Integrand size = 21, antiderivative size = 179 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {\left (b^2 (1+m)+a^2 (2+m)\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt {\sin ^2(c+d x)}} \] Output:
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)
Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.94 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=-\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \left (a^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )+b (1+m) \cos (c+d x) \left (2 a (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right )+b (2+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+m) (2+m) (3+m)} \] Input:
Integrate[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^2,x]
Output:
-((Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(a^2*(6 + 5*m + m^2)*Hypergeometric2F 1[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)))
Time = 0.52 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3268, 3042, 3122, 3493, 3042, 3122}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 3268 |
\(\displaystyle \int \cos ^m(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right )dx+2 a b \int \cos ^{m+1}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}}\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \int \cos ^m(c+d x)dx-\frac {2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx-\frac {2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {\left (a^2+\frac {b^2 (m+1)}{m+2}\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\) |
Input:
Int[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^2,x]
Output:
(b^2*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) - ((a^2 + (b^2*(1 + m) )/(2 + m))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/ 2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 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])
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]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)])^2, x_Symbol] :> Simp[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]
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] + Simp[(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]
\[\int \cos \left (d x +c \right )^{m} \left (a +\cos \left (d x +c \right ) b \right )^{2}d x\]
Input:
int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^2,x)
Output:
int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^2,x)
\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*cos(d*x + c)^m, x )
\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \cos ^{m}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**m*(a+b*cos(d*x+c))**2,x)
Output:
Integral((a + b*cos(c + d*x))**2*cos(c + d*x)**m, x)
\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)
\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)
Timed out. \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:
int(cos(c + d*x)^m*(a + b*cos(c + d*x))^2,x)
Output:
int(cos(c + d*x)^m*(a + b*cos(c + d*x))^2, x)
\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx=\left (\int \cos \left (d x +c \right )^{m}d x \right ) a^{2}+2 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )d x \right ) a b +\left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:
int(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,x)
Output:
int(cos(c + d*x)**m,x)*a**2 + 2*int(cos(c + d*x)**m*cos(c + d*x),x)*a*b + int(cos(c + d*x)**m*cos(c + d*x)**2,x)*b**2