3.769 \(\int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=330 \[ -\frac {4 a b \left (a^2 (m+3)+b^2 (m+2)\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+4)}-\frac {\left (a^4 \left (m^2+6 m+8\right )+6 a^2 b^2 \left (m^2+5 m+4\right )+b^4 \left (m^2+4 m+3\right )\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)} \]

[Out]

b^2*(b^2*(3+m)+a^2*(22+5*m))*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)/(4+m)+2*a*b^3*(5+m)*cos(d*x+c)^(2+m)*sin(d*x+
c)/d/(3+m)/(4+m)+b^2*cos(d*x+c)^(1+m)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/(4+m)-(b^4*(m^2+4*m+3)+6*a^2*b^2*(m^2+5*
m+4)+a^4*(m^2+6*m+8))*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)-4*a*b*(b^2*(2+m)+a^2*(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 = 0.67, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2793, 3033, 3023, 2748, 2643} \[ -\frac {\left (6 a^2 b^2 \left (m^2+5 m+4\right )+a^4 \left (m^2+6 m+8\right )+b^4 \left (m^2+4 m+3\right )\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}-\frac {4 a b \left (a^2 (m+3)+b^2 (m+2)\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+4)}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^4,x]

[Out]

(b^2*(b^2*(3 + m) + a^2*(22 + 5*m))*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(4 + m)) + (2*a*b^3*(5 + m)*
Cos[c + d*x]^(2 + m)*Sin[c + d*x])/(d*(3 + m)*(4 + m)) + (b^2*Cos[c + d*x]^(1 + m)*(a + b*Cos[c + d*x])^2*Sin[
c + d*x])/(d*(4 + m)) - ((b^4*(3 + 4*m + m^2) + 6*a^2*b^2*(4 + 5*m + m^2) + a^4*(8 + 6*m + m^2))*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]) - (4*a*b*(b^2*(2 + m) + a^2*(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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

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))^4 \, dx &=\frac {b^2 \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 \left (b^2 (1+m)+a^2 (4+m)\right )+b \left (b^2 (3+m)+3 a^2 (4+m)\right ) \cos (c+d x)+2 a b^2 (5+m) \cos ^2(c+d x)\right ) \, dx}{4+m}\\ &=\frac {2 a b^3 (5+m) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {b^2 \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) \left (b^2 (1+m)+a^2 (4+m)\right )+4 a b (4+m) \left (b^2 (2+m)+a^2 (3+m)\right ) \cos (c+d x)+b^2 (3+m) \left (b^2 (3+m)+a^2 (22+5 m)\right ) \cos ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac {b^2 \left (b^2 (3+m)+a^2 (22+5 m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b^3 (5+m) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {b^2 \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 (b^4 \left (3+4 m+m^2\right )+6 a^2 b^2 \left (4+5 m+m^2\right )+a^4 \left (8+6 m+m^2\right )\right )+4 a b (2+m) (4+m) \left (b^2 (2+m)+a^2 (3+m)\right ) \cos (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=\frac {b^2 \left (b^2 (3+m)+a^2 (22+5 m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b^3 (5+m) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\left (4 a b \left (a^2+\frac {b^2 (2+m)}{3+m}\right )\right ) \int \cos ^{1+m}(c+d x) \, dx+\frac {\left (b^4 \left (3+4 m+m^2\right )+6 a^2 b^2 \left (4+5 m+m^2\right )+a^4 \left (8+6 m+m^2\right )\right ) \int \cos ^m(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac {b^2 \left (b^2 (3+m)+a^2 (22+5 m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b^3 (5+m) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}-\frac {\left (b^4 \left (3+4 m+m^2\right )+6 a^2 b^2 \left (4+5 m+m^2\right )+a^4 \left (8+6 m+m^2\right )\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 {4 a b \left (a^2+\frac {b^2 (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*}

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Mathematica [A]  time = 1.78, size = 242, normalized size = 0.73 \[ \frac {\sqrt {\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (b \cos (c+d x) \left (b \cos (c+d x) \left (b \cos (c+d x) \left (-\frac {4 a \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};\cos ^2(c+d x)\right )}{m+4}-\frac {b \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+5}{2};\frac {m+7}{2};\cos ^2(c+d x)\right )}{m+5}\right )-\frac {6 a^2 \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\cos ^2(c+d x)\right )}{m+3}\right )-\frac {4 a^3 \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{m+2}\right )-\frac {a^4 \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{m+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^4,x]

[Out]

(Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((a^4*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2])/(1 +
m)) + b*Cos[c + d*x]*((-4*a^3*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m) + b*Cos[c
+ d*x]*((-6*a^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2])/(3 + m) + b*Cos[c + d*x]*((-4*a*
Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[c + d*x]^2])/(4 + m) - (b*Cos[c + d*x]*Hypergeometric2F1[1/2,
(5 + m)/2, (7 + m)/2, Cos[c + d*x]^2])/(5 + m)))))*Sqrt[Sin[c + d*x]^2])/d

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{4} \cos \left (d x + c\right )^{4} + 4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{3} b \cos \left (d x + c\right ) + a^{4}\right )} \cos \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((b^4*cos(d*x + c)^4 + 4*a*b^3*cos(d*x + c)^3 + 6*a^2*b^2*cos(d*x + c)^2 + 4*a^3*b*cos(d*x + c) + a^4)
*cos(d*x + c)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^4*cos(d*x + c)^m, x)

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +b \cos \left (d x +c \right )\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x)

[Out]

int(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^4*cos(d*x + c)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^4,x)

[Out]

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^4, 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))**4,x)

[Out]

Timed out

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