\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) [169]

Optimal result

Integrand size = 33, antiderivative size = 92 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{16 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/8*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(8/3)+3/16*(5*A+8*C)*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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {16, 3091, 2722} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{16 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

[In]

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

[Out]

(3*A*b^2*Sin[c + d*x])/(8*d*(b*Cos[c + d*x])^(8/3)) + (3*(5*A + 8*C)*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c +
d*x]^2]*Sin[c + d*x])/(16*d*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2])

Rule 16

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

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3091

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = b^3 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{11/3}} \, dx \\ & = \frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {1}{8} (b (5 A+8 C)) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx \\ & = \frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{16 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 \left (4 C \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)+A \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},-\frac {1}{3},\cos ^2(c+d x)\right ) \sec (c+d x) \tan (c+d x)\right )}{8 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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