, e^{\left (-2 \, d x - 2 \, c\right )} + 609 \, e^{\left (-4 \, d x - 4 \, c\right )} + 770 \, e^{\left (-6 \, d x - 6 \, c\right )} + 770 \, e^{\left (-8 \, d x - 8 \, c\right )} + 315 \, e^{\left (-10 \, d x - 10 \, c\right )} + 105 \, e^{\left (-12 \, d x - 12 \, c\right )} + 44\right )}}{d {\left (7 \, e^{\left (-2 \, d x - 2 \, c\right )} + 21 \, e^{\left (-4 \, d x - 4 \, c\right )} + 35 \, e^{\left (-6 \, d x - 6 \, c\right )} + 35 \, e^{\left (-8 \, d x - 8 \, c\right )} + 21 \, e^{\left (-10 \, d x - 10 \, c\right )} + 7 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )} + 1\right )}}\right )} + \frac {2}{15} \, a b {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {1}{3} \, a^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \] Input:
integrate(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/105*b^2*(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 - 14*c) + 1))) + 2/15*a*b*(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))) + 1/3*a^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)))
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (77) = 154\).
Time = 0.21 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.61 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {105 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} + \frac {4 \, {\left (105 \, a^{2} e^{\left (12 \, d x + 12 \, c\right )} + 315 \, a b e^{\left (12 \, d x + 12 \, c\right )} + 210 \, b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 525 \, a^{2} e^{\left (10 \, d x + 10 \, c\right )} + 1260 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 630 \, b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 1120 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2555 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 1540 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1330 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 3080 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 1540 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 945 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2121 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 1218 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 385 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 812 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 406 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 70 \, a^{2} + 161 \, a b + 88 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{7}}}{105 \, d} \] Input:
integrate(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/105*(105*(a^2 + 2*a*b + b^2)*(d*x + c) + 4*(105*a^2*e^(12*d*x + 12*c) + 315*a*b*e^(12*d*x + 12*c) + 210*b^2*e^(12*d*x + 12*c) + 525*a^2*e^(10*d*x + 10*c) + 1260*a*b*e^(10*d*x + 10*c) + 630*b^2*e^(10*d*x + 10*c) + 1120*a^ 2*e^(8*d*x + 8*c) + 2555*a*b*e^(8*d*x + 8*c) + 1540*b^2*e^(8*d*x + 8*c) + 1330*a^2*e^(6*d*x + 6*c) + 3080*a*b*e^(6*d*x + 6*c) + 1540*b^2*e^(6*d*x + 6*c) + 945*a^2*e^(4*d*x + 4*c) + 2121*a*b*e^(4*d*x + 4*c) + 1218*b^2*e^(4* d*x + 4*c) + 385*a^2*e^(2*d*x + 2*c) + 812*a*b*e^(2*d*x + 2*c) + 406*b^2*e ^(2*d*x + 2*c) + 70*a^2 + 161*a*b + 88*b^2)/(e^(2*d*x + 2*c) + 1)^7)/d
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,\left (a^2+2\,a\,b+b^2\right )-\frac {\mathrm {tanh}\left (c+d\,x\right )\,{\left (a+b\right )}^2}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^5\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^7}{7\,d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (a^2+2\,a\,b+b^2\right )}{3\,d} \] Input:
int(tanh(c + d*x)^4*(a + b*tanh(c + d*x)^2)^2,x)
Output:
x*(2*a*b + a^2 + b^2) - (tanh(c + d*x)*(a + b)^2)/d - (tanh(c + d*x)^5*(2* a*b + b^2))/(5*d) - (b^2*tanh(c + d*x)^7)/(7*d) - (tanh(c + d*x)^3*(2*a*b + a^2 + b^2))/(3*d)
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {-15 \tanh \left (d x +c \right )^{7} b^{2}-42 \tanh \left (d x +c \right )^{5} a b -21 \tanh \left (d x +c \right )^{5} b^{2}-35 \tanh \left (d x +c \right )^{3} a^{2}-70 \tanh \left (d x +c \right )^{3} a b -35 \tanh \left (d x +c \right )^{3} b^{2}-105 \tanh \left (d x +c \right ) a^{2}-210 \tanh \left (d x +c \right ) a b -105 \tanh \left (d x +c \right ) b^{2}+105 a^{2} d x +210 a b d x +105 b^{2} d x}{105 d} \] Input:
int(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x)
Output:
( - 15*tanh(c + d*x)**7*b**2 - 42*tanh(c + d*x)**5*a*b - 21*tanh(c + d*x)* *5*b**2 - 35*tanh(c + d*x)**3*a**2 - 70*tanh(c + d*x)**3*a*b - 35*tanh(c + d*x)**3*b**2 - 105*tanh(c + d*x)*a**2 - 210*tanh(c + d*x)*a*b - 105*tanh( c + d*x)*b**2 + 105*a**2*d*x + 210*a*b*d*x + 105*b**2*d*x)/(105*d)