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

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

[Out]

-96*(a+a*cos(d*x+c))^(1/2)/d^4+12*x^2*(a+a*cos(d*x+c))^(1/2)/d^2-48*x*(a+a*cos(d*x+c))^(1/2)*tan(1/2*d*x+1/2*c
)/d^3+2*x^3*(a+a*cos(d*x+c))^(1/2)*tan(1/2*d*x+1/2*c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 110, normalized size of antiderivative = 1.00, 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 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^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 = 0.22, 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

________________________________________________________________________________________

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^3*(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)

________________________________________________________________________________________

giac [A]  time = 0.47, size = 98, normalized size = 0.89 \[ 2 \, \sqrt {2} \sqrt {a} {\left (\frac {6 \, {\left (d^{2} x^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{4}} + \frac {{\left (d^{3} x^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 24 \, d x \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{4}}\right )} \]

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]

2*sqrt(2)*sqrt(a)*(6*(d^2*x^2*sgn(cos(1/2*d*x + 1/2*c)) - 8*sgn(cos(1/2*d*x + 1/2*c)))*cos(1/2*d*x + 1/2*c)/d^
4 + (d^3*x^3*sgn(cos(1/2*d*x + 1/2*c)) - 24*d*x*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d^4)

________________________________________________________________________________________

maple [C]  time = 0.10, size = 132, normalized size = 1.20 \[ -\frac {i \sqrt {2}\, \sqrt {a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2} {\mathrm e}^{-i \left (d x +c \right )}}\, \left (d^{3} x^{3} {\mathrm e}^{i \left (d x +c \right )}+6 i d^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}-d^{3} x^{3}+6 i d^{2} x^{2}-24 d x \,{\mathrm e}^{i \left (d x +c \right )}-48 i {\mathrm e}^{i \left (d x +c \right )}+24 d x -48 i\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+a*cos(d*x+c))^(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

________________________________________________________________________________________

maxima [B]  time = 1.71, size = 206, normalized size = 1.87 \[ -\frac {2 \, {\left (\sqrt {2} \sqrt {a} c^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (\sqrt {2} {\left (d x + c\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} c^{2} + 3 \, {\left (\sqrt {2} {\left (d x + c\right )}^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} {\left (d x + c\right )} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} c - {\left (\sqrt {2} {\left (d x + c\right )}^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, \sqrt {2} {\left (d x + c\right )}^{2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, \sqrt {2} {\left (d x + c\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}\right )}}{d^{4}} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.55, size = 83, normalized size = 0.75 \[ -\frac {2\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (48\,\cos \left (c+d\,x\right )-6\,d^2\,x^2-6\,d^2\,x^2\,\cos \left (c+d\,x\right )-d^3\,x^3\,\sin \left (c+d\,x\right )+24\,d\,x\,\sin \left (c+d\,x\right )+48\right )}{d^4\,\left (\cos \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + a*cos(c + d*x))^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}\, dx \]

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)

________________________________________________________________________________________