3.42 \(\int x (a+b x^2) \cosh (c+d x) \, dx\)

Optimal. Leaf size=79 \[ -\frac{a \cosh (c+d x)}{d^2}+\frac{a x \sinh (c+d x)}{d}-\frac{3 b x^2 \cosh (c+d x)}{d^2}+\frac{6 b x \sinh (c+d x)}{d^3}-\frac{6 b \cosh (c+d x)}{d^4}+\frac{b x^3 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.121082, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5287, 3296, 2638} \[ -\frac{a \cosh (c+d x)}{d^2}+\frac{a x \sinh (c+d x)}{d}-\frac{3 b x^2 \cosh (c+d x)}{d^2}+\frac{6 b x \sinh (c+d x)}{d^3}-\frac{6 b \cosh (c+d x)}{d^4}+\frac{b x^3 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)*Cosh[c + d*x],x]

[Out]

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

Rule 5287

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[Cosh[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.0997394, size = 57, normalized size = 0.72 \[ \frac{d x \left (a d^2+b \left (d^2 x^2+6\right )\right ) \sinh (c+d x)-\left (a d^2+3 b \left (d^2 x^2+2\right )\right ) \cosh (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 183, normalized size = 2.3 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{b \left ( \left ( dx+c \right ) ^{3}\sinh \left ( dx+c \right ) -3\, \left ( dx+c \right ) ^{2}\cosh \left ( dx+c \right ) +6\, \left ( dx+c \right ) \sinh \left ( dx+c \right ) -6\,\cosh \left ( dx+c \right ) \right ) }{{d}^{2}}}-3\,{\frac{cb \left ( \left ( dx+c \right ) ^{2}\sinh \left ( dx+c \right ) -2\, \left ( dx+c \right ) \cosh \left ( dx+c \right ) +2\,\sinh \left ( dx+c \right ) \right ) }{{d}^{2}}}+3\,{\frac{b{c}^{2} \left ( \left ( dx+c \right ) \sinh \left ( dx+c \right ) -\cosh \left ( dx+c \right ) \right ) }{{d}^{2}}}+a \left ( \left ( dx+c \right ) \sinh \left ( dx+c \right ) -\cosh \left ( dx+c \right ) \right ) -{\frac{b{c}^{3}\sinh \left ( dx+c \right ) }{{d}^{2}}}-ca\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.08556, size = 288, normalized size = 3.65 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{2} \cosh \left (d x + c\right )}{4 \, b} - \frac{{\left (\frac{a^{2} e^{\left (d x + c\right )}}{d} + \frac{a^{2} e^{\left (-d x - c\right )}}{d} + \frac{2 \,{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a b e^{\left (d x\right )}}{d^{3}} + \frac{2 \,{\left (d^{2} x^{2} + 2 \, d x + 2\right )} a b e^{\left (-d x - c\right )}}{d^{3}} + \frac{{\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} b^{2} e^{\left (d x\right )}}{d^{5}} + \frac{{\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} b^{2} e^{\left (-d x - c\right )}}{d^{5}}\right )} d}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2 + a)^2*cosh(d*x + c)/b - 1/8*(a^2*e^(d*x + c)/d + a^2*e^(-d*x - c)/d + 2*(d^2*x^2*e^c - 2*d*x*e^c +
2*e^c)*a*b*e^(d*x)/d^3 + 2*(d^2*x^2 + 2*d*x + 2)*a*b*e^(-d*x - c)/d^3 + (d^4*x^4*e^c - 4*d^3*x^3*e^c + 12*d^2
*x^2*e^c - 24*d*x*e^c + 24*e^c)*b^2*e^(d*x)/d^5 + (d^4*x^4 + 4*d^3*x^3 + 12*d^2*x^2 + 24*d*x + 24)*b^2*e^(-d*x
- c)/d^5)*d/b

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Fricas [A]  time = 1.98437, size = 132, normalized size = 1.67 \begin{align*} -\frac{{\left (3 \, b d^{2} x^{2} + a d^{2} + 6 \, b\right )} \cosh \left (d x + c\right ) -{\left (b d^{3} x^{3} +{\left (a d^{3} + 6 \, b d\right )} x\right )} \sinh \left (d x + c\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.31972, size = 99, normalized size = 1.25 \begin{align*} \begin{cases} \frac{a x \sinh{\left (c + d x \right )}}{d} - \frac{a \cosh{\left (c + d x \right )}}{d^{2}} + \frac{b x^{3} \sinh{\left (c + d x \right )}}{d} - \frac{3 b x^{2} \cosh{\left (c + d x \right )}}{d^{2}} + \frac{6 b x \sinh{\left (c + d x \right )}}{d^{3}} - \frac{6 b \cosh{\left (c + d x \right )}}{d^{4}} & \text{for}\: d \neq 0 \\\left (\frac{a x^{2}}{2} + \frac{b x^{4}}{4}\right ) \cosh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17752, size = 136, normalized size = 1.72 \begin{align*} \frac{{\left (b d^{3} x^{3} + a d^{3} x - 3 \, b d^{2} x^{2} - a d^{2} + 6 \, b d x - 6 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac{{\left (b d^{3} x^{3} + a d^{3} x + 3 \, b d^{2} x^{2} + a d^{2} + 6 \, b d x + 6 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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