\(\int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx\) [771]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {\left (b^2 (1+m)+a^2 (2+m)\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

b^2*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)-(b^2*(1+m)+a^2*(2+m))*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2
+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(1+m)/(2+m)/(sin(d*x+c)^2)^(1/2)-2*a*b*cos(d*x+c)^(2+m)*hypergeom([1/2, 1+1
/2*m],[2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(2+m)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2868, 2722, 3093} \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=-\frac {\left (a^2 (m+2)+b^2 (m+1)\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)} \]

[In]

Int[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^2,x]

[Out]

(b^2*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) - ((b^2*(1 + m) + a^2*(2 + m))*Cos[c + d*x]^(1 + m)*Hyperg
eometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m)*(2 + m)*Sqrt[Sin[c + d*x]^2])
- (2*a*b*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(2
+ m)*Sqrt[Sin[c + d*x]^2])

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 2868

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[2*c*(d/b)
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3093

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.94 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=-\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \left (a^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )+b (1+m) \cos (c+d x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right )+b (2+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+m) (2+m) (3+m)} \]

[In]

Integrate[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^2,x]

[Out]

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

Maple [F]

\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}d x\]

[In]

int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^2,x)

[Out]

int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^2,x)

Fricas [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*cos(d*x + c)^m, x)

Sympy [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \cos ^{m}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**m*(a+b*cos(d*x+c))**2,x)

[Out]

Integral((a + b*cos(c + d*x))**2*cos(c + d*x)**m, x)

Maxima [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)

Giac [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

[In]

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^2,x)

[Out]

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^2, x)