\(\int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx\) [1514]

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=-\frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2},-\frac {4}{3},\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (c+d \sin (e+f x))^{4/3}}{3 f \sqrt {1+\sin (e+f x)} \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{4/3}} \] Output:

-2/3*2^(1/2)*AppellF1(3/2,-4/3,-1/2,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin 
(f*x+e))*cos(f*x+e)*(1-sin(f*x+e))*(c+d*sin(f*x+e))^(4/3)/f/(1+sin(f*x+e)) 
^(1/2)/((c+d*sin(f*x+e))/(c+d))^(4/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(301\) vs. \(2(118)=236\).

Time = 2.45 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.55 \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=-\frac {3 \sec (e+f x) \sqrt [3]{c+d \sin (e+f x)} \left (12 \left (4 c^4+3 c^2 d^2-7 d^4\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sin (e+f x))}{c-d}}-3 c \left (4 c^2+51 d^2\right ) \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sin (e+f x))}{c-d}} (c+d \sin (e+f x))+4 d^2 \cos ^2(e+f x) \left (-4 c^2+7 d^2+14 d^2 \cos (2 (e+f x))-44 c d \sin (e+f x)\right )\right )}{1120 d^3 f} \] Input:

Integrate[Cos[e + f*x]^2*(c + d*Sin[e + f*x])^(4/3),x]
 

Output:

(-3*Sec[e + f*x]*(c + d*Sin[e + f*x])^(1/3)*(12*(4*c^4 + 3*c^2*d^2 - 7*d^4 
)*AppellF1[1/3, 1/2, 1/2, 4/3, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e 
+ f*x])/(c + d)]*Sqrt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[-((d*(1 + S 
in[e + f*x]))/(c - d))] - 3*c*(4*c^2 + 51*d^2)*AppellF1[4/3, 1/2, 1/2, 7/3 
, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + f*x])/(c + d)]*Sqrt[-((d*(- 
1 + Sin[e + f*x]))/(c + d))]*Sqrt[-((d*(1 + Sin[e + f*x]))/(c - d))]*(c + 
d*Sin[e + f*x]) + 4*d^2*Cos[e + f*x]^2*(-4*c^2 + 7*d^2 + 14*d^2*Cos[2*(e + 
 f*x)] - 44*c*d*Sin[e + f*x])))/(1120*d^3*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3183, 155}

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.

\(\displaystyle \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^2 (c+d \sin (e+f x))^{4/3}dx\)

\(\Big \downarrow \) 3183

\(\displaystyle \frac {\cos (e+f x) \int (c+d \sin (e+f x))^{4/3} \sqrt {-\frac {\sin (e+f x) d}{c-d}-\frac {d}{c-d}} \sqrt {\frac {d}{c+d}-\frac {d \sin (e+f x)}{c+d}}d\sin (e+f x)}{f \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {3 \cos (e+f x) (c+d \sin (e+f x))^{7/3} \operatorname {AppellF1}\left (\frac {7}{3},-\frac {1}{2},-\frac {1}{2},\frac {10}{3},\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )}{7 d f \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}}\)

Input:

Int[Cos[e + f*x]^2*(c + d*Sin[e + f*x])^(4/3),x]
 

Output:

(3*AppellF1[7/3, -1/2, -1/2, 10/3, (c + d*Sin[e + f*x])/(c - d), (c + d*Si 
n[e + f*x])/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^(7/3))/(7*d*f*Sqrt[ 
1 - (c + d*Sin[e + f*x])/(c - d)]*Sqrt[1 - (c + d*Sin[e + f*x])/(c + d)])
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3183

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin 
[e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p - 1) 
/2)))   Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*( 
x/(a + b)))^((p - 1)/2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, 
b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] &&  !IGtQ[m, 0]
 

Maple [F]

\[\int \cos \left (f x +e \right )^{2} \left (c +d \sin \left (f x +e \right )\right )^{\frac {4}{3}}d x\]

Input:

int(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x)
 

Output:

int(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x)
 

Fricas [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {4}{3}} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x, algorithm="fricas")
 

Output:

integral((d*cos(f*x + e)^2*sin(f*x + e) + c*cos(f*x + e)^2)*(d*sin(f*x + e 
) + c)^(1/3), x)
 

Sympy [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\int \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {4}{3}} \cos ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate(cos(f*x+e)**2*(c+d*sin(f*x+e))**(4/3),x)
 

Output:

Integral((c + d*sin(e + f*x))**(4/3)*cos(e + f*x)**2, x)
 

Maxima [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {4}{3}} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x, algorithm="maxima")
 

Output:

integrate((d*sin(f*x + e) + c)^(4/3)*cos(f*x + e)^2, x)
 

Giac [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {4}{3}} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x, algorithm="giac")
 

Output:

integrate((d*sin(f*x + e) + c)^(4/3)*cos(f*x + e)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{4/3} \,d x \] Input:

int(cos(e + f*x)^2*(c + d*sin(e + f*x))^(4/3),x)
 

Output:

int(cos(e + f*x)^2*(c + d*sin(e + f*x))^(4/3), x)
 

Reduce [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^{4/3} \, dx=\left (\int \left (\sin \left (f x +e \right ) d +c \right )^{\frac {1}{3}} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )d x \right ) d +\left (\int \left (\sin \left (f x +e \right ) d +c \right )^{\frac {1}{3}} \cos \left (f x +e \right )^{2}d x \right ) c \] Input:

int(cos(f*x+e)^2*(c+d*sin(f*x+e))^(4/3),x)
 

Output:

int((sin(e + f*x)*d + c)**(1/3)*cos(e + f*x)**2*sin(e + f*x),x)*d + int((s 
in(e + f*x)*d + c)**(1/3)*cos(e + f*x)**2,x)*c