3.11 \(\int \sinh ^3(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=85 \[ \frac {b (2 a-3 b) \cosh ^5(c+d x)}{5 d}+\frac {(a-3 b) (a-b) \cosh ^3(c+d x)}{3 d}-\frac {(a-b)^2 \cosh (c+d x)}{d}+\frac {b^2 \cosh ^7(c+d x)}{7 d} \]

[Out]

-(a-b)^2*cosh(d*x+c)/d+1/3*(a-3*b)*(a-b)*cosh(d*x+c)^3/d+1/5*(2*a-3*b)*b*cosh(d*x+c)^5/d+1/7*b^2*cosh(d*x+c)^7
/d

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3186, 373} \[ \frac {b (2 a-3 b) \cosh ^5(c+d x)}{5 d}+\frac {(a-3 b) (a-b) \cosh ^3(c+d x)}{3 d}-\frac {(a-b)^2 \cosh (c+d x)}{d}+\frac {b^2 \cosh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

-(((a - b)^2*Cosh[c + d*x])/d) + ((a - 3*b)*(a - b)*Cosh[c + d*x]^3)/(3*d) + ((2*a - 3*b)*b*Cosh[c + d*x]^5)/(
5*d) + (b^2*Cosh[c + d*x]^7)/(7*d)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3186

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

Rubi steps

\begin {align*} \int \sinh ^3(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \left (a-b+b x^2\right )^2 \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left ((a-b)^2+(a-3 b) (-a+b) x^2-(2 a-3 b) b x^4-b^2 x^6\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {(a-b)^2 \cosh (c+d x)}{d}+\frac {(a-3 b) (a-b) \cosh ^3(c+d x)}{3 d}+\frac {(2 a-3 b) b \cosh ^5(c+d x)}{5 d}+\frac {b^2 \cosh ^7(c+d x)}{7 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 154, normalized size = 1.81 \[ -\frac {3 a^2 \cosh (c+d x)}{4 d}+\frac {a^2 \cosh (3 (c+d x))}{12 d}+\frac {5 a b \cosh (c+d x)}{4 d}-\frac {5 a b \cosh (3 (c+d x))}{24 d}+\frac {a b \cosh (5 (c+d x))}{40 d}-\frac {35 b^2 \cosh (c+d x)}{64 d}+\frac {7 b^2 \cosh (3 (c+d x))}{64 d}-\frac {7 b^2 \cosh (5 (c+d x))}{320 d}+\frac {b^2 \cosh (7 (c+d x))}{448 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

(-3*a^2*Cosh[c + d*x])/(4*d) + (5*a*b*Cosh[c + d*x])/(4*d) - (35*b^2*Cosh[c + d*x])/(64*d) + (a^2*Cosh[3*(c +
d*x)])/(12*d) - (5*a*b*Cosh[3*(c + d*x)])/(24*d) + (7*b^2*Cosh[3*(c + d*x)])/(64*d) + (a*b*Cosh[5*(c + d*x)])/
(40*d) - (7*b^2*Cosh[5*(c + d*x)])/(320*d) + (b^2*Cosh[7*(c + d*x)])/(448*d)

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 213, normalized size = 2.51 \[ \frac {15 \, b^{2} \cosh \left (d x + c\right )^{7} + 105 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + 21 \, {\left (8 \, a b - 7 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 105 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} + {\left (8 \, a b - 7 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 35 \, {\left (16 \, a^{2} - 40 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 105 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{5} + 2 \, {\left (8 \, a b - 7 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (16 \, a^{2} - 40 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 105 \, {\left (48 \, a^{2} - 80 \, a b + 35 \, b^{2}\right )} \cosh \left (d x + c\right )}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/6720*(15*b^2*cosh(d*x + c)^7 + 105*b^2*cosh(d*x + c)*sinh(d*x + c)^6 + 21*(8*a*b - 7*b^2)*cosh(d*x + c)^5 +
105*(5*b^2*cosh(d*x + c)^3 + (8*a*b - 7*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 + 35*(16*a^2 - 40*a*b + 21*b^2)*co
sh(d*x + c)^3 + 105*(3*b^2*cosh(d*x + c)^5 + 2*(8*a*b - 7*b^2)*cosh(d*x + c)^3 + (16*a^2 - 40*a*b + 21*b^2)*co
sh(d*x + c))*sinh(d*x + c)^2 - 105*(48*a^2 - 80*a*b + 35*b^2)*cosh(d*x + c))/d

________________________________________________________________________________________

giac [B]  time = 0.15, size = 196, normalized size = 2.31 \[ \frac {b^{2} e^{\left (7 \, d x + 7 \, c\right )}}{896 \, d} + \frac {b^{2} e^{\left (-7 \, d x - 7 \, c\right )}}{896 \, d} + \frac {{\left (8 \, a b - 7 \, b^{2}\right )} e^{\left (5 \, d x + 5 \, c\right )}}{640 \, d} + \frac {{\left (16 \, a^{2} - 40 \, a b + 21 \, b^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{384 \, d} - \frac {{\left (48 \, a^{2} - 80 \, a b + 35 \, b^{2}\right )} e^{\left (d x + c\right )}}{128 \, d} - \frac {{\left (48 \, a^{2} - 80 \, a b + 35 \, b^{2}\right )} e^{\left (-d x - c\right )}}{128 \, d} + \frac {{\left (16 \, a^{2} - 40 \, a b + 21 \, b^{2}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{384 \, d} + \frac {{\left (8 \, a b - 7 \, b^{2}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/896*b^2*e^(7*d*x + 7*c)/d + 1/896*b^2*e^(-7*d*x - 7*c)/d + 1/640*(8*a*b - 7*b^2)*e^(5*d*x + 5*c)/d + 1/384*(
16*a^2 - 40*a*b + 21*b^2)*e^(3*d*x + 3*c)/d - 1/128*(48*a^2 - 80*a*b + 35*b^2)*e^(d*x + c)/d - 1/128*(48*a^2 -
80*a*b + 35*b^2)*e^(-d*x - c)/d + 1/384*(16*a^2 - 40*a*b + 21*b^2)*e^(-3*d*x - 3*c)/d + 1/640*(8*a*b - 7*b^2)
*e^(-5*d*x - 5*c)/d

