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

Optimal. Leaf size=123 \[ -\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}+\frac{4 d (c+d x) \sinh (a+b x)}{3 b^2}+\frac{2 d^2 \cosh ^3(a+b x)}{27 b^3}-\frac{14 d^2 \cosh (a+b x)}{9 b^3}-\frac{2 (c+d x)^2 \cosh (a+b x)}{3 b}+\frac{(c+d x)^2 \sinh ^2(a+b x) \cosh (a+b x)}{3 b} \]

[Out]

(-14*d^2*Cosh[a + b*x])/(9*b^3) - (2*(c + d*x)^2*Cosh[a + b*x])/(3*b) + (2*d^2*Cosh[a + b*x]^3)/(27*b^3) + (4*
d*(c + d*x)*Sinh[a + b*x])/(3*b^2) + ((c + d*x)^2*Cosh[a + b*x]*Sinh[a + b*x]^2)/(3*b) - (2*d*(c + d*x)*Sinh[a
+ b*x]^3)/(9*b^2)

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Rubi [A]  time = 0.131249, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 3296, 2638, 2633} \[ -\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}+\frac{4 d (c+d x) \sinh (a+b x)}{3 b^2}+\frac{2 d^2 \cosh ^3(a+b x)}{27 b^3}-\frac{14 d^2 \cosh (a+b x)}{9 b^3}-\frac{2 (c+d x)^2 \cosh (a+b x)}{3 b}+\frac{(c+d x)^2 \sinh ^2(a+b x) \cosh (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sinh[a + b*x]^3,x]

[Out]

(-14*d^2*Cosh[a + b*x])/(9*b^3) - (2*(c + d*x)^2*Cosh[a + b*x])/(3*b) + (2*d^2*Cosh[a + b*x]^3)/(27*b^3) + (4*
d*(c + d*x)*Sinh[a + b*x])/(3*b^2) + ((c + d*x)^2*Cosh[a + b*x]*Sinh[a + b*x]^2)/(3*b) - (2*d*(c + d*x)*Sinh[a
+ b*x]^3)/(9*b^2)

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int (c+d x)^2 \sinh ^3(a+b x) \, dx &=\frac{(c+d x)^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}-\frac{2}{3} \int (c+d x)^2 \sinh (a+b x) \, dx+\frac{\left (2 d^2\right ) \int \sinh ^3(a+b x) \, dx}{9 b^2}\\ &=-\frac{2 (c+d x)^2 \cosh (a+b x)}{3 b}+\frac{(c+d x)^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}+\frac{(4 d) \int (c+d x) \cosh (a+b x) \, dx}{3 b}-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (a+b x)\right )}{9 b^3}\\ &=-\frac{2 d^2 \cosh (a+b x)}{9 b^3}-\frac{2 (c+d x)^2 \cosh (a+b x)}{3 b}+\frac{2 d^2 \cosh ^3(a+b x)}{27 b^3}+\frac{4 d (c+d x) \sinh (a+b x)}{3 b^2}+\frac{(c+d x)^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}-\frac{\left (4 d^2\right ) \int \sinh (a+b x) \, dx}{3 b^2}\\ &=-\frac{14 d^2 \cosh (a+b x)}{9 b^3}-\frac{2 (c+d x)^2 \cosh (a+b x)}{3 b}+\frac{2 d^2 \cosh ^3(a+b x)}{27 b^3}+\frac{4 d (c+d x) \sinh (a+b x)}{3 b^2}+\frac{(c+d x)^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b}-\frac{2 d (c+d x) \sinh ^3(a+b x)}{9 b^2}\\ \end{align*}

Mathematica [A]  time = 0.403865, size = 86, normalized size = 0.7 \[ \frac{-81 \cosh (a+b x) \left (b^2 (c+d x)^2+2 d^2\right )+\cosh (3 (a+b x)) \left (9 b^2 (c+d x)^2+2 d^2\right )-6 b d (c+d x) (\sinh (3 (a+b x))-27 \sinh (a+b x))}{108 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-81*(2*d^2 + b^2*(c + d*x)^2)*Cosh[a + b*x] + (2*d^2 + 9*b^2*(c + d*x)^2)*Cosh[3*(a + b*x)] - 6*b*d*(c + d*x)
*(-27*Sinh[a + b*x] + Sinh[3*(a + b*x)]))/(108*b^3)

