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]

-1/256*(128*a^2+160*a*b+63*b^2)*x+1/256*(128*a^2+352*a*b+193*b^2)*cosh(d*x+c)*sinh(d*x+c)/d-1/384*b*(416*a+447
*b)*cosh(d*x+c)^3*sinh(d*x+c)/d+1/480*b*(160*a+513*b)*cosh(d*x+c)^5*sinh(d*x+c)/d-41/80*b^2*cosh(d*x+c)^7*sinh
(d*x+c)/d+1/10*b^2*cosh(d*x+c)^9*sinh(d*x+c)/d

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Rubi [A]  time = 0.28, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.34, 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]

-1/30720*(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*Sinh[6*(c + d*x)] + 75*b^2*Sinh[8*(c + d*x)] - 6*b^2*Sinh[10*(c + d*x)])/d

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fricas [B]  time = 0.61, size = 305, normalized size = 1.89 \[ \frac {15 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{9} + 30 \, {\left (6 \, b^{2} \cosh \left (d x + c\right )^{3} - 5 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{7} + 3 \, {\left (126 \, b^{2} \cosh \left (d x + c\right )^{5} - 350 \, b^{2} \cosh \left (d x + c\right )^{3} + 5 \, {\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 10 \, {\left (18 \, b^{2} \cosh \left (d x + c\right )^{7} - 105 \, b^{2} \cosh \left (d x + c\right )^{5} + 5 \, {\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 36 \, {\left (8 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 30 \, {\left (128 \, a^{2} + 160 \, a b + 63 \, b^{2}\right )} d x + 15 \, {\left (b^{2} \cosh \left (d x + c\right )^{9} - 10 \, b^{2} \cosh \left (d x + c\right )^{7} + {\left (32 \, a b + 45 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 24 \, {\left (8 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 2 \, {\left (128 \, a^{2} + 240 \, a b + 105 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{7680 \, d} \]

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

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giac [A]  time = 0.21, size = 241, normalized size = 1.50 \[ -\frac {1}{256} \, {\left (128 \, a^{2} + 160 \, a b + 63 \, b^{2}\right )} x + \frac {b^{2} e^{\left (10 \, d x + 10 \, c\right )}}{10240 \, d} - \frac {5 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )}}{4096 \, d} + \frac {5 \, b^{2} e^{\left (-8 \, d x - 8 \, c\right )}}{4096 \, d} - \frac {b^{2} e^{\left (-10 \, d x - 10 \, c\right )}}{10240 \, d} + \frac {{\left (32 \, a b + 45 \, b^{2}\right )} e^{\left (6 \, d x + 6 \, c\right )}}{6144 \, d} - \frac {3 \, {\left (8 \, a b + 5 \, b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{512 \, d} + \frac {{\left (128 \, a^{2} + 240 \, a b + 105 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{1024 \, d} - \frac {{\left (128 \, a^{2} + 240 \, a b + 105 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{1024 \, d} + \frac {3 \, {\left (8 \, a b + 5 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{512 \, d} - \frac {{\left (32 \, a b + 45 \, b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{6144 \, d} \]

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]

-1/256*(128*a^2 + 160*a*b + 63*b^2)*x + 1/10240*b^2*e^(10*d*x + 10*c)/d - 5/4096*b^2*e^(8*d*x + 8*c)/d + 5/409
6*b^2*e^(-8*d*x - 8*c)/d - 1/10240*b^2*e^(-10*d*x - 10*c)/d + 1/6144*(32*a*b + 45*b^2)*e^(6*d*x + 6*c)/d - 3/5
12*(8*a*b + 5*b^2)*e^(4*d*x + 4*c)/d + 1/1024*(128*a^2 + 240*a*b + 105*b^2)*e^(2*d*x + 2*c)/d - 1/1024*(128*a^
2 + 240*a*b + 105*b^2)*e^(-2*d*x - 2*c)/d + 3/512*(8*a*b + 5*b^2)*e^(-4*d*x - 4*c)/d - 1/6144*(32*a*b + 45*b^2
)*e^(-6*d*x - 6*c)/d