________________________________________________________________________________________

maple [A]  time = 0.04, size = 102, normalized size = 1.20 \[ \frac {b^{2} \left (-\frac {16}{35}+\frac {\left (\sinh ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sinh ^{4}\left (d x +c \right )\right )}{35}+\frac {8 \left (\sinh ^{2}\left (d x +c \right )\right )}{35}\right ) \cosh \left (d x +c \right )+2 a b \left (\frac {8}{15}+\frac {\left (\sinh ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sinh ^{2}\left (d x +c \right )\right )}{15}\right ) \cosh \left (d x +c \right )+a^{2} \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (d x +c \right )\right )}{3}\right ) \cosh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

1/d*(b^2*(-16/35+1/7*sinh(d*x+c)^6-6/35*sinh(d*x+c)^4+8/35*sinh(d*x+c)^2)*cosh(d*x+c)+2*a*b*(8/15+1/5*sinh(d*x
+c)^4-4/15*sinh(d*x+c)^2)*cosh(d*x+c)+a^2*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c))

________________________________________________________________________________________

maxima [B]  time = 0.38, size = 247, normalized size = 2.91 \[ -\frac {1}{4480} \, b^{2} {\left (\frac {{\left (49 \, e^{\left (-2 \, d x - 2 \, c\right )} - 245 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1225 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5\right )} e^{\left (7 \, d x + 7 \, c\right )}}{d} + \frac {1225 \, e^{\left (-d x - c\right )} - 245 \, e^{\left (-3 \, d x - 3 \, c\right )} + 49 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d}\right )} + \frac {1}{240} \, a b {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} + \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac {1}{24} \, a^{2} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/4480*b^2*((49*e^(-2*d*x - 2*c) - 245*e^(-4*d*x - 4*c) + 1225*e^(-6*d*x - 6*c) - 5)*e^(7*d*x + 7*c)/d + (122
5*e^(-d*x - c) - 245*e^(-3*d*x - 3*c) + 49*e^(-5*d*x - 5*c) - 5*e^(-7*d*x - 7*c))/d) + 1/240*a*b*(3*e^(5*d*x +
5*c)/d - 25*e^(3*d*x + 3*c)/d + 150*e^(d*x + c)/d + 150*e^(-d*x - c)/d - 25*e^(-3*d*x - 3*c)/d + 3*e^(-5*d*x
- 5*c)/d) + 1/24*a^2*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d + e^(-3*d*x - 3*c)/d)

________________________________________________________________________________________

mupad [B]  time = 0.23, size = 112, normalized size = 1.32 \[ \frac {\frac {a^2\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{3}-a^2\,\mathrm {cosh}\left (c+d\,x\right )+\frac {2\,a\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}-\frac {4\,a\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{3}+2\,a\,b\,\mathrm {cosh}\left (c+d\,x\right )+\frac {b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^7}{7}-\frac {3\,b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}+b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^3-b^2\,\mathrm {cosh}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^3*(a + b*sinh(c + d*x)^2)^2,x)

[Out]

((a^2*cosh(c + d*x)^3)/3 - b^2*cosh(c + d*x) - a^2*cosh(c + d*x) + b^2*cosh(c + d*x)^3 - (3*b^2*cosh(c + d*x)^
5)/5 + (b^2*cosh(c + d*x)^7)/7 + 2*a*b*cosh(c + d*x) - (4*a*b*cosh(c + d*x)^3)/3 + (2*a*b*cosh(c + d*x)^5)/5)/
d

________________________________________________________________________________________

sympy [A]  time = 5.45, size = 204, normalized size = 2.40 \[ \begin {cases} \frac {a^{2} \sinh ^{2}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {2 a^{2} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 a b \sinh ^{4}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {8 a b \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {16 a b \cosh ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sinh ^{6}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {2 b^{2} \sinh ^{4}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{d} + \frac {8 b^{2} \sinh ^{2}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{5 d} - \frac {16 b^{2} \cosh ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\relax (c )}\right )^{2} \sinh ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**3*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Piecewise((a**2*sinh(c + d*x)**2*cosh(c + d*x)/d - 2*a**2*cosh(c + d*x)**3/(3*d) + 2*a*b*sinh(c + d*x)**4*cosh
(c + d*x)/d - 8*a*b*sinh(c + d*x)**2*cosh(c + d*x)**3/(3*d) + 16*a*b*cosh(c + d*x)**5/(15*d) + b**2*sinh(c + d
*x)**6*cosh(c + d*x)/d - 2*b**2*sinh(c + d*x)**4*cosh(c + d*x)**3/d + 8*b**2*sinh(c + d*x)**2*cosh(c + d*x)**5
/(5*d) - 16*b**2*cosh(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**2*sinh(c)**3, True))

________________________________________________________________________________________