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Maple [B]  time = 0.007, size = 320, normalized size = 2.6 \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2}}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) }{3}}-{\frac{2\, \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) }{3}}-{\frac{ \left ( 2\,bx+2\,a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{9}}+{\frac{ \left ( 14\,bx+14\,a \right ) \sinh \left ( bx+a \right ) }{9}}+{\frac{2\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) }{27}}-{\frac{40\,\cosh \left ( bx+a \right ) }{27}} \right ) }-2\,{\frac{a{d}^{2}}{{b}^{2}} \left ( -2/3\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) +1/3\, \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) +{\frac{7\,\sinh \left ( bx+a \right ) }{9}}-1/9\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) }+{\frac{{a}^{2}{d}^{2}\cosh \left ( bx+a \right ) }{{b}^{2}} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) }+2\,{\frac{cd}{b} \left ( -2/3\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) +1/3\, \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) +{\frac{7\,\sinh \left ( bx+a \right ) }{9}}-1/9\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) }-2\,{\frac{cda \left ( -2/3+1/3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \right ) \cosh \left ( bx+a \right ) }{b}}+{c}^{2} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b^2*d^2*(1/3*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)-2/3*(b*x+a)^2*cosh(b*x+a)-2/9*(b*x+a)*sinh(b*x+a)*cosh
(b*x+a)^2+14/9*(b*x+a)*sinh(b*x+a)+2/27*sinh(b*x+a)^2*cosh(b*x+a)-40/27*cosh(b*x+a))-2/b^2*d^2*a*(-2/3*(b*x+a)
*cosh(b*x+a)+1/3*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)+7/9*sinh(b*x+a)-1/9*sinh(b*x+a)*cosh(b*x+a)^2)+1/b^2*d^2*a^
2*(-2/3+1/3*sinh(b*x+a)^2)*cosh(b*x+a)+2/b*c*d*(-2/3*(b*x+a)*cosh(b*x+a)+1/3*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)
+7/9*sinh(b*x+a)-1/9*sinh(b*x+a)*cosh(b*x+a)^2)-2/b*c*d*a*(-2/3+1/3*sinh(b*x+a)^2)*cosh(b*x+a)+c^2*(-2/3+1/3*s
inh(b*x+a)^2)*cosh(b*x+a))

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Maxima [B]  time = 1.35653, size = 363, normalized size = 2.95 \begin{align*} \frac{1}{36} \, c d{\left (\frac{{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{b^{2}} - \frac{27 \,{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2}} - \frac{27 \,{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{b^{2}} + \frac{{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{b^{2}}\right )} + \frac{1}{24} \, c^{2}{\left (\frac{e^{\left (3 \, b x + 3 \, a\right )}}{b} - \frac{9 \, e^{\left (b x + a\right )}}{b} - \frac{9 \, e^{\left (-b x - a\right )}}{b} + \frac{e^{\left (-3 \, b x - 3 \, a\right )}}{b}\right )} + \frac{1}{216} \, d^{2}{\left (\frac{{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{b^{3}} - \frac{81 \,{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{b^{3}} - \frac{81 \,{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{b^{3}} + \frac{{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*c*d*((3*b*x*e^(3*a) - e^(3*a))*e^(3*b*x)/b^2 - 27*(b*x*e^a - e^a)*e^(b*x)/b^2 - 27*(b*x + 1)*e^(-b*x - a)
/b^2 + (3*b*x + 1)*e^(-3*b*x - 3*a)/b^2) + 1/24*c^2*(e^(3*b*x + 3*a)/b - 9*e^(b*x + a)/b - 9*e^(-b*x - a)/b +
e^(-3*b*x - 3*a)/b) + 1/216*d^2*((9*b^2*x^2*e^(3*a) - 6*b*x*e^(3*a) + 2*e^(3*a))*e^(3*b*x)/b^3 - 81*(b^2*x^2*e
^a - 2*b*x*e^a + 2*e^a)*e^(b*x)/b^3 - 81*(b^2*x^2 + 2*b*x + 2)*e^(-b*x - a)/b^3 + (9*b^2*x^2 + 6*b*x + 2)*e^(-
3*b*x - 3*a)/b^3)

