3.197 \(\int \sinh ^2(c+d x) (a+b \sinh ^4(c+d x))^2 \, dx\)

Optimal. Leaf size=161 \[ \frac{\left (128 a^2+352 a b+193 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{256 d}-\frac{1}{256} x \left (128 a^2+160 a b+63 b^2\right )+\frac{b (160 a+513 b) \sinh (c+d x) \cosh ^5(c+d x)}{480 d}-\frac{b (416 a+447 b) \sinh (c+d x) \cosh ^3(c+d x)}{384 d}+\frac{b^2 \sinh (c+d x) \cosh ^9(c+d x)}{10 d}-\frac{41 b^2 \sinh (c+d x) \cosh ^7(c+d x)}{80 d} \]

[Out]

-((128*a^2 + 160*a*b + 63*b^2)*x)/256 + ((128*a^2 + 352*a*b + 193*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(256*d) -
(b*(416*a + 447*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(384*d) + (b*(160*a + 513*b)*Cosh[c + d*x]^5*Sinh[c + d*x])/
(480*d) - (41*b^2*Cosh[c + d*x]^7*Sinh[c + d*x])/(80*d) + (b^2*Cosh[c + d*x]^9*Sinh[c + d*x])/(10*d)

________________________________________________________________________________________

Rubi [A]  time = 0.280985, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3217, 1257, 1814, 1157, 385, 206} \[ \frac{\left (128 a^2+352 a b+193 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{256 d}-\frac{1}{256} x \left (128 a^2+160 a b+63 b^2\right )+\frac{b (160 a+513 b) \sinh (c+d x) \cosh ^5(c+d x)}{480 d}-\frac{b (416 a+447 b) \sinh (c+d x) \cosh ^3(c+d x)}{384 d}+\frac{b^2 \sinh (c+d x) \cosh ^9(c+d x)}{10 d}-\frac{41 b^2 \sinh (c+d x) \cosh ^7(c+d x)}{80 d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^2*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

-((128*a^2 + 160*a*b + 63*b^2)*x)/256 + ((128*a^2 + 352*a*b + 193*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(256*d) -
(b*(416*a + 447*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(384*d) + (b*(160*a + 513*b)*Cosh[c + d*x]^5*Sinh[c + d*x])/
(480*d) - (41*b^2*Cosh[c + d*x]^7*Sinh[c + d*x])/(80*d) + (b^2*Cosh[c + d*x]^9*Sinh[c + d*x])/(10*d)

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(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] + Dist[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] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sinh ^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a-2 a x^2+(a+b) x^4\right )^2}{\left (1-x^2\right )^6} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}+\frac{\operatorname{Subst}\left (\int \frac{-b^2+10 \left (a^2-b^2\right ) x^2-10 \left (3 a^2+b^2\right ) x^4+10 (3 a-b) (a+b) x^6-10 (a+b)^2 x^8}{\left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=-\frac{41 b^2 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac{\operatorname{Subst}\left (\int \frac{-33 b^2-80 \left (a^2+3 b^2\right ) x^2+160 \left (a^2-b^2\right ) x^4-80 (a+b)^2 x^6}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{80 d}\\ &=\frac{b (160 a+513 b) \cosh ^5(c+d x) \sinh (c+d x)}{480 d}-\frac{41 b^2 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}+\frac{\operatorname{Subst}\left (\int \frac{-5 b (32 a+63 b)+480 (a-3 b) (a+b) x^2-480 (a+b)^2 x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{480 d}\\ &=-\frac{b (416 a+447 b) \cosh ^3(c+d x) \sinh (c+d x)}{384 d}+\frac{b (160 a+513 b) \cosh ^5(c+d x) \sinh (c+d x)}{480 d}-\frac{41 b^2 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac{\operatorname{Subst}\left (\int \frac{-15 b (96 a+65 b)-1920 (a+b)^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{1920 d}\\ &=\frac{\left (128 a^2+352 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac{b (416 a+447 b) \cosh ^3(c+d x) \sinh (c+d x)}{384 d}+\frac{b (160 a+513 b) \cosh ^5(c+d x) \sinh (c+d x)}{480 d}-\frac{41 b^2 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac{\left (128 a^2+160 a b+63 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{256 d}\\ &=-\frac{1}{256} \left (128 a^2+160 a b+63 b^2\right ) x+\frac{\left (128 a^2+352 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac{b (416 a+447 b) \cosh ^3(c+d x) \sinh (c+d x)}{384 d}+\frac{b (160 a+513 b) \cosh ^5(c+d x) \sinh (c+d x)}{480 d}-\frac{41 b^2 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac{b^2 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}\\ \end{align*}

