3.1156 \(\int \cos ^m(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=367 \[ -\frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (m+4) (A (m+2)+C (m+1))+2 a b B \left (m^2+5 m+4\right )+b^2 (m+1) (A (m+4)+C (m+3))\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}-\frac {\sin (c+d x) \cos ^{m+2}(c+d x) \left (a^2 B (m+3)+2 a b (A (m+3)+C (m+2))+b^2 B (m+2)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (2 a^2 C+2 a b B (m+4)+A b^2 (m+4)+b^2 C (m+3)\right )}{d (m+2) (m+4)}+\frac {b \sin (c+d x) (2 a C+b B (m+4)) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)}+\frac {C \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)} \]

[Out]

(2*a^2*C+b^2*C*(3+m)+A*b^2*(4+m)+2*a*b*B*(4+m))*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)/(4+m)+b*(2*a*C+b*B*(4+m))*
cos(d*x+c)^(2+m)*sin(d*x+c)/d/(3+m)/(4+m)+C*cos(d*x+c)^(1+m)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/(4+m)-(2*a*b*B*(m
^2+5*m+4)+a^2*(4+m)*(C*(1+m)+A*(2+m))+b^2*(1+m)*(C*(3+m)+A*(4+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/(4+m)/(m^2+3*m+2)/(sin(d*x+c)^2)^(1/2)-(b^2*B*(2+m)+a^2*B*(3+m)+2*a*b*
(C*(2+m)+A*(3+m)))*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)/(3+m)/
(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.94, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3049, 3033, 3023, 2748, 2643} \[ -\frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (m+4) (A (m+2)+C (m+1))+2 a b B \left (m^2+5 m+4\right )+b^2 (m+1) (A (m+4)+C (m+3))\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}-\frac {\sin (c+d x) \cos ^{m+2}(c+d x) \left (a^2 B (m+3)+2 a b (A (m+3)+C (m+2))+b^2 B (m+2)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (2 a^2 C+2 a b B (m+4)+A b^2 (m+4)+b^2 C (m+3)\right )}{d (m+2) (m+4)}+\frac {b \sin (c+d x) (2 a C+b B (m+4)) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)}+\frac {C \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps

\begin {align*} \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) (a+b \cos (c+d x)) \left (a (C (1+m)+A (4+m))+(b C (3+m)+(A b+a B) (4+m)) \cos (c+d x)+(2 a C+b B (4+m)) \cos ^2(c+d x)\right ) \, dx}{4+m}\\ &=\frac {b (2 a C+b B (4+m)) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) \left (a^2 (3+m) (C (1+m)+A (4+m))+(4+m) \left (b^2 B (2+m)+a^2 B (3+m)+2 a b (C (2+m)+A (3+m))\right ) \cos (c+d x)+(3+m) \left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)+2 a b B (4+m)\right ) \cos ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)+2 a b B (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {b (2 a C+b B (4+m)) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) \left ((3+m) \left (2 a b B \left (4+5 m+m^2\right )+a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+m))\right )+(2+m) (4+m) \left (b^2 B (2+m)+a^2 B (3+m)+2 a b (C (2+m)+A (3+m))\right ) \cos (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)+2 a b B (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {b (2 a C+b B (4+m)) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\left (b^2 B (2+m)+a^2 B (3+m)+2 a b (C (2+m)+A (3+m))\right ) \int \cos ^{1+m}(c+d x) \, dx}{3+m}+\frac {\left (2 a b B \left (4+5 m+m^2\right )+a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+m))\right ) \int \cos ^m(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)+2 a b B (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {b (2 a C+b B (4+m)) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}-\frac {\left (2 a b B \left (4+5 m+m^2\right )+a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+m))\right ) \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) \left (8+6 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}-\frac {\left (b^2 B (2+m)+a^2 B (3+m)+2 a b (C (2+m)+A (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) (3+m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.22, size = 268, normalized size = 0.73 \[ \frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (\cos (c+d x) \left (\cos (c+d x) \left (b \cos (c+d x) \left (-\frac {(2 a C+b B) \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};\cos ^2(c+d x)\right )}{m+4}-\frac {b C \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+5}{2};\frac {m+7}{2};\cos ^2(c+d x)\right )}{m+5}\right )-\frac {\left (a (a C+2 b B)+A b^2\right ) \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\cos ^2(c+d x)\right )}{m+3}\right )-\frac {a (a B+2 A b) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{m+2}\right )-\frac {a^2 A \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{m+1}\right )}{d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(1 + m)*(-((a^2*A*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2])/(1 + m)) + Cos[c
+ d*x]*(-((a*(2*A*b + a*B)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m)) + Cos[c + d
*x]*(-(((A*b^2 + a*(2*b*B + a*C))*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2])/(3 + m)) + b*C
os[c + d*x]*(-(((b*B + 2*a*C)*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[c + d*x]^2])/(4 + m)) - (b*C*Co
s[c + d*x]*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, Cos[c + d*x]^2])/(5 + m)))))*Sin[c + d*x])/(d*Sqrt[Sin
[c + d*x]^2])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 2.90, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +b \cos \left (d x +c \right )\right )^{2} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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