\(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [350]

Optimal result

Integrand size = 41, antiderivative size = 152 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b (2 A+5 C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}} \] Output:

3/5*A*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(5/3)+3/2*b^2*B*hypergeom([-1/3, 1/2 
],[2/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/ 
2)-3/5*b*(2*A+5*C)*(b*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[7/6],cos(d*x 
+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.82 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \csc (c+d x) \left (-2 A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (-B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )+2 C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right )\right ) \sec ^3(c+d x) \sqrt {\sin ^2(c+d x)}}{10 d} \] Input:

Integrate[(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S 
ec[c + d*x]^4,x]
 

Output:

(-3*(b*Cos[c + d*x])^(4/3)*Csc[c + d*x]*(-2*A*Hypergeometric2F1[-5/6, 1/2, 
 1/6, Cos[c + d*x]^2] + 5*Cos[c + d*x]*(-(B*Hypergeometric2F1[-1/3, 1/2, 2 
/3, Cos[c + d*x]^2]) + 2*C*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, C 
os[c + d*x]^2]))*Sec[c + d*x]^3*Sqrt[Sin[c + d*x]^2])/(10*d)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3042, 2030, 3500, 27, 3042, 3227, 3042, 3122}

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.

\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^4 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\)

\(\Big \downarrow \) 3500

\(\displaystyle b^4 \left (\frac {3 \int \frac {5 B b^2+(2 A+5 C) \cos (c+d x) b^2}{3 (b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle b^4 \left (\frac {\int \frac {5 B b^2+(2 A+5 C) \cos (c+d x) b^2}{(b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {\int \frac {5 B b^2+(2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle b^4 \left (\frac {b (2 A+5 C) \int \frac {1}{(b \cos (c+d x))^{2/3}}dx+5 b^2 B \int \frac {1}{(b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {b (2 A+5 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+5 b^2 B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle b^4 \left (\frac {\frac {15 b B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 (2 A+5 C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\)

Input:

Int[(b*Cos[c + d*x])^(4/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + 
 d*x]^4,x]
 

Output:

b^4*((3*A*Sin[c + d*x])/(5*b*d*(b*Cos[c + d*x])^(5/3)) + ((15*b*B*Hypergeo 
metric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Cos[c + d* 
x])^(2/3)*Sqrt[Sin[c + d*x]^2]) - (3*(2*A + 5*C)*(b*Cos[c + d*x])^(1/3)*Hy 
pergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqrt[Sin[c 
 + d*x]^2]))/(5*b^3))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right ) \sec \left (d x +c \right )^{4}d x\]

Input:

int((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Output:

int((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Fricas [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="fricas")
 

Output:

integral((C*b*cos(d*x + c)^3 + B*b*cos(d*x + c)^2 + A*b*cos(d*x + c))*(b*c 
os(d*x + c))^(1/3)*sec(d*x + c)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c 
)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*s 
ec(d*x + c)^4, x)
 

Giac [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(4/3)*s 
ec(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int(((b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^4,x)
 

Output:

int(((b*cos(c + d*x))^(4/3)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^4, x)
 

Reduce [F]

\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=b^{\frac {4}{3}} \left (\left (\int \cos \left (d x +c \right )^{\frac {4}{3}} \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \cos \left (d x +c \right )^{\frac {10}{3}} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \cos \left (d x +c \right )^{\frac {7}{3}} \sec \left (d x +c \right )^{4}d x \right ) b \right ) \] Input:

int((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
                                                                                    
                                                                                    
 

Output:

b**(1/3)*b*(int(cos(c + d*x)**(1/3)*cos(c + d*x)*sec(c + d*x)**4,x)*a + in 
t(cos(c + d*x)**(1/3)*cos(c + d*x)**3*sec(c + d*x)**4,x)*c + int(cos(c + d 
*x)**(1/3)*cos(c + d*x)**2*sec(c + d*x)**4,x)*b)