Integrand size = 27, antiderivative size = 135 \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {9 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac {3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac {(28 A+13 C) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{14 \sqrt [6]{2} d (1+\cos (c+d x))^{5/6}} \] Output:
-9/28*C*(a+a*cos(d*x+c))^(1/3)*sin(d*x+c)/d+3/7*C*(a+a*cos(d*x+c))^(4/3)*s in(d*x+c)/a/d+1/28*(28*A+13*C)*(a+a*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2] ,[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(5/6)/d/(1+cos(d*x+c))^(5/6)
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(135)=270\).
Time = 4.93 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.14 \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt [3]{a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (-2 (28 A+13 C) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )+\frac {1}{2} \left (5 (28 A+13 C) \cos \left (\frac {1}{2} \left (c-d x-2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+(28 A+13 C) \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+6 \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec ^2\left (\frac {c}{2}\right )} \left (-\left ((28 A+13 C) \cot \left (\frac {c}{2}\right )\right )+C (\sin (c+d x)+2 \sin (2 (c+d x)))\right )\right ) \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}\right )}{28 d \sqrt {\sec ^2\left (\frac {c}{2}\right )} \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}} \] Input:
Integrate[(a + a*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
Output:
((a*(1 + Cos[c + d*x]))^(1/3)*Sec[(c + d*x)/2]*(-2*(28*A + 13*C)*Hypergeom etricPFQ[{-1/2, -1/6}, {5/6}, Cos[(d*x)/2 + ArcTan[Tan[c/2]]]^2]*Sec[c/2]* Sin[(d*x)/2 + ArcTan[Tan[c/2]]] + ((5*(28*A + 13*C)*Cos[(c - d*x - 2*ArcTa n[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2] + (28*A + 13*C)*Cos[(c + d*x + 2*ArcTan[ Tan[c/2]])/2]*Csc[c/2]*Sec[c/2] + 6*Cos[(c + d*x)/2]*Sqrt[Sec[c/2]^2]*(-(( 28*A + 13*C)*Cot[c/2]) + C*(Sin[c + d*x] + 2*Sin[2*(c + d*x)])))*Sqrt[Sin[ (d*x)/2 + ArcTan[Tan[c/2]]]^2])/2))/(28*d*Sqrt[Sec[c/2]^2]*Sqrt[Sin[(d*x)/ 2 + ArcTan[Tan[c/2]]]^2])
Time = 0.61 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3503, 27, 3042, 3230, 3042, 3131, 3042, 3130}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt [3]{a \cos (c+d x)+a} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt [3]{a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3503 |
| \(\displaystyle \frac {3 \int \frac {1}{3} \sqrt [3]{\cos (c+d x) a+a} (a (7 A+4 C)-3 a C \cos (c+d x))dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt [3]{\cos (c+d x) a+a} (a (7 A+4 C)-3 a C \cos (c+d x))dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 A+4 C)-3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {1}{4} a (28 A+13 C) \int \sqrt [3]{\cos (c+d x) a+a}dx-\frac {9 a C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} a (28 A+13 C) \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {9 a C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3131 |
| \(\displaystyle \frac {\frac {a (28 A+13 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\cos (c+d x)+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}-\frac {9 a C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (28 A+13 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}-\frac {9 a C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
| \(\Big \downarrow \) 3130 |
| \(\displaystyle \frac {\frac {a (28 A+13 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}-\frac {9 a C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
Input:
Int[(a + a*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
Output:
(3*C*(a + a*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*a*d) + ((-9*a*C*(a + a*Co s[c + d*x])^(1/3)*Sin[c + d*x])/(4*d) + (a*(28*A + 13*C)*(a + a*Cos[c + d* x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d *x])/(2*2^(1/6)*d*(1 + Cos[c + d*x])^(5/6)))/(7*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] ) Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ (m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^ m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a , b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \left (a +a \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \cos \left (d x +c \right )^{2}\right )d x\]
Input:
int((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x)
Output:
int((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^(1/3), x)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt [3]{a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \] Input:
integrate((a+a*cos(d*x+c))**(1/3)*(A+C*cos(d*x+c)**2),x)
Output:
Integral((a*(cos(c + d*x) + 1))**(1/3)*(A + C*cos(c + d*x)**2), x)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^(1/3), x)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^(1/3), x)
Timed out. \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3} \,d x \] Input:
int((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(1/3),x)
Output:
int((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(1/3), x)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=a^{\frac {1}{3}} \left (\left (\int \left (\cos \left (d x +c \right )+1\right )^{\frac {1}{3}}d x \right ) a +\left (\int \left (\cos \left (d x +c \right )+1\right )^{\frac {1}{3}} \cos \left (d x +c \right )^{2}d x \right ) c \right ) \] Input:
int((a+a*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x)
Output:
a**(1/3)*(int((cos(c + d*x) + 1)**(1/3),x)*a + int((cos(c + d*x) + 1)**(1/ 3)*cos(c + d*x)**2,x)*c)