Mathematica [A]  time = 0.352497, size = 139, normalized size = 0.86 \[ -\frac{-60 \left (128 a^2+240 a b+105 b^2\right ) \sinh (2 (c+d x))+15360 a^2 c+15360 a^2 d x-320 a b \sinh (6 (c+d x))+360 b (8 a+5 b) \sinh (4 (c+d x))+19200 a b c+19200 a b d x-450 b^2 \sinh (6 (c+d x))+75 b^2 \sinh (8 (c+d x))-6 b^2 \sinh (10 (c+d x))+7560 b^2 c+7560 b^2 d x}{30720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^2*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

-(15360*a^2*c + 19200*a*b*c + 7560*b^2*c + 15360*a^2*d*x + 19200*a*b*d*x + 7560*b^2*d*x - 60*(128*a^2 + 240*a*
b + 105*b^2)*Sinh[2*(c + d*x)] + 360*b*(8*a + 5*b)*Sinh[4*(c + d*x)] - 320*a*b*Sinh[6*(c + d*x)] - 450*b^2*Sin
h[6*(c + d*x)] + 75*b^2*Sinh[8*(c + d*x)] - 6*b^2*Sinh[10*(c + d*x)])/(30720*d)

________________________________________________________________________________________

Maple [A]  time = 0.022, size = 148, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{9}}{10}}-{\frac{9\, \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{80}}+{\frac{21\, \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{160}}-{\frac{21\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{128}}+{\frac{63\,\sinh \left ( dx+c \right ) }{256}} \right ) \cosh \left ( dx+c \right ) -{\frac{63\,dx}{256}}-{\frac{63\,c}{256}} \right ) +2\,ab \left ( \left ( 1/6\, \left ( \sinh \left ( dx+c \right ) \right ) ^{5}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \right ) +{a}^{2} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^2*(a+b*sinh(d*x+c)^4)^2,x)

[Out]

1/d*(b^2*((1/10*sinh(d*x+c)^9-9/80*sinh(d*x+c)^7+21/160*sinh(d*x+c)^5-21/128*sinh(d*x+c)^3+63/256*sinh(d*x+c))
*cosh(d*x+c)-63/256*d*x-63/256*c)+2*a*b*((1/6*sinh(d*x+c)^5-5/24*sinh(d*x+c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5
/16*d*x-5/16*c)+a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 1.10221, size = 351, normalized size = 2.18 \begin{align*} -\frac{1}{8} \, a^{2}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{20480} \, b^{2}{\left (\frac{{\left (25 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 \, e^{\left (-4 \, d x - 4 \, c\right )} + 600 \, e^{\left (-6 \, d x - 6 \, c\right )} - 2100 \, e^{\left (-8 \, d x - 8 \, c\right )} - 2\right )} e^{\left (10 \, d x + 10 \, c\right )}}{d} + \frac{5040 \,{\left (d x + c\right )}}{d} + \frac{2100 \, e^{\left (-2 \, d x - 2 \, c\right )} - 600 \, e^{\left (-4 \, d x - 4 \, c\right )} + 150 \, e^{\left (-6 \, d x - 6 \, c\right )} - 25 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, e^{\left (-10 \, d x - 10 \, c\right )}}{d}\right )} - \frac{1}{192} \, a b{\left (\frac{{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac{120 \,{\left (d x + c\right )}}{d} + \frac{45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*sinh(d*x+c)^4)^2,x, algorithm="maxima")

[Out]

