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

Optimal. Leaf size=95 \[ -\frac{d (c+d x) \sinh ^2(a+b x)}{2 b^2}+\frac{d^2 \sinh (a+b x) \cosh (a+b x)}{4 b^3}+\frac{(c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{d^2 x}{4 b^2}-\frac{(c+d x)^3}{6 d} \]

[Out]

-(d^2*x)/(4*b^2) - (c + d*x)^3/(6*d) + (d^2*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^3) + ((c + d*x)^2*Cosh[a + b*x]*
Sinh[a + b*x])/(2*b) - (d*(c + d*x)*Sinh[a + b*x]^2)/(2*b^2)

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Rubi [A]  time = 0.0543412, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 32, 2635, 8} \[ -\frac{d (c+d x) \sinh ^2(a+b x)}{2 b^2}+\frac{d^2 \sinh (a+b x) \cosh (a+b x)}{4 b^3}+\frac{(c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{d^2 x}{4 b^2}-\frac{(c+d x)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sinh[a + b*x]^2,x]

[Out]

-(d^2*x)/(4*b^2) - (c + d*x)^3/(6*d) + (d^2*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^3) + ((c + d*x)^2*Cosh[a + b*x]*
Sinh[a + b*x])/(2*b) - (d*(c + d*x)*Sinh[a + b*x]^2)/(2*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.310909, size = 75, normalized size = 0.79 \[ \frac{3 \sinh (2 (a+b x)) \left (2 b^2 (c+d x)^2+d^2\right )-6 b d (c+d x) \cosh (2 (a+b x))-4 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^3*x*(3*c^2 + 3*c*d*x + d^2*x^2) - 6*b*d*(c + d*x)*Cosh[2*(a + b*x)] + 3*(d^2 + 2*b^2*(c + d*x)^2)*Sinh[2
*(a + b*x)])/(24*b^3)

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Maple [B]  time = 0.007, size = 262, normalized size = 2.8 \begin{align*}{\frac{1}{b} \left ({\frac{{d}^{2}}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{ \left ( bx+a \right ) ^{3}}{6}}-{\frac{ \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{4}}+{\frac{bx}{4}}+{\frac{a}{4}} \right ) }-2\,{\frac{{d}^{2}a \left ( 1/2\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/4\, \left ( bx+a \right ) ^{2}-1/4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) }{{b}^{2}}}+{\frac{{a}^{2}{d}^{2}}{{b}^{2}} \left ({\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{bx}{2}}-{\frac{a}{2}} \right ) }+2\,{\frac{cd \left ( 1/2\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/4\, \left ( bx+a \right ) ^{2}-1/4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) }{b}}-2\,{\frac{cda \left ( 1/2\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/2\,bx-a/2 \right ) }{b}}+{c}^{2} \left ({\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{bx}{2}}-{\frac{a}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b^2*d^2*(1/2*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)-1/6*(b*x+a)^3-1/2*(b*x+a)*cosh(b*x+a)^2+1/4*cosh(b*x+a)*
sinh(b*x+a)+1/4*b*x+1/4*a)-2/b^2*d^2*a*(1/2*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-1/4*(b*x+a)^2-1/4*cosh(b*x+a)^2)+1
/b^2*d^2*a^2*(1/2*cosh(b*x+a)*sinh(b*x+a)-1/2*b*x-1/2*a)+2/b*c*d*(1/2*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-1/4*(b*x
+a)^2-1/4*cosh(b*x+a)^2)-2/b*c*d*a*(1/2*cosh(b*x+a)*sinh(b*x+a)-1/2*b*x-1/2*a)+c^2*(1/2*cosh(b*x+a)*sinh(b*x+a
)-1/2*b*x-1/2*a))

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Maxima [A]  time = 1.14048, size = 223, normalized size = 2.35 \begin{align*} -\frac{1}{8} \,{\left (4 \, x^{2} - \frac{{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2}} + \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{2}}\right )} c d - \frac{1}{48} \,{\left (8 \, x^{3} - \frac{3 \,{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{3}} + \frac{3 \,{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{3}}\right )} d^{2} - \frac{1}{8} \, c^{2}{\left (4 \, x - \frac{e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*x^2 - (2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 + (2*b*x + 1)*e^(-2*b*x - 2*a)/b^2)*c*d - 1/48*(8*x^3 -
3*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x)/b^3 + 3*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3
)*d^2 - 1/8*c^2*(4*x - e^(2*b*x + 2*a)/b + e^(-2*b*x - 2*a)/b)

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Fricas [A]  time = 2.69329, size = 288, normalized size = 3.03 \begin{align*} -\frac{2 \, b^{3} d^{2} x^{3} + 6 \, b^{3} c d x^{2} + 6 \, b^{3} c^{2} x + 3 \,{\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right )^{2} - 3 \,{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} + d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 3 \,{\left (b d^{2} x + b c d\right )} \sinh \left (b x + a\right )^{2}}{12 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*b^3*d^2*x^3 + 6*b^3*c*d*x^2 + 6*b^3*c^2*x + 3*(b*d^2*x + b*c*d)*cosh(b*x + a)^2 - 3*(2*b^2*d^2*x^2 +
4*b^2*c*d*x + 2*b^2*c^2 + d^2)*cosh(b*x + a)*sinh(b*x + a) + 3*(b*d^2*x + b*c*d)*sinh(b*x + a)^2)/b^3

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Sympy [A]  time = 1.77054, size = 264, normalized size = 2.78 \begin{align*} \begin{cases} \frac{c^{2} x \sinh ^{2}{\left (a + b x \right )}}{2} - \frac{c^{2} x \cosh ^{2}{\left (a + b x \right )}}{2} + \frac{c d x^{2} \sinh ^{2}{\left (a + b x \right )}}{2} - \frac{c d x^{2} \cosh ^{2}{\left (a + b x \right )}}{2} + \frac{d^{2} x^{3} \sinh ^{2}{\left (a + b x \right )}}{6} - \frac{d^{2} x^{3} \cosh ^{2}{\left (a + b x \right )}}{6} + \frac{c^{2} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2 b} + \frac{c d x \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{b} + \frac{d^{2} x^{2} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2 b} - \frac{c d \sinh ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac{d^{2} x \sinh ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac{d^{2} x \cosh ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac{d^{2} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4 b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sinh ^{2}{\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)**2,x)

[Out]

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

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Giac [A]  time = 1.17973, size = 184, normalized size = 1.94 \begin{align*} -\frac{1}{6} \, d^{2} x^{3} - \frac{1}{2} \, c d x^{2} - \frac{1}{2} \, c^{2} x + \frac{{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - 2 \, b d^{2} x - 2 \, b c d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{3}} - \frac{{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} + 2 \, b d^{2} x + 2 \, b c d + d^{2}\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*d^2*x^3 - 1/2*c*d*x^2 - 1/2*c^2*x + 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - 2*b*d^2*x - 2*b*c*d +
d^2)*e^(2*b*x + 2*a)/b^3 - 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 + 2*b*d^2*x + 2*b*c*d + d^2)*e^(-2*b
*x - 2*a)/b^3