3.123 \(\int x \sqrt {a+a \cosh (c+d x)} \, dx\)

Optimal. Leaf size=53 \[ \frac {2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}}{d}-\frac {4 \sqrt {a \cosh (c+d x)+a}}{d^2} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3319, 3296, 2638} \[ \frac {2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}}{d}-\frac {4 \sqrt {a \cosh (c+d x)+a}}{d^2} \]

Antiderivative was successfully verified.

[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 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 3319

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*} \int x \sqrt {a+a \cosh (c+d x)} \, dx &=\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*}

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Mathematica [A]  time = 0.13, size = 34, normalized size = 0.64 \[ \frac {2 \left (d x \tanh \left (\frac {1}{2} (c+d x)\right )-2\right ) \sqrt {a (\cosh (c+d x)+1)}}{d^2} \]

Antiderivative was successfully verified.

[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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.13, size = 67, normalized size = 1.26 \[ \frac {\sqrt {2} {\left (\sqrt {a} d x e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \sqrt {a} d x e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}\right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.10, size = 64, normalized size = 1.21 \[ \frac {\sqrt {2}\, \sqrt {a \left ({\mathrm e}^{d x +c}+1\right )^{2} {\mathrm e}^{-d x -c}}\, \left (d x \,{\mathrm e}^{d x +c}-d x -2 \,{\mathrm e}^{d x +c}-2\right )}{\left ({\mathrm e}^{d x +c}+1\right ) d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 0.45, size = 60, normalized size = 1.13 \[ -\frac {{\left (\sqrt {2} \sqrt {a} d x - {\left (\sqrt {2} \sqrt {a} d x e^{c} - 2 \, \sqrt {2} \sqrt {a} e^{c}\right )} e^{\left (d x\right )} + 2 \, \sqrt {2} \sqrt {a}\right )} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.91, size = 56, normalized size = 1.06 \[ \frac {2\,x\,\mathrm {sinh}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}}{d\,\mathrm {cosh}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {a \left (\cosh {\left (c + d x \right )} + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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