-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/20480*b^2*((25*e^(-2*d*x - 2*c) - 150*e^(-4*d*x -
4*c) + 600*e^(-6*d*x - 6*c) - 2100*e^(-8*d*x - 8*c) - 2)*e^(10*d*x + 10*c)/d + 5040*(d*x + c)/d + (2100*e^(-2*
d*x - 2*c) - 600*e^(-4*d*x - 4*c) + 150*e^(-6*d*x - 6*c) - 25*e^(-8*d*x - 8*c) + 2*e^(-10*d*x - 10*c))/d) - 1/
192*a*b*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x -
2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d)

________________________________________________________________________________________

Fricas [B]  time = 1.65095, size = 802, normalized size = 4.98 \begin{align*} \frac{15 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{9} + 30 \,{\left (6 \, b^{2} \cosh \left (d x + c\right )^{3} - 5 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{7} + 3 \,{\left (126 \, b^{2} \cosh \left (d x + c\right )^{5} - 350 \, b^{2} \cosh \left (d x + c\right )^{3} + 5 \,{\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 10 \,{\left (18 \, b^{2} \cosh \left (d x + c\right )^{7} - 105 \, b^{2} \cosh \left (d x + c\right )^{5} + 5 \,{\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 36 \,{\left (8 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 30 \,{\left (128 \, a^{2} + 160 \, a b + 63 \, b^{2}\right )} d x + 15 \,{\left (b^{2} \cosh \left (d x + c\right )^{9} - 10 \, b^{2} \cosh \left (d x + c\right )^{7} +{\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 24 \,{\left (8 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 2 \,{\left (128 \, a^{2} + 240 \, a b + 105 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{7680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*sinh(d*x+c)^4)^2,x, algorithm="fricas")

[Out]

1/7680*(15*b^2*cosh(d*x + c)*sinh(d*x + c)^9 + 30*(6*b^2*cosh(d*x + c)^3 - 5*b^2*cosh(d*x + c))*sinh(d*x + c)^
7 + 3*(126*b^2*cosh(d*x + c)^5 - 350*b^2*cosh(d*x + c)^3 + 5*(32*a*b + 45*b^2)*cosh(d*x + c))*sinh(d*x + c)^5
+ 10*(18*b^2*cosh(d*x + c)^7 - 105*b^2*cosh(d*x + c)^5 + 5*(32*a*b + 45*b^2)*cosh(d*x + c)^3 - 36*(8*a*b + 5*b
^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 30*(128*a^2 + 160*a*b + 63*b^2)*d*x + 15*(b^2*cosh(d*x + c)^9 - 10*b^2*co
sh(d*x + c)^7 + (32*a*b + 45*b^2)*cosh(d*x + c)^5 - 24*(8*a*b + 5*b^2)*cosh(d*x + c)^3 + 2*(128*a^2 + 240*a*b
+ 105*b^2)*cosh(d*x + c))*sinh(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 60.8387, size = 484, normalized size = 3.01 \begin{align*} \begin{cases} \frac{a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac{a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 d} + \frac{5 a b x \sinh ^{6}{\left (c + d x \right )}}{8} - \frac{15 a b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{8} + \frac{15 a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{8} - \frac{5 a b x \cosh ^{6}{\left (c + d x \right )}}{8} + \frac{11 a b \sinh ^{5}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{8 d} - \frac{5 a b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac{5 a b \sinh{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{8 d} + \frac{63 b^{2} x \sinh ^{10}{\left (c + d x \right )}}{256} - \frac{315 b^{2} x \sinh ^{8}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{256} + \frac{315 b^{2} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{128} - \frac{315 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{128} + \frac{315 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{8}{\left (c + d x \right )}}{256} - \frac{63 b^{2} x \cosh ^{10}{\left (c + d x \right )}}{256} + \frac{193 b^{2} \sinh ^{9}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{256 d} - \frac{237 b^{2} \sinh ^{7}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{128 d} + \frac{21 b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{10 d} - \frac{147 b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} + \frac{63 b^{2} \sinh{\left (c + d x \right )} \cosh ^{9}{\left (c + d x \right )}}{256 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{4}{\left (c \right )}\right )^{2} \sinh ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**2*(a+b*sinh(d*x+c)**4)**2,x)

[Out]

