Integrand size = 23, antiderivative size = 92 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3}{8} b \left (8 a^2-12 a b+5 b^2\right ) x+\frac {3 (4 a-3 b) b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {b^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {(a-b)^3 \tanh (c+d x)}{d} \] Output:
3/8*b*(8*a^2-12*a*b+5*b^2)*x+3/8*(4*a-3*b)*b^2*cosh(d*x+c)*sinh(d*x+c)/d+1 /4*b^3*cosh(d*x+c)^3*sinh(d*x+c)/d+(a-b)^3*tanh(d*x+c)/d
Time = 1.62 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {12 b \left (8 a^2-12 a b+5 b^2\right ) (c+d x)+8 (3 a-2 b) b^2 \sinh (2 (c+d x))+b^3 \sinh (4 (c+d x))+32 (a-b)^3 \tanh (c+d x)}{32 d} \] Input:
Integrate[Sech[c + d*x]^2*(a + b*Sinh[c + d*x]^2)^3,x]
Output:
(12*b*(8*a^2 - 12*a*b + 5*b^2)*(c + d*x) + 8*(3*a - 2*b)*b^2*Sinh[2*(c + d *x)] + b^3*Sinh[4*(c + d*x)] + 32*(a - b)^3*Tanh[c + d*x])/(32*d)
Time = 0.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 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.
\(\displaystyle \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \sin (i c+i d x)^2\right )^3}{\cos (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {\left (a-(a-b) \tanh ^2(c+d x)\right )^3}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left ((a-b)^3+\frac {3 (a-b)^2 b \tanh ^4(c+d x)-3 (a-b) (2 a-b) b \tanh ^2(c+d x)+b \left (3 a^2-3 b a+b^2\right )}{\left (1-\tanh ^2(c+d x)\right )^3}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3}{8} b \left (8 a^2-12 a b+5 b^2\right ) \text {arctanh}(\tanh (c+d x))+\frac {3 b^2 (4 a-3 b) \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )}+(a-b)^3 \tanh (c+d x)+\frac {b^3 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
Input:
Int[Sech[c + d*x]^2*(a + b*Sinh[c + d*x]^2)^3,x]
Output:
((3*b*(8*a^2 - 12*a*b + 5*b^2)*ArcTanh[Tanh[c + d*x]])/8 + (a - b)^3*Tanh[ c + d*x] + (b^3*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + (3*(4*a - 3*b )*b^2*Tanh[c + d*x])/(8*(1 - Tanh[c + d*x]^2)))/d
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.42
\[\frac {\tanh \left (d x +c \right ) a^{3}+3 a^{2} b \left (d x +c -\tanh \left (d x +c \right )\right )+3 b^{2} a \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\]
Input:
int(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x)
Output:
1/d*(tanh(d*x+c)*a^3+3*a^2*b*(d*x+c-tanh(d*x+c))+3*b^2*a*(1/2*sinh(d*x+c)^ 3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+b^3*(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)))
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (86) = 172\).
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.93 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^{3} \sinh \left (d x + c\right )^{5} + {\left (10 \, b^{3} \cosh \left (d x + c\right )^{2} + 24 \, a b^{2} - 15 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 8 \, {\left (8 \, a^{3} - 24 \, a^{2} b + 24 \, a b^{2} - 8 \, b^{3} - 3 \, {\left (8 \, a^{2} b - 12 \, a b^{2} + 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 64 \, a^{3} - 192 \, a^{2} b + 216 \, a b^{2} - 80 \, b^{3} + 9 \, {\left (8 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \] Input:
integrate(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")
Output:
1/64*(b^3*sinh(d*x + c)^5 + (10*b^3*cosh(d*x + c)^2 + 24*a*b^2 - 15*b^3)*s inh(d*x + c)^3 - 8*(8*a^3 - 24*a^2*b + 24*a*b^2 - 8*b^3 - 3*(8*a^2*b - 12* a*b^2 + 5*b^3)*d*x)*cosh(d*x + c) + (5*b^3*cosh(d*x + c)^4 + 64*a^3 - 192* a^2*b + 216*a*b^2 - 80*b^3 + 9*(8*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c))
Timed out. \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:
integrate(sech(d*x+c)**2*(a+b*sinh(d*x+c)**2)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (86) = 172\).
