3.1157 \(\int \cos ^m(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=235 \[ -\frac {\sin (c+d x) \cos ^{m+1}(c+d x) (a A (m+2)+(m+1) (a C+b B)) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt {\sin ^2(c+d x)}}-\frac {\sin (c+d x) \cos ^{m+2}(c+d x) (a B (m+3)+A b (m+3)+b C (m+2)) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {(a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3033, 3023, 2748, 2643} \[ -\frac {\sin (c+d x) \cos ^{m+1}(c+d x) (a A (m+2)+(m+1) (a C+b B)) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt {\sin ^2(c+d x)}}-\frac {\sin (c+d x) \cos ^{m+2}(c+d x) (a B (m+3)+A b (m+3)+b C (m+2)) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {(a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + a*C)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) + (b*C*Cos[c + d*x]^(2 + m)*Sin[c + d*x])/(d*(3 +
m)) - (((b*B + a*C)*(1 + m) + a*A*(2 + m))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, C
os[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m)*(2 + m)*Sqrt[Sin[c + d*x]^2]) - ((b*C*(2 + m) + A*b*(3 + m) + a*B*(3 +
m))*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
)*(3 + m)*Sqrt[Sin[c + d*x]^2])

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

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 3023

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]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^m(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m)}+\frac {\int \cos ^m(c+d x) \left (a A (3+m)+(b C (2+m)+A b (3+m)+a B (3+m)) \cos (c+d x)+(b B+a C) (3+m) \cos ^2(c+d x)\right ) \, dx}{3+m}\\ &=\frac {(b B+a C) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}+\frac {b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m)}+\frac {\int \cos ^m(c+d x) ((3+m) ((b B+a C) (1+m)+a A (2+m))+(2+m) (b C (2+m)+A b (3+m)+a B (3+m)) \cos (c+d x)) \, dx}{6+5 m+m^2}\\ &=\frac {(b B+a C) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}+\frac {b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m)}+\frac {(b B (1+m)+a C (1+m)+a A (2+m)) \int \cos ^m(c+d x) \, dx}{2+m}+\left (A b+a B+\frac {b C (2+m)}{3+m}\right ) \int \cos ^{1+m}(c+d x) \, dx\\ &=\frac {(b B+a C) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}+\frac {b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m)}-\frac {(b B (1+m)+a C (1+m)+a A (2+m)) \cos ^{1+m}(c+d x) \, _2F_1\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 {\left (A b+a B+\frac {b C (2+m)}{3+m}\right ) \cos ^{2+m}(c+d x) \, _2F_1\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]  time = 1.76, size = 205, normalized size = 0.87 \[ \frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (\cos (c+d x) \left (\cos (c+d x) \left (-\frac {(a C+b B) \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\cos ^2(c+d x)\right )}{m+3}-\frac {b C \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};\cos ^2(c+d x)\right )}{m+4}\right )-\frac {(a B+A b) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{m+2}\right )-\frac {a A \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{m+1}\right )}{d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(1 + m)*(-((a*A*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2])/(1 + m)) + Cos[c +
d*x]*(-(((A*b + a*B)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m)) + Cos[c + d*x]*(-
(((b*B + a*C)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2])/(3 + m)) - (b*C*Cos[c + d*x]*Hyper
geometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[c + d*x]^2])/(4 + m))))*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2])

________________________________________________________________________________________

fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + {\left (C a + B b\right )} \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(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) + a)*cos(d*x + c)^m, x)

________________________________________________________________________________________

maple [F]  time = 14.56, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +b \cos \left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(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) + a)*cos(d*x + c)^m, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^m\,\left (a+b\,\cos \left (c+d\,x\right )\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________