\(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [173]

Optimal result

Integrand size = 31, antiderivative size = 90 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/4*A*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^2/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3091, 2722} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*A*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1
/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^2*d*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 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/3}} \, dx \\ & = \frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac {(A+4 C) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}} \, dx}{4 b} \\ & = \frac {3 A \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^2 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \csc (c+d x) \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )+2 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{4/3}} \]

[In]

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

[Out]

(-3*Csc[c + d*x]*(-(A*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]) + 2*C*Cos[c + d*x]^2*Hypergeometric2F
1[1/3, 1/2, 4/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(4/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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