Time = 0.08 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.34 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=3 \, a^{2} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac {1}{64} \, b^{3} {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} - \frac {3}{8} \, a b^{2} {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \] Input:
integrate(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")
Output:
3*a^2*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 1/64*b^3*(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) + 1 44*e^(-4*d*x - 4*c) - 1)/(d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c)))) - 3/8* a*b^2*(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)))) + 2*a^3/(d*(e^(-2*d*x - 2*c) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (86) = 172\).
Time = 0.18 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.14 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (8 \, a^{2} b - 12 \, a b^{2} + 5 \, b^{3}\right )} {\left (d x + c\right )} - {\left (144 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 216 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {128 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \] Input:
integrate(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")
Output:
1/64*(b^3*e^(4*d*x + 4*c) + 24*a*b^2*e^(2*d*x + 2*c) - 16*b^3*e^(2*d*x + 2 *c) + 24*(8*a^2*b - 12*a*b^2 + 5*b^3)*(d*x + c) - (144*a^2*b*e^(4*d*x + 4* c) - 216*a*b^2*e^(4*d*x + 4*c) + 90*b^3*e^(4*d*x + 4*c) + 24*a*b^2*e^(2*d* x + 2*c) - 16*b^3*e^(2*d*x + 2*c) + b^3)*e^(-4*d*x - 4*c) - 128*(a^3 - 3*a ^2*b + 3*a*b^2 - b^3)/(e^(2*d*x + 2*c) + 1))/d
Time = 1.91 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {b^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}-\frac {2\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {3\,b\,x\,\left (8\,a^2-12\,a\,b+5\,b^2\right )}{8}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (3\,a-2\,b\right )}{8\,d}+\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a-2\,b\right )}{8\,d} \] Input:
int((a + b*sinh(c + d*x)^2)^3/cosh(c + d*x)^2,x)
Output:
(b^3*exp(4*c + 4*d*x))/(64*d) - (b^3*exp(- 4*c - 4*d*x))/(64*d) - (2*(3*a* b^2 - 3*a^2*b + a^3 - b^3))/(d*(exp(2*c + 2*d*x) + 1)) + (3*b*x*(8*a^2 - 1 2*a*b + 5*b^2))/8 - (b^2*exp(- 2*c - 2*d*x)*(3*a - 2*b))/(8*d) + (b^2*exp( 2*c + 2*d*x)*(3*a - 2*b))/(8*d)
Time = 0.17 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.05 \[ \int \text {sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {e^{10 d x +10 c} b^{3}+24 e^{8 d x +8 c} a \,b^{2}-15 e^{8 d x +8 c} b^{3}+128 e^{6 d x +6 c} a^{3}+192 e^{6 d x +6 c} a^{2} b d x -384 e^{6 d x +6 c} a^{2} b -288 e^{6 d x +6 c} a \,b^{2} d x +432 e^{6 d x +6 c} a \,b^{2}+120 e^{6 d x +6 c} b^{3} d x -160 e^{6 d x +6 c} b^{3}+192 e^{4 d x +4 c} a^{2} b d x -288 e^{4 d x +4 c} a \,b^{2} d x +120 e^{4 d x +4 c} b^{3} d x -24 e^{2 d x +2 c} a \,b^{2}+15 e^{2 d x +2 c} b^{3}-b^{3}}{64 e^{4 d x +4 c} d \left (e^{2 d x +2 c}+1\right )} \] Input:
int(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x)
Output:
(e**(10*c + 10*d*x)*b**3 + 24*e**(8*c + 8*d*x)*a*b**2 - 15*e**(8*c + 8*d*x )*b**3 + 128*e**(6*c + 6*d*x)*a**3 + 192*e**(6*c + 6*d*x)*a**2*b*d*x - 384 *e**(6*c + 6*d*x)*a**2*b - 288*e**(6*c + 6*d*x)*a*b**2*d*x + 432*e**(6*c + 6*d*x)*a*b**2 + 120*e**(6*c + 6*d*x)*b**3*d*x - 160*e**(6*c + 6*d*x)*b**3 + 192*e**(4*c + 4*d*x)*a**2*b*d*x - 288*e**(4*c + 4*d*x)*a*b**2*d*x + 120 *e**(4*c + 4*d*x)*b**3*d*x - 24*e**(2*c + 2*d*x)*a*b**2 + 15*e**(2*c + 2*d *x)*b**3 - b**3)/(64*e**(4*c + 4*d*x)*d*(e**(2*c + 2*d*x) + 1))