\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [905]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 A (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^4 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/5*A*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)-3/
8*B*(b*cos(d*x+c))^(8/3)*hypergeom([1/2, 4/3],[7/3],cos(d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 119, 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} \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 A \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )}{8 b^4 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(-3*A*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^3*d*Sqrt[Sin
[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*
b^4*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}& = \frac {\int (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx}{b^2} \\ & = \frac {A \int (b \cos (c+d x))^{2/3} \, dx}{b^2}+\frac {B \int (b \cos (c+d x))^{5/3} \, dx}{b^3} \\ & = -\frac {3 A (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^4 d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cos ^2(c+d x) \cot (c+d x) \left (8 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )+5 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{40 d (b \cos (c+d x))^{4/3}} \]

[In]

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

[Out]

(-3*Cos[c + d*x]^2*Cot[c + d*x]*(8*A*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2] + 5*B*Cos[c + d*x]*Hype
rgeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(40*d*(b*Cos[c + d*x])^(4/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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