\(\int (a+b x^3) \cosh (c+d x) \, dx\) [82]

Optimal result

Integrand size = 14, antiderivative size = 66 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {6 b \cosh (c+d x)}{d^4}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {a \sinh (c+d x)}{d}+\frac {6 b x \sinh (c+d x)}{d^3}+\frac {b x^3 \sinh (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5385, 2717, 3377, 2718} \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {a \sinh (c+d x)}{d}-\frac {6 b \cosh (c+d x)}{d^4}+\frac {6 b x \sinh (c+d x)}{d^3}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {b x^3 \sinh (c+d x)}{d} \]

[In]

Int[(a + b*x^3)*Cosh[c + d*x],x]

[Out]

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

Rule 2717

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

Rule 2718

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

Rule 3377

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 5385

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {-3 b \left (2+d^2 x^2\right ) \cosh (c+d x)+d \left (a d^2+b x \left (6+d^2 x^2\right )\right ) \sinh (c+d x)}{d^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23

method	result	size
parallelrisch	\(\frac {3 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{2} d^{2}-2 d \left (\left (b \,x^{3}+a \right ) d^{2}+6 b x \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+3 b \,d^{2} x^{2}+12 b}{d^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(81\)
risch	\(\frac {\left (b \,d^{3} x^{3}-3 b \,d^{2} x^{2}+d^{3} a +6 d x b -6 b \right ) {\mathrm e}^{d x +c}}{2 d^{4}}-\frac {\left (b \,d^{3} x^{3}+3 b \,d^{2} x^{2}+d^{3} a +6 d x b +6 b \right ) {\mathrm e}^{-d x -c}}{2 d^{4}}\)	\(89\)
parts	\(\frac {b \,x^{3} \sinh \left (d x +c \right )}{d}+\frac {a \sinh \left (d x +c \right )}{d}-\frac {3 b \left (c^{2} \cosh \left (d x +c \right )-2 c \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )+\left (d x +c \right )^{2} \cosh \left (d x +c \right )-2 \left (d x +c \right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )\right )}{d^{4}}\)	\(103\)
meijerg	\(\frac {8 b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (\frac {3 x^{2} d^{2}}{2}+3\right ) \cosh \left (d x \right )}{4 \sqrt {\pi }}+\frac {d x \left (\frac {x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}-\frac {8 i b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \left (\frac {5 x^{2} d^{2}}{2}+15\right ) \cosh \left (d x \right )}{20 \sqrt {\pi }}-\frac {i \left (\frac {15 x^{2} d^{2}}{2}+15\right ) \sinh \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {a \cosh \left (c \right ) \sinh \left (d x \right )}{d}-\frac {a \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (d x \right )}{\sqrt {\pi }}\right )}{d}\)	\(149\)
derivativedivides	\(\frac {-\frac {b \,c^{3} \sinh \left (d x +c \right )}{d^{3}}+\frac {3 b \,c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}-\frac {3 b c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+a \sinh \left (d x +c \right )}{d}\)	\(158\)
default	\(\frac {-\frac {b \,c^{3} \sinh \left (d x +c \right )}{d^{3}}+\frac {3 b \,c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}-\frac {3 b c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+a \sinh \left (d x +c \right )}{d}\)	\(158\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {3 \, {\left (b d^{2} x^{2} + 2 \, b\right )} \cosh \left (d x + c\right ) - {\left (b d^{3} x^{3} + a d^{3} + 6 \, b d x\right )} \sinh \left (d x + c\right )}{d^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.24 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\begin {cases} \frac {a \sinh {\left (c + d x \right )}}{d} + \frac {b x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 b x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 b \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (a x + \frac {b x^{4}}{4}\right ) \cosh {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*sinh(c + d*x)/d + 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 + b*x**4/4)*cosh(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.58 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {a e^{\left (d x + c\right )}}{2 \, d} - \frac {a e^{\left (-d x - c\right )}}{2 \, d} + \frac {{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} b e^{\left (d x\right )}}{2 \, d^{4}} - \frac {{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} b e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {{\left (b d^{3} x^{3} - 3 \, b d^{2} x^{2} + a d^{3} + 6 \, b d x - 6 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac {{\left (b d^{3} x^{3} + 3 \, b d^{2} x^{2} + a d^{3} + 6 \, b d x + 6 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {a\,\mathrm {sinh}\left (c+d\,x\right )+b\,x^3\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {6\,b\,\mathrm {cosh}\left (c+d\,x\right )}{d^4}+\frac {6\,b\,x\,\mathrm {sinh}\left (c+d\,x\right )}{d^3}-\frac {3\,b\,x^2\,\mathrm {cosh}\left (c+d\,x\right )}{d^2} \]

[In]

int(cosh(c + d*x)*(a + b*x^3),x)

[Out]

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