\(\int (a+b x^2) \cosh (c+d x) \, dx\) [43]

Optimal result

Integrand size = 14, antiderivative size = 51 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=-\frac {2 b x \cosh (c+d x)}{d^2}+\frac {2 b \sinh (c+d x)}{d^3}+\frac {a \sinh (c+d x)}{d}+\frac {b x^2 \sinh (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5385, 2717, 3377} \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\frac {a \sinh (c+d x)}{d}+\frac {2 b \sinh (c+d x)}{d^3}-\frac {2 b x \cosh (c+d x)}{d^2}+\frac {b x^2 \sinh (c+d x)}{d} \]

[In]

Int[(a + b*x^2)*Cosh[c + d*x],x]

[Out]

(-2*b*x*Cosh[c + d*x])/d^2 + (2*b*Sinh[c + d*x])/d^3 + (a*Sinh[c + d*x])/d + (b*x^2*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 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^2 \cosh (c+d x)\right ) \, dx \\ & = a \int \cosh (c+d x) \, dx+b \int x^2 \cosh (c+d x) \, dx \\ & = \frac {a \sinh (c+d x)}{d}+\frac {b x^2 \sinh (c+d x)}{d}-\frac {(2 b) \int x \sinh (c+d x) \, dx}{d} \\ & = -\frac {2 b x \cosh (c+d x)}{d^2}+\frac {a \sinh (c+d x)}{d}+\frac {b x^2 \sinh (c+d x)}{d}+\frac {(2 b) \int \cosh (c+d x) \, dx}{d^2} \\ & = -\frac {2 b x \cosh (c+d x)}{d^2}+\frac {2 b \sinh (c+d x)}{d^3}+\frac {a \sinh (c+d x)}{d}+\frac {b x^2 \sinh (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\frac {-2 b d x \cosh (c+d x)+\left (a d^2+b \left (2+d^2 x^2\right )\right ) \sinh (c+d x)}{d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
parts	\(\frac {b \,x^{2} \sinh \left (d x +c \right )}{d}+\frac {a \sinh \left (d x +c \right )}{d}-\frac {2 b \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )-c \cosh \left (d x +c \right )\right )}{d^{3}}\)	\(63\)
parallelrisch	\(\frac {2 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b d +2 \left (\left (-b \,x^{2}-a \right ) d^{2}-2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 d x b}{d^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(69\)
risch	\(\frac {\left (b \,d^{2} x^{2}+a \,d^{2}-2 d x b +2 b \right ) {\mathrm e}^{d x +c}}{2 d^{3}}-\frac {\left (b \,d^{2} x^{2}+a \,d^{2}+2 d x b +2 b \right ) {\mathrm e}^{-d x -c}}{2 d^{3}}\)	\(71\)
derivativedivides	\(\frac {\frac {b \,c^{2} \sinh \left (d x +c \right )}{d^{2}}-\frac {2 b c \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{2}}+\frac {b \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{2}}+a \sinh \left (d x +c \right )}{d}\)	\(97\)
default	\(\frac {\frac {b \,c^{2} \sinh \left (d x +c \right )}{d^{2}}-\frac {2 b c \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{2}}+\frac {b \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{2}}+a \sinh \left (d x +c \right )}{d}\)	\(97\)
meijerg	\(\frac {4 i b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {i \left (\frac {3 x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{6 \sqrt {\pi }}\right )}{d^{3}}+\frac {4 b \sinh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (\frac {x^{2} d^{2}}{2}+1\right ) \cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+\frac {a \cosh \left (c \right ) \sinh \left (d x \right )}{d}-\frac {a \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (d x \right )}{\sqrt {\pi }}\right )}{d}\)	\(129\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=-\frac {2 \, b d x \cosh \left (d x + c\right ) - {\left (b d^{2} x^{2} + a d^{2} + 2 \, b\right )} \sinh \left (d x + c\right )}{d^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\begin {cases} \frac {a \sinh {\left (c + d x \right )}}{d} + \frac {b x^{2} \sinh {\left (c + d x \right )}}{d} - \frac {2 b x \cosh {\left (c + d x \right )}}{d^{2}} + \frac {2 b \sinh {\left (c + d x \right )}}{d^{3}} & \text {for}\: d \neq 0 \\\left (a x + \frac {b x^{3}}{3}\right ) \cosh {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.69 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\frac {a e^{\left (d x + c\right )}}{2 \, d} - \frac {a e^{\left (-d x - c\right )}}{2 \, d} + \frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} b e^{\left (d x\right )}}{2 \, d^{3}} - \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} b e^{\left (-d x - c\right )}}{2 \, d^{3}} \]

[In]

integrate((b*x^2+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^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*b*e^(d*x)/d^3 - 1/2*(d^2*x^
2 + 2*d*x + 2)*b*e^(-d*x - c)/d^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.37 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\frac {{\left (b d^{2} x^{2} + a d^{2} - 2 \, b d x + 2 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{3}} - \frac {{\left (b d^{2} x^{2} + a d^{2} + 2 \, b d x + 2 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int \left (a+b x^2\right ) \cosh (c+d x) \, dx=\frac {\mathrm {sinh}\left (c+d\,x\right )\,\left (a\,d^2+2\,b\right )}{d^3}-\frac {2\,b\,x\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}+\frac {b\,x^2\,\mathrm {sinh}\left (c+d\,x\right )}{d} \]

[In]

int(cosh(c + d*x)*(a + b*x^2),x)

[Out]

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