\(\int x \sqrt {a+a \cosh (c+d x)} \, dx\) [123]

Optimal result

Integrand size = 16, antiderivative size = 53 \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=-\frac {4 \sqrt {a+a \cosh (c+d x)}}{d^2}+\frac {2 x \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \]

[Out]

-4*(a+a*cosh(d*x+c))^(1/2)/d^2+2*x*(a+a*cosh(d*x+c))^(1/2)*tanh(1/2*d*x+1/2*c)/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3400, 3377, 2718} \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\frac {2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}}{d}-\frac {4 \sqrt {a \cosh (c+d x)+a}}{d^2} \]

[In]

Int[x*Sqrt[a + a*Cosh[c + d*x]],x]

[Out]

(-4*Sqrt[a + a*Cosh[c + d*x]])/d^2 + (2*x*Sqrt[a + a*Cosh[c + d*x]]*Tanh[c/2 + (d*x)/2])/d

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 3400

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(2*a)^IntPart[n]
*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n])), Int[(c + d*x)^m*Sin[e/2
+ a*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int x \sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right ) \, dx \\ & = \frac {2 x \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {\left (2 \sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \sinh \left (\frac {c}{2}+\frac {d x}{2}\right ) \, dx}{d} \\ & = -\frac {4 \sqrt {a+a \cosh (c+d x)}}{d^2}+\frac {2 x \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.64 \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\frac {2 \sqrt {a (1+\cosh (c+d x))} \left (-2+d x \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{d^2} \]

[In]

Integrate[x*Sqrt[a + a*Cosh[c + d*x]],x]

[Out]

(2*Sqrt[a*(1 + Cosh[c + d*x])]*(-2 + d*x*Tanh[(c + d*x)/2]))/d^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {\sqrt {2}\, \sqrt {a \left ({\mathrm e}^{d x +c}+1\right )^{2} {\mathrm e}^{-d x -c}}\, \left (d x \,{\mathrm e}^{d x +c}-d x -2 \,{\mathrm e}^{d x +c}-2\right )}{\left ({\mathrm e}^{d x +c}+1\right ) d^{2}}\)	\(64\)

[In]

int(x*(a+a*cosh(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2^(1/2)*(a*(exp(d*x+c)+1)^2*exp(-d*x-c))^(1/2)/(exp(d*x+c)+1)*(d*x*exp(d*x+c)-d*x-2*exp(d*x+c)-2)/d^2

Fricas [F(-2)]

Exception generated. \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*(a+a*cosh(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\int x \sqrt {a \left (\cosh {\left (c + d x \right )} + 1\right )}\, dx \]

[In]

integrate(x*(a+a*cosh(d*x+c))**(1/2),x)

[Out]

Integral(x*sqrt(a*(cosh(c + d*x) + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13 \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=-\frac {{\left (\sqrt {2} \sqrt {a} d x - {\left (\sqrt {2} \sqrt {a} d x e^{c} - 2 \, \sqrt {2} \sqrt {a} e^{c}\right )} e^{\left (d x\right )} + 2 \, \sqrt {2} \sqrt {a}\right )} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{d^{2}} \]

[In]

integrate(x*(a+a*cosh(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26 \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\frac {\sqrt {2} {\left (\sqrt {a} d x e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \sqrt {a} d x e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}\right )}}{d^{2}} \]

[In]

integrate(x*(a+a*cosh(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.81 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int x \sqrt {a+a \cosh (c+d x)} \, dx=\frac {2\,x\,\mathrm {sinh}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}}{d\,\mathrm {cosh}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}}{d^2} \]

[In]

int(x*(a + a*cosh(c + d*x))^(1/2),x)

[Out]

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