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

Optimal. Leaf size=83 \[ \frac{(8 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} x (8 a+5 b)+\frac{b \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

[Out]

-((8*a + 5*b)*x)/16 + ((8*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b*Cosh[c + d*x]^3*Sinh[c + d*x])
/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*d)

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Rubi [A]  time = 0.0976626, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3217, 1257, 1157, 385, 206} \[ \frac{(8 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} x (8 a+5 b)+\frac{b \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a + 5*b)*x)/16 + ((8*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b*Cosh[c + d*x]^3*Sinh[c + d*x])
/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*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 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 ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a-2 a x^2+(a+b) x^4\right )}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac{\operatorname{Subst}\left (\int \frac{-b+6 (a-b) x^2-6 (a+b) x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=-\frac{13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-9 b-24 (a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{24 d}\\ &=\frac{(8 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{(8 a+5 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{1}{16} (8 a+5 b) x+\frac{(8 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.0771824, size = 63, normalized size = 0.76 \[ \frac{(48 a+45 b) \sinh (2 (c+d x))-96 a c-96 a d x-9 b \sinh (4 (c+d x))+b \sinh (6 (c+d x))-60 b c-60 b d x}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-96*a*c - 60*b*c - 96*a*d*x - 60*b*d*x + (48*a + 45*b)*Sinh[2*(c + d*x)] - 9*b*Sinh[4*(c + d*x)] + b*Sinh[6*(
c + d*x)])/(192*d)

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Maple [A]  time = 0.014, size = 76, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( b \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\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 \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),x)

[Out]

1/d*(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*(1/2*cosh(d*x+c
)*sinh(d*x+c)-1/2*d*x-1/2*c))

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Maxima [A]  time = 1.11428, size = 165, normalized size = 1.99 \begin{align*} -\frac{1}{8} \, a{\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}{384} \, 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),x, algorithm="maxima")

[Out]

-1/8*a*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/384*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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Fricas [A]  time = 1.60365, size = 292, normalized size = 3.52 \begin{align*} \frac{3 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \,{\left (5 \, b \cosh \left (d x + c\right )^{3} - 9 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 6 \,{\left (8 \, a + 5 \, b\right )} d x + 3 \,{\left (b \cosh \left (d x + c\right )^{5} - 6 \, b \cosh \left (d x + c\right )^{3} +{\left (16 \, a + 15 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, 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),x, algorithm="fricas")

[Out]

1/96*(3*b*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b*cosh(d*x + c)^3 - 9*b*cosh(d*x + c))*sinh(d*x + c)^3 - 6*(8*a
+ 5*b)*d*x + 3*(b*cosh(d*x + c)^5 - 6*b*cosh(d*x + c)^3 + (16*a + 15*b)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [A]  time = 4.93975, size = 206, normalized size = 2.48 \begin{align*} \begin{cases} \frac{a x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac{a x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{a \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 d} + \frac{5 b x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac{15 b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac{15 b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac{5 b x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac{11 b \sinh ^{5}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{16 d} - \frac{5 b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac{5 b \sinh{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{4}{\left (c \right )}\right ) \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),x)

[Out]

Piecewise((a*x*sinh(c + d*x)**2/2 - a*x*cosh(c + d*x)**2/2 + a*sinh(c + d*x)*cosh(c + d*x)/(2*d) + 5*b*x*sinh(
c + d*x)**6/16 - 15*b*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 15*b*x*sinh(c + d*x)**2*cosh(c + d*x)**4/16 - 5
*b*x*cosh(c + d*x)**6/16 + 11*b*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) - 5*b*sinh(c + d*x)**3*cosh(c + d*x)**3/
(6*d) + 5*b*sinh(c + d*x)*cosh(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sinh(c)**4)*sinh(c)**2, True))

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Giac [A]  time = 1.16648, size = 193, normalized size = 2.33 \begin{align*} -\frac{24 \,{\left (d x + c\right )}{\left (8 \, a + 5 \, b\right )} - b e^{\left (6 \, d x + 6 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 48 \, a e^{\left (2 \, d x + 2 \, c\right )} - 45 \, b e^{\left (2 \, d x + 2 \, c\right )} -{\left (176 \, a e^{\left (6 \, d x + 6 \, c\right )} + 110 \, b e^{\left (6 \, d x + 6 \, c\right )} - 48 \, a e^{\left (4 \, d x + 4 \, c\right )} - 45 \, b e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b e^{\left (2 \, d x + 2 \, c\right )} - b\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, 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),x, algorithm="giac")

[Out]

-1/384*(24*(d*x + c)*(8*a + 5*b) - b*e^(6*d*x + 6*c) + 9*b*e^(4*d*x + 4*c) - 48*a*e^(2*d*x + 2*c) - 45*b*e^(2*
d*x + 2*c) - (176*a*e^(6*d*x + 6*c) + 110*b*e^(6*d*x + 6*c) - 48*a*e^(4*d*x + 4*c) - 45*b*e^(4*d*x + 4*c) + 9*
b*e^(2*d*x + 2*c) - b)*e^(-6*d*x - 6*c))/d