3.4.62 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [362]

3.4.62.1 Optimal result

Integrand size = 41, antiderivative size = 149 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 (4 A+7 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

output

3/7*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(7/3)+3/4*B*hypergeom([-2/3, 1/2],[1/3 
],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)+3/7 
*(4*A+7*C)*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*cos 
(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
 

3.4.62.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 b^2 \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )+7 \cos (c+d x) \left (B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )+4 C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{28 d (b \cos (c+d x))^{10/3}} \]

input

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

output

(3*b^2*Cot[c + d*x]*(4*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2 
] + 7*Cos[c + d*x]*(B*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2] + 
4*C*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]))*Sqrt[ 
Sin[c + d*x]^2])/(28*d*(b*Cos[c + d*x])^(10/3))
 

3.4.62.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^2 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{10/3}}dx\)

\(\Big \downarrow \) 3500

\(\displaystyle b^2 \left (\frac {3 \int \frac {7 B b^2+(4 A+7 C) \cos (c+d x) b^2}{3 (b \cos (c+d x))^{7/3}}dx}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle b^2 \left (\frac {\int \frac {7 B b^2+(4 A+7 C) \cos (c+d x) b^2}{(b \cos (c+d x))^{7/3}}dx}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {\int \frac {7 B b^2+(4 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}}dx}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle b^2 \left (\frac {b (4 A+7 C) \int \frac {1}{(b \cos (c+d x))^{4/3}}dx+7 b^2 B \int \frac {1}{(b \cos (c+d x))^{7/3}}dx}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {b (4 A+7 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx+7 b^2 B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}}dx}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle b^2 \left (\frac {\frac {3 (4 A+7 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac {21 b B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}}}{7 b^3}+\frac {3 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/3}}\right )\)

input

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

output

b^2*((3*A*Sin[c + d*x])/(7*b*d*(b*Cos[c + d*x])^(7/3)) + ((21*b*B*Hypergeo 
metric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Cos[c + d* 
x])^(4/3)*Sqrt[Sin[c + d*x]^2]) + (3*(4*A + 7*C)*Hypergeometric2F1[-1/6, 1 
/2, 5/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[ 
c + d*x]^2]))/(7*b^3))
 

3.4.62.3.1 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 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]
 

3.4.62.4 Maple [F]

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

input

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

3.4.62.5 Fricas [F]

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

input

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

output

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

3.4.62.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.62.7 Maxima [F]

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

input

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

output

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

3.4.62.8 Giac [F]

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

input

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

output

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

3.4.62.9 Mupad [F(-1)]

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

input

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