Integrand size = 21, antiderivative size = 70 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {3}{8} (a-4 b) x-\frac {(5 a-4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b \tanh (c+d x)}{d} \]
3/8*(a-4*b)*x-1/8*(5*a-4*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*a*cosh(d*x+c)^3* sinh(d*x+c)/d+b*tanh(d*x+c)/d
Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.77 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {12 (a-4 b) (c+d x)-8 (a-b) \sinh (2 (c+d x))+a \sinh (4 (c+d x))+32 b \tanh (c+d x)}{32 d} \]
(12*(a - 4*b)*(c + d*x) - 8*(a - b)*Sinh[2*(c + d*x)] + a*Sinh[4*(c + d*x) ] + 32*b*Tanh[c + d*x])/(32*d)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4620, 360, 25, 1471, 299, 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.
\(\displaystyle \int \sinh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i c+i d x)^4 \left (a+b \sec (i c+i d x)^2\right )dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {\frac {1}{4} \int -\frac {-4 b \tanh ^4(c+d x)+4 a \tanh ^2(c+d x)+a}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {a \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {a \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {-4 b \tanh ^4(c+d x)+4 a \tanh ^2(c+d x)+a}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {-8 b \tanh ^2(c+d x)+3 a-4 b}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {(5 a-4 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 (a-4 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+8 b \tanh (c+d x)\right )-\frac {(5 a-4 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} (3 (a-4 b) \text {arctanh}(\tanh (c+d x))+8 b \tanh (c+d x))-\frac {(5 a-4 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
((a*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + ((3*(a - 4*b)*ArcTanh[Tan h[c + d*x]] + 8*b*Tanh[c + d*x])/2 - ((5*a - 4*b)*Tanh[c + d*x])/(2*(1 - T anh[c + d*x]^2)))/4)/d
3.1.1.3.1 Defintions of rubi rules used
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[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 10.95 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(78\) |
default | \(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(78\) |
parts | \(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(80\) |
risch | \(\frac {3 a x}{8}-\frac {3 b x}{2}+\frac {a \,{\mathrm e}^{4 d x +4 c}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}+\frac {{\mathrm e}^{2 d x +2 c} b}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} b}{8 d}-\frac {a \,{\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {2 b}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(119\) |
1/d*(a*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+b*( 1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c)))
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.63 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {a \sinh \left (d x + c\right )^{5} + {\left (10 \, a \cosh \left (d x + c\right )^{2} - 7 \, a + 8 \, b\right )} \sinh \left (d x + c\right )^{3} + 8 \, {\left (3 \, {\left (a - 4 \, b\right )} d x - 8 \, b\right )} \cosh \left (d x + c\right ) + {\left (5 \, a \cosh \left (d x + c\right )^{4} - 3 \, {\left (7 \, a - 8 \, b\right )} \cosh \left (d x + c\right )^{2} - 8 \, a + 72 \, b\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]
1/64*(a*sinh(d*x + c)^5 + (10*a*cosh(d*x + c)^2 - 7*a + 8*b)*sinh(d*x + c) ^3 + 8*(3*(a - 4*b)*d*x - 8*b)*cosh(d*x + c) + (5*a*cosh(d*x + c)^4 - 3*(7 *a - 8*b)*cosh(d*x + c)^2 - 8*a + 72*b)*sinh(d*x + c))/(d*cosh(d*x + c))
\[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \sinh ^{4}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (64) = 128\).
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.84 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{8} \, b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]
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) - 1/8*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - ( 17*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (64) = 128\).
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.86 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {24 \, {\left (d x + c\right )} {\left (a - 4 \, b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} - 72 \, b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {128 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \]
1/64*(24*(d*x + c)*(a - 4*b) + a*e^(4*d*x + 4*c) - 8*a*e^(2*d*x + 2*c) + 8 *b*e^(2*d*x + 2*c) - (18*a*e^(4*d*x + 4*c) - 72*b*e^(4*d*x + 4*c) - 8*a*e^ (2*d*x + 2*c) + 8*b*e^(2*d*x + 2*c) + a)*e^(-4*d*x - 4*c) - 128*b/(e^(2*d* x + 2*c) + 1))/d
Time = 2.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \sinh ^4(c+d x) \, dx=\frac {3\,a\,x}{8}-\frac {3\,b\,x}{2}-\frac {a\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{32\,d}+\frac {b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4\,d}+\frac {b\,\mathrm {sinh}\left (c+d\,x\right )}{d\,\mathrm {cosh}\left (c+d\,x\right )} \]