3.81 \(\int x (a+b x^3) \cosh (c+d x) \, dx\)

Optimal. Leaf size=94 \[ -\frac {a \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {24 b \sinh (c+d x)}{d^5}-\frac {24 b x \cosh (c+d x)}{d^4}+\frac {12 b x^2 \sinh (c+d x)}{d^3}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {b x^4 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5287, 3296, 2638, 2637} \[ -\frac {a \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {24 b \sinh (c+d x)}{d^5}-\frac {24 b x \cosh (c+d x)}{d^4}+\frac {b x^4 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^3)*Cosh[c + d*x],x]

[Out]

-((a*Cosh[c + d*x])/d^2) - (24*b*x*Cosh[c + d*x])/d^4 - (4*b*x^3*Cosh[c + d*x])/d^2 + (24*b*Sinh[c + d*x])/d^5
+ (a*x*Sinh[c + d*x])/d + (12*b*x^2*Sinh[c + d*x])/d^3 + (b*x^4*Sinh[c + d*x])/d

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

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 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]

Rubi steps

\begin {align*} \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx &=\int \left (a x \cosh (c+d x)+b x^4 \cosh (c+d x)\right ) \, dx\\ &=a \int x \cosh (c+d x) \, dx+b \int x^4 \cosh (c+d x) \, dx\\ &=\frac {a x \sinh (c+d x)}{d}+\frac {b x^4 \sinh (c+d x)}{d}-\frac {a \int \sinh (c+d x) \, dx}{d}-\frac {(4 b) \int x^3 \sinh (c+d x) \, dx}{d}\\ &=-\frac {a \cosh (c+d x)}{d^2}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {b x^4 \sinh (c+d x)}{d}+\frac {(12 b) \int x^2 \cosh (c+d x) \, dx}{d^2}\\ &=-\frac {a \cosh (c+d x)}{d^2}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d}-\frac {(24 b) \int x \sinh (c+d x) \, dx}{d^3}\\ &=-\frac {a \cosh (c+d x)}{d^2}-\frac {24 b x \cosh (c+d x)}{d^4}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d}+\frac {(24 b) \int \cosh (c+d x) \, dx}{d^4}\\ &=-\frac {a \cosh (c+d x)}{d^2}-\frac {24 b x \cosh (c+d x)}{d^4}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {24 b \sinh (c+d x)}{d^5}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 66, normalized size = 0.70 \[ \frac {\left (a d^4 x+b \left (d^4 x^4+12 d^2 x^2+24\right )\right ) \sinh (c+d x)-d \left (a d^2+4 b x \left (d^2 x^2+6\right )\right ) \cosh (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*(a*d^2 + 4*b*x*(6 + d^2*x^2))*Cosh[c + d*x]) + (a*d^4*x + b*(24 + 12*d^2*x^2 + d^4*x^4))*Sinh[c + d*x])/d
^5

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fricas [A]  time = 0.60, size = 68, normalized size = 0.72 \[ -\frac {{\left (4 \, b d^{3} x^{3} + a d^{3} + 24 \, b d x\right )} \cosh \left (d x + c\right ) - {\left (b d^{4} x^{4} + a d^{4} x + 12 \, b d^{2} x^{2} + 24 \, b\right )} \sinh \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 119, normalized size = 1.27 \[ \frac {{\left (b d^{4} x^{4} - 4 \, b d^{3} x^{3} + a d^{4} x + 12 \, b d^{2} x^{2} - a d^{3} - 24 \, b d x + 24 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{5}} - \frac {{\left (b d^{4} x^{4} + 4 \, b d^{3} x^{3} + a d^{4} x + 12 \, b d^{2} x^{2} + a d^{3} + 24 \, b d x + 24 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 257, normalized size = 2.73 \[ \frac {\frac {b \left (\left (d x +c \right )^{4} \sinh \left (d x +c \right )-4 \left (d x +c \right )^{3} \cosh \left (d x +c \right )+12 \left (d x +c \right )^{2} \sinh \left (d x +c \right )-24 \left (d x +c \right ) \cosh \left (d x +c \right )+24 \sinh \left (d x +c \right )\right )}{d^{3}}-\frac {4 b c \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+\frac {6 b \,c^{2} \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}-\frac {4 b \,c^{3} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}+a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )+\frac {b \,c^{4} \sinh \left (d x +c \right )}{d^{3}}-a c \sinh \left (d x +c \right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(1/d^3*b*((d*x+c)^4*sinh(d*x+c)-4*(d*x+c)^3*cosh(d*x+c)+12*(d*x+c)^2*sinh(d*x+c)-24*(d*x+c)*cosh(d*x+c)+
24*sinh(d*x+c))-4/d^3*b*c*((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))+
6/d^3*b*c^2*((d*x+c)^2*sinh(d*x+c)-2*(d*x+c)*cosh(d*x+c)+2*sinh(d*x+c))-4/d^3*b*c^3*((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^3*b*c^4*sinh(d*x+c)-a*c*sinh(d*x+c))