Piecewise((a**2*x*sinh(c + d*x)**2/2 - a**2*x*cosh(c + d*x)**2/2 + a**2*sinh(c + d*x)*cosh(c + d*x)/(2*d) + 5*
a*b*x*sinh(c + d*x)**6/8 - 15*a*b*x*sinh(c + d*x)**4*cosh(c + d*x)**2/8 + 15*a*b*x*sinh(c + d*x)**2*cosh(c + d
*x)**4/8 - 5*a*b*x*cosh(c + d*x)**6/8 + 11*a*b*sinh(c + d*x)**5*cosh(c + d*x)/(8*d) - 5*a*b*sinh(c + d*x)**3*c
osh(c + d*x)**3/(3*d) + 5*a*b*sinh(c + d*x)*cosh(c + d*x)**5/(8*d) + 63*b**2*x*sinh(c + d*x)**10/256 - 315*b**
2*x*sinh(c + d*x)**8*cosh(c + d*x)**2/256 + 315*b**2*x*sinh(c + d*x)**6*cosh(c + d*x)**4/128 - 315*b**2*x*sinh
(c + d*x)**4*cosh(c + d*x)**6/128 + 315*b**2*x*sinh(c + d*x)**2*cosh(c + d*x)**8/256 - 63*b**2*x*cosh(c + d*x)
**10/256 + 193*b**2*sinh(c + d*x)**9*cosh(c + d*x)/(256*d) - 237*b**2*sinh(c + d*x)**7*cosh(c + d*x)**3/(128*d
) + 21*b**2*sinh(c + d*x)**5*cosh(c + d*x)**5/(10*d) - 147*b**2*sinh(c + d*x)**3*cosh(c + d*x)**7/(128*d) + 63
*b**2*sinh(c + d*x)*cosh(c + d*x)**9/(256*d), Ne(d, 0)), (x*(a + b*sinh(c)**4)**2*sinh(c)**2, True))

________________________________________________________________________________________

Giac [B]  time = 1.34792, size = 428, normalized size = 2.66 \begin{align*} \frac{6 \, b^{2} e^{\left (10 \, d x + 10 \, c\right )} - 75 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 320 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 450 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 2880 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 1800 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 7680 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14400 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6300 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 240 \,{\left (128 \, a^{2} + 160 \, a b + 63 \, b^{2}\right )}{\left (d x + c\right )} +{\left (35072 \, a^{2} e^{\left (10 \, d x + 10 \, c\right )} + 43840 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 17262 \, b^{2} e^{\left (10 \, d x + 10 \, c\right )} - 7680 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} - 14400 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 6300 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2880 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 1800 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 320 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 450 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 75 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b^{2}\right )} e^{\left (-10 \, d x - 10 \, c\right )}}{61440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*sinh(d*x+c)^4)^2,x, algorithm="giac")

[Out]

1/61440*(6*b^2*e^(10*d*x + 10*c) - 75*b^2*e^(8*d*x + 8*c) + 320*a*b*e^(6*d*x + 6*c) + 450*b^2*e^(6*d*x + 6*c)
- 2880*a*b*e^(4*d*x + 4*c) - 1800*b^2*e^(4*d*x + 4*c) + 7680*a^2*e^(2*d*x + 2*c) + 14400*a*b*e^(2*d*x + 2*c) +
6300*b^2*e^(2*d*x + 2*c) - 240*(128*a^2 + 160*a*b + 63*b^2)*(d*x + c) + (35072*a^2*e^(10*d*x + 10*c) + 43840*
a*b*e^(10*d*x + 10*c) + 17262*b^2*e^(10*d*x + 10*c) - 7680*a^2*e^(8*d*x + 8*c) - 14400*a*b*e^(8*d*x + 8*c) - 6
300*b^2*e^(8*d*x + 8*c) + 2880*a*b*e^(6*d*x + 6*c) + 1800*b^2*e^(6*d*x + 6*c) - 320*a*b*e^(4*d*x + 4*c) - 450*
b^2*e^(4*d*x + 4*c) + 75*b^2*e^(2*d*x + 2*c) - 6*b^2)*e^(-10*d*x - 10*c))/d