Integrand size = 21, antiderivative size = 63 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 a-b) x+\frac {(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \]
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {-4 b d x+12 a (c+d x)+8 a \sinh (2 (c+d x))+(a+b) \sinh (4 (c+d x))}{32 d} \]
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4158, 298, 215, 219}
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.
\(\displaystyle \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-b \tan (i c+i d x)^2}{\sec (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {b \tanh ^2(c+d x)+a}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{4} (3 a-b) \int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{4} (3 a-b) \left (\frac {1}{2} \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} (3 a-b) \left (\frac {1}{2} \text {arctanh}(\tanh (c+d x))+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
(((a + b)*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + ((3*a - b)*(ArcTanh [Tanh[c + d*x]]/2 + Tanh[c + d*x]/(2*(1 - Tanh[c + d*x]^2))))/4)/d
3.1.81.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 4.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(\frac {a \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(82\) |
default | \(\frac {a \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(82\) |
risch | \(\frac {3 a x}{8}-\frac {b x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b}{64 d}+\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} a}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} b}{64 d}\) | \(100\) |
1/d*(a*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c)+b*( 1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c))
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (3 \, a - b\right )} d x + {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + 4 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \]
1/8*((a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a - b)*d*x + ((a + b)*cosh (d*x + c)^3 + 4*a*cosh(d*x + c))*sinh(d*x + c))/d
\[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \cosh ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.65 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{64} \, b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \]
1/64*a*(24*x + e^(4*d*x + 4*c)/d + 8*e^(2*d*x + 2*c)/d - 8*e^(-2*d*x - 2*c )/d - e^(-4*d*x - 4*c)/d) - 1/64*b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^ (-4*d*x - 4*c)/d)
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.65 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} {\left (3 \, a - b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]
1/64*(8*(d*x + c)*(3*a - b) + a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 8*a* e^(2*d*x + 2*c) - (18*a*e^(4*d*x + 4*c) - 6*b*e^(4*d*x + 4*c) + 8*a*e^(2*d *x + 2*c) + a + b)*e^(-4*d*x - 4*c))/d
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.17 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=x\,\left (\frac {3\,a}{8}-\frac {b}{8}\right )-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{64\,d}-\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d} \]