3.2.44 \(\int x^2 \sqrt {a+a \cos (c+d x)} \, dx\) [144]

Optimal. Leaf size=88 \[ \frac {8 x \sqrt {a+a \cos (c+d x)}}{d^2}-\frac {16 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3}+\frac {2 x^2 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \]

[Out]

8*x*(a+a*cos(d*x+c))^(1/2)/d^2-16*(a+a*cos(d*x+c))^(1/2)*tan(1/2*d*x+1/2*c)/d^3+2*x^2*(a+a*cos(d*x+c))^(1/2)*t
an(1/2*d*x+1/2*c)/d

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Rubi [A]
time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3400, 3377, 2717} \begin {gather*} -\frac {16 \tan \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}}{d^3}+\frac {8 x \sqrt {a \cos (c+d x)+a}}{d^2}+\frac {2 x^2 \tan \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(8*x*Sqrt[a + a*Cos[c + d*x]])/d^2 - (16*Sqrt[a + a*Cos[c + d*x]]*Tan[c/2 + (d*x)/2])/d^3 + (2*x^2*Sqrt[a + a*
Cos[c + d*x]]*Tan[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*} \int x^2 \sqrt {a+a \cos (c+d x)} \, dx &=\left (\sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x^2 \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx\\ &=\frac {2 x^2 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {\left (4 \sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \, dx}{d}\\ &=\frac {8 x \sqrt {a+a \cos (c+d x)}}{d^2}+\frac {2 x^2 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {\left (8 \sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \, dx}{d^2}\\ &=\frac {8 x \sqrt {a+a \cos (c+d x)}}{d^2}-\frac {16 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3}+\frac {2 x^2 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 44, normalized size = 0.50 \begin {gather*} \frac {2 \sqrt {a (1+\cos (c+d x))} \left (4 d x+\left (-8+d^2 x^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.06, size = 105, normalized size = 1.19

method	result	size
risch	\(-\frac {i \sqrt {2}\, \sqrt {a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2} {\mathrm e}^{-i \left (d x +c \right )}}\, \left (d^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}+4 i d x \,{\mathrm e}^{i \left (d x +c \right )}-d^{2} x^{2}+4 i d x -8 \,{\mathrm e}^{i \left (d x +c \right )}+8\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) d^{3}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 122, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (\sqrt {2} \sqrt {a} c^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, {\left (\sqrt {2} {\left (d x + c\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} c + {\left (\sqrt {2} {\left (d x + c\right )}^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} {\left (d x + c\right )} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}\right )}}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+a*cos(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 77, normalized size = 0.88 \begin {gather*} 2 \, \sqrt {2} \sqrt {a} {\left (\frac {4 \, x \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d^{2}} + \frac {{\left (d^{2} x^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.43, size = 63, normalized size = 0.72 \begin {gather*} \frac {2\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (4\,d\,x-8\,\sin \left (c+d\,x\right )+d^2\,x^2\,\sin \left (c+d\,x\right )+4\,d\,x\,\cos \left (c+d\,x\right )\right )}{d^3\,\left (\cos \left (c+d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + a*cos(c + d*x))^(1/2),x)

[Out]

(2*(a*(cos(c + d*x) + 1))^(1/2)*(4*d*x - 8*sin(c + d*x) + d^2*x^2*sin(c + d*x) + 4*d*x*cos(c + d*x)))/(d^3*(co
s(c + d*x) + 1))

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