\(\int (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [374]

Optimal result

Integrand size = 39, antiderivative size = 173 \[ \int (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3102, 2827, 2722} \[ \int (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {b (C (1-n)-A n) \sin (c+d x) (b \cos (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n-1}{2},\frac {n+1}{2},\cos ^2(c+d x)\right )}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {b C \sin (c+d x) (b \cos (c+d x))^{n-1}}{d n} \]

[In]

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

[Out]

(b*C*(b*Cos[c + d*x])^(-1 + n)*Sin[c + d*x])/(d*n) - (b*(C*(1 - n) - A*n)*(b*Cos[c + d*x])^(-1 + n)*Hypergeome
tric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - n)*n*Sqrt[Sin[c + d*x]^2]) - (B*(b*C
os[c + d*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*n*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]

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}& = b^2 \int (b \cos (c+d x))^{-2+n} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}+\frac {b \int (b \cos (c+d x))^{-2+n} (-b (C (1-n)-A n)+b B n \cos (c+d x)) \, dx}{n} \\ & = \frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}+(b B) \int (b \cos (c+d x))^{-1+n} \, dx-\frac {\left (b^2 (C (1-n)-A n)\right ) \int (b \cos (c+d x))^{-2+n} \, dx}{n} \\ & = \frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.75 \[ \int (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {(b \cos (c+d x))^n \left ((C (-1+n)+A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right )+(-1+n) \left (B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right )-C \sqrt {\sin ^2(c+d x)}\right )\right ) \tan (c+d x)}{d (-1+n) n \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

-(((b*Cos[c + d*x])^n*((C*(-1 + n) + A*n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2] + (-1
+ n)*(B*Cos[c + d*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[c + d*x]^2] - C*Sqrt[Sin[c + d*x]^2]))*Tan[c +
d*x])/(d*(-1 + n)*n*Sqrt[Sin[c + d*x]^2]))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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