3.4.60 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [360]

3.4.60.1 Optimal result

Integrand size = 33, antiderivative size = 152 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}-\frac {3 B (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^3 d \sqrt {\sin ^2(c+d x)}} \]

output

3*A*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/3)-3/2*B*(b*cos(d*x+c))^(2/3)*hyperge 
om([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)+3/ 
5*(2*A-C)*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*s 
in(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)
 

3.4.60.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (-10 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )+\cos (c+d x) \left (5 B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )+2 C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{10 d (b \cos (c+d x))^{4/3}} \]

input

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

output

(-3*Cot[c + d*x]*(-10*A*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2] 
+ Cos[c + d*x]*(5*B*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2] + 2*C 
*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]))*Sqrt[Sin 
[c + d*x]^2])/(10*d*(b*Cos[c + d*x])^(4/3))
 

3.4.60.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {3 \int \frac {b^2 B-b^2 (2 A-C) \cos (c+d x)}{3 \sqrt [3]{b \cos (c+d x)}}dx}{b^3}+\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b^2 B-b^2 (2 A-C) \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}}dx}{b^3}+\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {b^2 B-b^2 (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^3}+\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {b^2 B \int \frac {1}{\sqrt [3]{b \cos (c+d x)}}dx-b (2 A-C) \int (b \cos (c+d x))^{2/3}dx}{b^3}+\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)}}}{b^3}+\frac {3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}\)

input

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

output

(3*A*Sin[c + d*x])/(b*d*(b*Cos[c + d*x])^(1/3)) + ((-3*b*B*(b*Cos[c + d*x] 
)^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2* 
d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeometr 
ic2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*Sqrt[Sin[c + d*x] 
^2]))/b^3
 

3.4.60.3.1 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 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]
 

3.4.60.4 Maple [F]

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

input

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

output

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

3.4.60.5 Fricas [F]

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

input

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

output

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

3.4.60.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.60.7 Maxima [F]

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

input

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

output

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

3.4.60.8 Giac [F]

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

input

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

output

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

3.4.60.9 Mupad [F(-1)]

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

input

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

output

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