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maple [A]  time = 0.05, size = 148, normalized size = 0.92 \[ \frac {b^{2} \left (\left (\frac {\left (\sinh ^{9}\left (d x +c \right )\right )}{10}-\frac {9 \left (\sinh ^{7}\left (d x +c \right )\right )}{80}+\frac {21 \left (\sinh ^{5}\left (d x +c \right )\right )}{160}-\frac {21 \left (\sinh ^{3}\left (d x +c \right )\right )}{128}+\frac {63 \sinh \left (d x +c \right )}{256}\right ) \cosh \left (d x +c \right )-\frac {63 d x}{256}-\frac {63 c}{256}\right )+2 a b \left (\left (\frac {\left (\sinh ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sinh ^{3}\left (d x +c \right )\right )}{24}+\frac {5 \sinh \left (d x +c \right )}{16}\right ) \cosh \left (d x +c \right )-\frac {5 d x}{16}-\frac {5 c}{16}\right )+a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d} \]

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))

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maxima [A]  time = 0.35, size = 260, normalized size = 1.61 \[ -\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{20480} \, b^{2} {\left (\frac {{\left (25 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 \, e^{\left (-4 \, d x - 4 \, c\right )} + 600 \, e^{\left (-6 \, d x - 6 \, c\right )} - 2100 \, e^{\left (-8 \, d x - 8 \, c\right )} - 2\right )} e^{\left (10 \, d x + 10 \, c\right )}}{d} + \frac {5040 \, {\left (d x + c\right )}}{d} + \frac {2100 \, e^{\left (-2 \, d x - 2 \, c\right )} - 600 \, e^{\left (-4 \, d x - 4 \, c\right )} + 150 \, e^{\left (-6 \, d x - 6 \, c\right )} - 25 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, e^{\left (-10 \, d x - 10 \, c\right )}}{d}\right )} - \frac {1}{192} \, a b {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} \]

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)

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mupad [B]  time = 0.41, size = 149, normalized size = 0.93 \[ \frac {960\,a^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+\frac {1575\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{2}-225\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+\frac {225\,b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}-\frac {75\,b^2\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{8}+\frac {3\,b^2\,\mathrm {sinh}\left (10\,c+10\,d\,x\right )}{4}+1800\,a\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-360\,a\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+40\,a\,b\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )-1920\,a^2\,d\,x-945\,b^2\,d\,x-2400\,a\,b\,d\,x}{3840\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(960*a^2*sinh(2*c + 2*d*x) + (1575*b^2*sinh(2*c + 2*d*x))/2 - 225*b^2*sinh(4*c + 4*d*x) + (225*b^2*sinh(6*c +
6*d*x))/4 - (75*b^2*sinh(8*c + 8*d*x))/8 + (3*b^2*sinh(10*c + 10*d*x))/4 + 1800*a*b*sinh(2*c + 2*d*x) - 360*a*
b*sinh(4*c + 4*d*x) + 40*a*b*sinh(6*c + 6*d*x) - 1920*a^2*d*x - 945*b^2*d*x - 2400*a*b*d*x)/(3840*d)

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sympy [A]  time = 21.71, size = 484, normalized size = 3.01 \[ \begin {cases} \frac {a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} + \frac {5 a b x \sinh ^{6}{\left (c + d x \right )}}{8} - \frac {15 a b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{8} + \frac {15 a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{8} - \frac {5 a b x \cosh ^{6}{\left (c + d x \right )}}{8} + \frac {11 a b \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {5 a b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 a b \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{8 d} + \frac {63 b^{2} x \sinh ^{10}{\left (c + d x \right )}}{256} - \frac {315 b^{2} x \sinh ^{8}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{256} + \frac {315 b^{2} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{128} - \frac {315 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{128} + \frac {315 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{8}{\left (c + d x \right )}}{256} - \frac {63 b^{2} x \cosh ^{10}{\left (c + d x \right )}}{256} + \frac {193 b^{2} \sinh ^{9}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{256 d} - \frac {237 b^{2} \sinh ^{7}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{128 d} + \frac {21 b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{10 d} - \frac {147 b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} + \frac {63 b^{2} \sinh {\left (c + d x \right )} \cosh ^{9}{\left (c + d x \right )}}{256 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{4}{\relax (c )}\right )^{2} \sinh ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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))

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