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Fricas [A]  time = 2.62494, size = 466, normalized size = 3.79 \begin{align*} \frac{{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 2 \, d^{2}\right )} \cosh \left (b x + a\right )^{3} + 3 \,{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 2 \, d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 6 \,{\left (b d^{2} x + b c d\right )} \sinh \left (b x + a\right )^{3} - 81 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + 2 \, d^{2}\right )} \cosh \left (b x + a\right ) + 18 \,{\left (9 \, b d^{2} x + 9 \, b c d -{\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right )^{2}\right )} \sinh \left (b x + a\right )}{108 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*((9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 + 2*d^2)*cosh(b*x + a)^3 + 3*(9*b^2*d^2*x^2 + 18*b^2*c*d*x +
9*b^2*c^2 + 2*d^2)*cosh(b*x + a)*sinh(b*x + a)^2 - 6*(b*d^2*x + b*c*d)*sinh(b*x + a)^3 - 81*(b^2*d^2*x^2 + 2*b
^2*c*d*x + b^2*c^2 + 2*d^2)*cosh(b*x + a) + 18*(9*b*d^2*x + 9*b*c*d - (b*d^2*x + b*c*d)*cosh(b*x + a)^2)*sinh(
b*x + a))/b^3

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Sympy [A]  time = 3.83939, size = 284, normalized size = 2.31 \begin{align*} \begin{cases} \frac{c^{2} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{b} - \frac{2 c^{2} \cosh ^{3}{\left (a + b x \right )}}{3 b} + \frac{2 c d x \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{b} - \frac{4 c d x \cosh ^{3}{\left (a + b x \right )}}{3 b} + \frac{d^{2} x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{b} - \frac{2 d^{2} x^{2} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac{14 c d \sinh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{4 c d \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{2}} - \frac{14 d^{2} x \sinh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac{4 d^{2} x \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{2}} + \frac{14 d^{2} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{9 b^{3}} - \frac{40 d^{2} \cosh ^{3}{\left (a + b x \right )}}{27 b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sinh ^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**2*sinh(a + b*x)**2*cosh(a + b*x)/b - 2*c**2*cosh(a + b*x)**3/(3*b) + 2*c*d*x*sinh(a + b*x)**2*co
sh(a + b*x)/b - 4*c*d*x*cosh(a + b*x)**3/(3*b) + d**2*x**2*sinh(a + b*x)**2*cosh(a + b*x)/b - 2*d**2*x**2*cosh
(a + b*x)**3/(3*b) - 14*c*d*sinh(a + b*x)**3/(9*b**2) + 4*c*d*sinh(a + b*x)*cosh(a + b*x)**2/(3*b**2) - 14*d**
2*x*sinh(a + b*x)**3/(9*b**2) + 4*d**2*x*sinh(a + b*x)*cosh(a + b*x)**2/(3*b**2) + 14*d**2*sinh(a + b*x)**2*co
sh(a + b*x)/(9*b**3) - 40*d**2*cosh(a + b*x)**3/(27*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sinh(
a)**3, True))

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Giac [B]  time = 1.18285, size = 311, normalized size = 2.53 \begin{align*} \frac{{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 6 \, b d^{2} x - 6 \, b c d + 2 \, d^{2}\right )} e^{\left (3 \, b x + 3 \, a\right )}}{216 \, b^{3}} - \frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, b d^{2} x - 2 \, b c d + 2 \, d^{2}\right )} e^{\left (b x + a\right )}}{8 \, b^{3}} - \frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + 2 \, b d^{2} x + 2 \, b c d + 2 \, d^{2}\right )} e^{\left (-b x - a\right )}}{8 \, b^{3}} + \frac{{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 6 \, b d^{2} x + 6 \, b c d + 2 \, d^{2}\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 - 6*b*d^2*x - 6*b*c*d + 2*d^2)*e^(3*b*x + 3*a)/b^3 - 3/8*(b^2*
d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*b*d^2*x - 2*b*c*d + 2*d^2)*e^(b*x + a)/b^3 - 3/8*(b^2*d^2*x^2 + 2*b^2*c*d*
x + b^2*c^2 + 2*b*d^2*x + 2*b*c*d + 2*d^2)*e^(-b*x - a)/b^3 + 1/216*(9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2
+ 6*b*d^2*x + 6*b*c*d + 2*d^2)*e^(-3*b*x - 3*a)/b^3