Integrand size = 21, antiderivative size = 73 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {3}{8} (a+5 b) x-\frac {(5 a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac {b \tanh (c+d x)}{d} \]
3/8*(a+5*b)*x-1/8*(5*a+9*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*(a+b)*cosh(d*x+c )^3*sinh(d*x+c)/d-b*tanh(d*x+c)/d
Time = 0.69 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {12 (a+5 b) (c+d x)-8 (a+2 b) \sinh (2 (c+d x))+(a+b) \sinh (4 (c+d x))-32 b \tanh (c+d x)}{32 d} \]
(12*(a + 5*b)*(c + d*x) - 8*(a + 2*b)*Sinh[2*(c + d*x)] + (a + b)*Sinh[4*( c + d*x)] - 32*b*Tanh[c + d*x])/(32*d)
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4146, 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.
\(\displaystyle \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i c+i d x)^4 \left (a-b \tan (i c+i d x)^2\right )dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (b \tanh ^2(c+d x)+a\right )}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {\frac {1}{4} \int -\frac {4 b \tanh ^4(c+d x)+4 (a+b) \tanh ^2(c+d x)+a+b}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {4 b \tanh ^4(c+d x)+4 (a+b) \tanh ^2(c+d x)+a+b}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {8 b \tanh ^2(c+d x)+3 a+7 b}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {(5 a+9 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 (a+5 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-8 b \tanh (c+d x)\right )-\frac {(5 a+9 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} (3 (a+5 b) \text {arctanh}(\tanh (c+d x))-8 b \tanh (c+d x))-\frac {(5 a+9 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {(a+b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
(((a + b)*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + ((3*(a + 5*b)*ArcTa nh[Tanh[c + d*x]] - 8*b*Tanh[c + d*x])/2 - ((5*a + 9*b)*Tanh[c + d*x])/(2* (1 - Tanh[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[sin[(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[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 1.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(96\) |
default | \(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(96\) |
risch | \(\frac {3 a x}{8}+\frac {15 b x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b}{4 d}+\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b}{4 d}-\frac {{\mathrm e}^{-4 d x -4 c} a}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} b}{64 d}+\frac {2 b}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(149\) |
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*( 1/4*sinh(d*x+c)^5/cosh(d*x+c)-5/8*sinh(d*x+c)^3/cosh(d*x+c)+15/8*d*x+15/8* c-15/8*tanh(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.64 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {{\left (a + b\right )} \sinh \left (d x + c\right )^{5} + {\left (10 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} - 7 \, a - 15 \, b\right )} \sinh \left (d x + c\right )^{3} + 8 \, {\left (3 \, {\left (a + 5 \, b\right )} d x + 8 \, b\right )} \cosh \left (d x + c\right ) + {\left (5 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{4} - 3 \, {\left (7 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{2} - 8 \, a - 80 \, b\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]
1/64*((a + b)*sinh(d*x + c)^5 + (10*(a + b)*cosh(d*x + c)^2 - 7*a - 15*b)* sinh(d*x + c)^3 + 8*(3*(a + 5*b)*d*x + 8*b)*cosh(d*x + c) + (5*(a + b)*cos h(d*x + c)^4 - 3*(7*a + 15*b)*cosh(d*x + c)^2 - 8*a - 80*b)*sinh(d*x + c)) /(d*cosh(d*x + c))
\[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \sinh ^{4}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (67) = 134\).
Time = 0.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.11 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{64} \, b {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \]
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) + 1/64*b*(120*(d*x + c)/d + (16*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (15*e^(-2*d*x - 2*c) + 144*e^(-4*d*x - 4*c) - 1)/ (d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))))
Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (67) = 134\).
Time = 0.32 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.95 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (d x + c\right )} {\left (a + 5 \, b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + \frac {128 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \]
1/64*(24*(d*x + c)*(a + 5*b) + a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) - 8*a *e^(2*d*x + 2*c) - 16*b*e^(2*d*x + 2*c) - (18*a*e^(4*d*x + 4*c) + 90*b*e^( 4*d*x + 4*c) - 8*a*e^(2*d*x + 2*c) - 16*b*e^(2*d*x + 2*c) + a + b)*e^(-4*d *x - 4*c) + 128*b/(e^(2*d*x + 2*c) + 1))/d
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.38 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=x\,\left (\frac {3\,a}{8}+\frac {15\,b}{8}\right )+\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+2\,b\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{8\,d} \]