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maxima [B]  time = 0.35, size = 196, normalized size = 2.09 \[ -\frac {1}{20} \, d {\left (\frac {5 \, {\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a e^{\left (d x\right )}}{d^{3}} + \frac {5 \, {\left (d^{2} x^{2} + 2 \, d x + 2\right )} a e^{\left (-d x - c\right )}}{d^{3}} + \frac {2 \, {\left (d^{5} x^{5} e^{c} - 5 \, d^{4} x^{4} e^{c} + 20 \, d^{3} x^{3} e^{c} - 60 \, d^{2} x^{2} e^{c} + 120 \, d x e^{c} - 120 \, e^{c}\right )} b e^{\left (d x\right )}}{d^{6}} + \frac {2 \, {\left (d^{5} x^{5} + 5 \, d^{4} x^{4} + 20 \, d^{3} x^{3} + 60 \, d^{2} x^{2} + 120 \, d x + 120\right )} b e^{\left (-d x - c\right )}}{d^{6}}\right )} + \frac {1}{10} \, {\left (2 \, b x^{5} + 5 \, a x^{2}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*d*(5*(d^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*a*e^(d*x)/d^3 + 5*(d^2*x^2 + 2*d*x + 2)*a*e^(-d*x - c)/d^3 + 2*(d
^5*x^5*e^c - 5*d^4*x^4*e^c + 20*d^3*x^3*e^c - 60*d^2*x^2*e^c + 120*d*x*e^c - 120*e^c)*b*e^(d*x)/d^6 + 2*(d^5*x
^5 + 5*d^4*x^4 + 20*d^3*x^3 + 60*d^2*x^2 + 120*d*x + 120)*b*e^(-d*x - c)/d^6) + 1/10*(2*b*x^5 + 5*a*x^2)*cosh(
d*x + c)

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mupad [B]  time = 0.92, size = 92, normalized size = 0.98 \[ \frac {b\,x^4\,\mathrm {sinh}\left (c+d\,x\right )+a\,x\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {a\,\mathrm {cosh}\left (c+d\,x\right )+4\,b\,x^3\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}+\frac {24\,b\,\mathrm {sinh}\left (c+d\,x\right )}{d^5}-\frac {24\,b\,x\,\mathrm {cosh}\left (c+d\,x\right )}{d^4}+\frac {12\,b\,x^2\,\mathrm {sinh}\left (c+d\,x\right )}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(c + d*x)*(a + b*x^3),x)

[Out]

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

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sympy [A]  time = 1.92, size = 116, normalized size = 1.23 \[ \begin {cases} \frac {a x \sinh {\left (c + d x \right )}}{d} - \frac {a \cosh {\left (c + d x \right )}}{d^{2}} + \frac {b x^{4} \sinh {\left (c + d x \right )}}{d} - \frac {4 b x^{3} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {12 b x^{2} \sinh {\left (c + d x \right )}}{d^{3}} - \frac {24 b x \cosh {\left (c + d x \right )}}{d^{4}} + \frac {24 b \sinh {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{2}}{2} + \frac {b x^{5}}{5}\right ) \cosh {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x*sinh(c + d*x)/d - a*cosh(c + d*x)/d**2 + b*x**4*sinh(c + d*x)/d - 4*b*x**3*cosh(c + d*x)/d**2 +
12*b*x**2*sinh(c + d*x)/d**3 - 24*b*x*cosh(c + d*x)/d**4 + 24*b*sinh(c + d*x)/d**5, Ne(d, 0)), ((a*x**2/2 + b
*x**5/5)*cosh(c), True))

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