\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [162]

Optimal result

Integrand size = 33, antiderivative size = 91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b d \sqrt {\sin ^2(c+d x)}} \] Output:

3/4*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)-3/8*(A+4*C)*(b*cos(d*x+c))^(2/3) 
*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1 
/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 b \csc (c+d x) \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )+2 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{4/3}} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(b*Cos[c + d*x])^(1/3),x 
]
 

Output:

(-3*b*Csc[c + d*x]*(-(A*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]) 
 + 2*C*Cos[c + d*x]^2*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2])*Sq 
rt[Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(4/3))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3042, 2030, 3491, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3491

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b^2 \left (\frac {3 A \sin (c+d x)}{4 b d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{8 b^3 d \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(b*Cos[c + d*x])^(1/3),x]
 

Output:

b^2*((3*A*Sin[c + d*x])/(4*b*d*(b*Cos[c + d*x])^(4/3)) - (3*(A + 4*C)*(b*C 
os[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c 
+ d*x])/(8*b^3*d*Sqrt[Sin[c + d*x]^2]))
 

Defintions of rubi rules used

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 3491

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

Maple [F]

\[\int \frac {\left (A +C \cos \left (d x +c \right )^{2}\right ) \sec \left (d x +c \right )^{2}}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x)
 

Output:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x)
 

Fricas [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x, algorith 
m="fricas")
 

Output:

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

Sympy [F]

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

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(b*cos(d*x+c))**(1/3),x)
 

Output:

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**2/(b*cos(c + d*x))**(1/3), 
x)
 

Maxima [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x, algorith 
m="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^2/(b*cos(d*x + c))^(1/3), x)
 

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^2/(b*cos(d*x + c))^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(b*cos(c + d*x))^(1/3)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(b*cos(c + d*x))^(1/3)), x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {\left (\int \frac {\sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \cos \left (d x +c \right )^{\frac {5}{3}} \sec \left (d x +c \right )^{2}d x \right ) c}{b^{\frac {1}{3}}} \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(b*cos(d*x+c))^(1/3),x)
 

Output:

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