\(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [898]

Optimal result

Integrand size = 31, antiderivative size = 115 \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3 A b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 2827, 2722} \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3 A b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 b B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*A*b^2*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[
c + d*x]^2]) - (3*b*B*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps \begin{align*} \text {integral}& = b^3 \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{5/3}} \, dx \\ & = \left (A b^3\right ) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx+\left (b^2 B\right ) \int \frac {1}{(b \cos (c+d x))^{2/3}} \, dx \\ & = \frac {3 A b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3 b^2 \csc (c+d x) \left (A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )-2 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{2 d (b \cos (c+d x))^{2/3}} \]

[In]

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

[Out]

(3*b^2*Csc[c + d*x]*(A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2] - 2*B*Cos[c + d*x]*Hypergeometric2F1[
1/6, 1/2, 7/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(2*d*(b*Cos[c + d*x])^(2/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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