\(\int \frac {\cos ^m(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [368]

Optimal result

Integrand size = 41, antiderivative size = 235 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}-\frac {3 (C (1-3 m)-A (2+3 m)) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*cos(d*x+c)^m*sin(d*x+c)/b/d/(2+3*m)/(b*cos(d*x+c))^(1/3)-3*(C*(1-3*m)-A*(2+3*m))*cos(d*x+c)^m*hypergeom([1
/2, -1/6+1/2*m],[5/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/b/d/(-9*m^2-3*m+2)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1
/2)-3*B*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/b/d/(2+3*m)/(b*cos(d*
x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {20, 3102, 2827, 2722} \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 \left (\frac {A}{1-3 m}-\frac {C}{3 m+2}\right ) \sin (c+d x) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m-1),\frac {1}{6} (3 m+5),\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+2),\frac {1}{6} (3 m+8),\cos ^2(c+d x)\right )}{b d (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac {3 C \sin (c+d x) \cos ^m(c+d x)}{b d (3 m+2) \sqrt [3]{b \cos (c+d x)}} \]

[In]

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

[Out]

(3*C*Cos[c + d*x]^m*Sin[c + d*x])/(b*d*(2 + 3*m)*(b*Cos[c + d*x])^(1/3)) + (3*(A/(1 - 3*m) - C/(2 + 3*m))*Cos[
c + d*x]^m*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(b*Cos[c + d*x
])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Co
s[c + d*x]^2]*Sin[c + d*x])/(b*d*(2 + 3*m)*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])

Rule 20

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

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]

Rule 3102

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac {4}{3}+m}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \cos (c+d x)}} \\ & = \frac {3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}+\frac {\left (3 \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {4}{3}+m}(c+d x) \left (\frac {1}{3} \left (-3 C \left (\frac {1}{3}-m\right )+3 A \left (\frac {2}{3}+m\right )\right )+\frac {1}{3} B (2+3 m) \cos (c+d x)\right ) \, dx}{b (2+3 m) \sqrt [3]{b \cos (c+d x)}} \\ & = \frac {3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}+\frac {\left (B \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}+\frac {\left ((-C (1-3 m)+A (2+3 m)) \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {4}{3}+m}(c+d x) \, dx}{b (2+3 m) \sqrt [3]{b \cos (c+d x)}} \\ & = \frac {3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}+\frac {3 \left (\frac {A}{1-3 m}-\frac {C}{2+3 m}\right ) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left ((C (-1+3 m)+A (2+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-(-1+3 m) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (-1+3 m) (2+3 m) (b \cos (c+d x))^{4/3}} \]

[In]

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

[Out]

(-3*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*((C*(-1 + 3*m) + A*(2 + 3*m))*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 +
3*m)/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] - (-1 + 3*m)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hypergeometric2F1
[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(d*(-1 + 3*m)*(2 + 3*m)*(b*Cos[c + d*x
])^(4/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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