\(\int x^2 \sqrt {a+a \cosh (c+d x)} \, dx\) [122]

Optimal result

Integrand size = 18, antiderivative size = 88 \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=-\frac {8 x \sqrt {a+a \cosh (c+d x)}}{d^2}+\frac {16 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3}+\frac {2 x^2 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \]

[Out]

-8*x*(a+a*cosh(d*x+c))^(1/2)/d^2+16*(a+a*cosh(d*x+c))^(1/2)*tanh(1/2*d*x+1/2*c)/d^3+2*x^2*(a+a*cosh(d*x+c))^(1
/2)*tanh(1/2*d*x+1/2*c)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3400, 3377, 2717} \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=\frac {16 \tanh \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}}{d^3}-\frac {8 x \sqrt {a \cosh (c+d x)+a}}{d^2}+\frac {2 x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}}{d} \]

[In]

Int[x^2*Sqrt[a + a*Cosh[c + d*x]],x]

[Out]

(-8*x*Sqrt[a + a*Cosh[c + d*x]])/d^2 + (16*Sqrt[a + a*Cosh[c + d*x]]*Tanh[c/2 + (d*x)/2])/d^3 + (2*x^2*Sqrt[a
+ a*Cosh[c + d*x]]*Tanh[c/2 + (d*x)/2])/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 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^2 \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^2 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {\left (4 \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 \sinh \left (\frac {c}{2}+\frac {d x}{2}\right ) \, dx}{d} \\ & = -\frac {8 x \sqrt {a+a \cosh (c+d x)}}{d^2}+\frac {2 x^2 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {\left (8 \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 \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \, dx}{d^2} \\ & = -\frac {8 x \sqrt {a+a \cosh (c+d x)}}{d^2}+\frac {16 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3}+\frac {2 x^2 \sqrt {a+a \cosh (c+d x)} \tanh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=\frac {2 \sqrt {a (1+\cosh (c+d x))} \left (-4 d x+\left (8+d^2 x^2\right ) \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{d^3} \]

[In]

Integrate[x^2*Sqrt[a + a*Cosh[c + d*x]],x]

[Out]

(2*Sqrt[a*(1 + Cosh[c + d*x])]*(-4*d*x + (8 + d^2*x^2)*Tanh[(c + d*x)/2]))/d^3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {\sqrt {2}\, \sqrt {a \left ({\mathrm e}^{d x +c}+1\right )^{2} {\mathrm e}^{-d x -c}}\, \left (d^{2} x^{2} {\mathrm e}^{d x +c}-x^{2} d^{2}-4 d x \,{\mathrm e}^{d x +c}-4 d x +8 \,{\mathrm e}^{d x +c}-8\right )}{\left ({\mathrm e}^{d x +c}+1\right ) d^{3}}\)	\(86\)

[In]

int(x^2*(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^2*x^2*exp(d*x+c)-x^2*d^2-4*d*x*exp(d*x+c)-4*d
*x+8*exp(d*x+c)-8)/d^3

Fricas [F(-2)]

Exception generated. \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^2*(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]

\[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=\int x^{2} \sqrt {a \left (\cosh {\left (c + d x \right )} + 1\right )}\, dx \]

[In]

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

[Out]

Integral(x**2*sqrt(a*(cosh(c + d*x) + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=-\frac {{\left (\sqrt {2} \sqrt {a} d^{2} x^{2} + 4 \, \sqrt {2} \sqrt {a} d x - {\left (\sqrt {2} \sqrt {a} d^{2} x^{2} e^{c} - 4 \, \sqrt {2} \sqrt {a} d x e^{c} + 8 \, \sqrt {2} \sqrt {a} e^{c}\right )} e^{\left (d x\right )} + 8 \, \sqrt {2} \sqrt {a}\right )} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}}{d^{3}} \]

[In]

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

[Out]

-(sqrt(2)*sqrt(a)*d^2*x^2 + 4*sqrt(2)*sqrt(a)*d*x - (sqrt(2)*sqrt(a)*d^2*x^2*e^c - 4*sqrt(2)*sqrt(a)*d*x*e^c +
8*sqrt(2)*sqrt(a)*e^c)*e^(d*x) + 8*sqrt(2)*sqrt(a))*e^(-1/2*d*x - 1/2*c)/d^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.22 \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=\frac {\sqrt {2} {\left (\sqrt {a} d^{2} x^{2} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \sqrt {a} d^{2} x^{2} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 4 \, \sqrt {a} d x e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 4 \, \sqrt {a} d x e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} + 8 \, \sqrt {a} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 8 \, \sqrt {a} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}\right )}}{d^{3}} \]

[In]

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

[Out]

sqrt(2)*(sqrt(a)*d^2*x^2*e^(1/2*d*x + 1/2*c) - sqrt(a)*d^2*x^2*e^(-1/2*d*x - 1/2*c) - 4*sqrt(a)*d*x*e^(1/2*d*x
+ 1/2*c) - 4*sqrt(a)*d*x*e^(-1/2*d*x - 1/2*c) + 8*sqrt(a)*e^(1/2*d*x + 1/2*c) - 8*sqrt(a)*e^(-1/2*d*x - 1/2*c
))/d^3

Mupad [B] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int x^2 \sqrt {a+a \cosh (c+d x)} \, dx=-\frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{c+d\,x}}{2}+\frac {{\mathrm {e}}^{-c-d\,x}}{2}\right )}\,\left (\frac {8\,x}{d^2}-\frac {16\,{\mathrm {e}}^{c+d\,x}}{d^3}+\frac {16}{d^3}+\frac {2\,x^2}{d}-\frac {2\,x^2\,{\mathrm {e}}^{c+d\,x}}{d}+\frac {8\,x\,{\mathrm {e}}^{c+d\,x}}{d^2}\right )}{{\mathrm {e}}^{c+d\,x}+1} \]

[In]

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

[Out]

-((a + a*(exp(c + d*x)/2 + exp(- c - d*x)/2))^(1/2)*((8*x)/d^2 - (16*exp(c + d*x))/d^3 + 16/d^3 + (2*x^2)/d -
(2*x^2*exp(c + d*x))/d + (8*x*exp(c + d*x))/d^2))/(exp(c + d*x) + 1)