3.143 \(\int x^3 \sqrt{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=110 \[ \frac{12 x^2 \sqrt{a \cos (c+d x)+a}}{d^2}-\frac{96 \sqrt{a \cos (c+d x)+a}}{d^4}-\frac{48 x \tan \left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a \cos (c+d x)+a}}{d^3}+\frac{2 x^3 \tan \left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a \cos (c+d x)+a}}{d} \]

[Out]

(-96*Sqrt[a + a*Cos[c + d*x]])/d^4 + (12*x^2*Sqrt[a + a*Cos[c + d*x]])/d^2 - (48*x*Sqrt[a + a*Cos[c + d*x]]*Ta
n[c/2 + (d*x)/2])/d^3 + (2*x^3*Sqrt[a + a*Cos[c + d*x]]*Tan[c/2 + (d*x)/2])/d

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Rubi [A]  time = 0.133953, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3319, 3296, 2638} \[ \frac{12 x^2 \sqrt{a \cos (c+d x)+a}}{d^2}-\frac{96 \sqrt{a \cos (c+d x)+a}}{d^4}-\frac{48 x \tan \left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a \cos (c+d x)+a}}{d^3}+\frac{2 x^3 \tan \left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a \cos (c+d x)+a}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(-96*Sqrt[a + a*Cos[c + d*x]])/d^4 + (12*x^2*Sqrt[a + a*Cos[c + d*x]])/d^2 - (48*x*Sqrt[a + a*Cos[c + d*x]]*Ta
n[c/2 + (d*x)/2])/d^3 + (2*x^3*Sqrt[a + a*Cos[c + d*x]]*Tan[c/2 + (d*x)/2])/d

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])

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \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^3 \sin \left (\frac{1}{2} \left (c+\frac{\pi }{2}\right )+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=\frac{2 x^3 \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{\left (6 \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{c}{2}+\frac{d x}{2}\right ) \, dx}{d}\\ &=\frac{12 x^2 \sqrt{a+a \cos (c+d x)}}{d^2}+\frac{2 x^3 \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{\left (24 \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 \cos \left (\frac{c}{2}+\frac{d x}{2}\right ) \, dx}{d^2}\\ &=\frac{12 x^2 \sqrt{a+a \cos (c+d x)}}{d^2}-\frac{48 x \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d^3}+\frac{2 x^3 \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{\left (48 \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 \sin \left (\frac{c}{2}+\frac{d x}{2}\right ) \, dx}{d^3}\\ &=-\frac{96 \sqrt{a+a \cos (c+d x)}}{d^4}+\frac{12 x^2 \sqrt{a+a \cos (c+d x)}}{d^2}-\frac{48 x \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d^3}+\frac{2 x^3 \sqrt{a+a \cos (c+d x)} \tan \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.2103, size = 53, normalized size = 0.48 \[ \frac{2 \left (d x \left (d^2 x^2-24\right ) \tan \left (\frac{1}{2} (c+d x)\right )+6 \left (d^2 x^2-8\right )\right ) \sqrt{a (\cos (c+d x)+1)}}{d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(6*(-8 + d^2*x^2) + d*x*(-24 + d^2*x^2)*Tan[(c + d*x)/2]))/d^4

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Maple [C]  time = 0.29, size = 132, normalized size = 1.2 \begin{align*}{\frac{-i\sqrt{2} \left ({d}^{3}{x}^{3}{{\rm e}^{i \left ( dx+c \right ) }}+6\,i{d}^{2}{x}^{2}{{\rm e}^{i \left ( dx+c \right ) }}-{d}^{3}{x}^{3}+6\,i{d}^{2}{x}^{2}-24\,dx{{\rm e}^{i \left ( dx+c \right ) }}-48\,i{{\rm e}^{i \left ( dx+c \right ) }}+24\,dx-48\,i \right ) }{ \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ){d}^{4}}\sqrt{a \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ) ^{2}{{\rm e}^{-i \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+cos(d*x+c)*a)^(1/2),x)

[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^3*x^3*exp(I*(d*x+c))+6*I*d^2*x
^2*exp(I*(d*x+c))-d^3*x^3+6*I*d^2*x^2-24*d*x*exp(I*(d*x+c))-48*I*exp(I*(d*x+c))+24*d*x-48*I)/d^4

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Maxima [B]  time = 2.86868, size = 278, normalized size = 2.53 \begin{align*} -\frac{2 \,{\left (\sqrt{2} \sqrt{a} c^{3} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \,{\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^{2} + 3 \,{\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} c -{\left (\sqrt{2}{\left (d x + c\right )}^{3} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, \sqrt{2}{\left (d x + c\right )}^{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, \sqrt{2}{\left (d x + c\right )} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, \sqrt{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}\right )}}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2*(sqrt(2)*sqrt(a)*c^3*sin(1/2*d*x + 1/2*c) - 3*(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^2 + 3*(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)*c - (sqrt(2)*(d*x + c)^3*sin(1/2*d*x + 1/2*c) + 6*sqrt(2)*(d*x +
c)^2*cos(1/2*d*x + 1/2*c) - 24*sqrt(2)*(d*x + c)*sin(1/2*d*x + 1/2*c) - 48*sqrt(2)*cos(1/2*d*x + 1/2*c))*sqrt
(a))/d^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Integral(x**3*sqrt(a*(cos(c + d*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \cos \left (d x + c\right ) + a} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*cos(d*x + c) + a)*x^3, x)