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

Optimal. Leaf size=302 \[ -\frac{a^4 \left (8 m^2+40 m+35\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt{\sin ^2(c+d x)}}-\frac{4 a^4 (2 m+5) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a^4 \left (4 m^2+29 m+55\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac{\sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2 \cos ^{m+1}(c+d x)}{d (m+4)}+\frac{2 (m+5) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right ) \cos ^{m+1}(c+d x)}{d (m+3) (m+4)} \]

[Out]

(a^4*(55 + 29*m + 4*m^2)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)*(4 + m)) + (Cos[c + d*x]^(1 + m
)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*(4 + m)) + (2*(5 + m)*Cos[c + d*x]^(1 + m)*(a^4 + a^4*Cos[c + d*
x])*Sin[c + d*x])/(d*(3 + m)*(4 + m)) - (a^4*(35 + 40*m + 8*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^4*(
5 + 2*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])

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Rubi [A]  time = 0.531488, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2763, 2976, 2968, 3023, 2748, 2643} \[ -\frac{a^4 \left (8 m^2+40 m+35\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt{\sin ^2(c+d x)}}-\frac{4 a^4 (2 m+5) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a^4 \left (4 m^2+29 m+55\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac{\sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2 \cos ^{m+1}(c+d x)}{d (m+4)}+\frac{2 (m+5) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right ) \cos ^{m+1}(c+d x)}{d (m+3) (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(55 + 29*m + 4*m^2)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)*(4 + m)) + (Cos[c + d*x]^(1 + m
)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*(4 + m)) + (2*(5 + m)*Cos[c + d*x]^(1 + m)*(a^4 + a^4*Cos[c + d*
x])*Sin[c + d*x])/(d*(3 + m)*(4 + m)) - (a^4*(35 + 40*m + 8*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^4*(
5 + 2*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 2763

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

Rule 2976

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

Rule 2968

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

Rubi steps

\begin{align*} \int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx &=\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{\int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \left (a^2 (5+2 m)+2 a^2 (5+m) \cos (c+d x)\right ) \, dx}{4+m}\\ &=\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac{\int \cos ^m(c+d x) (a+a \cos (c+d x)) \left (a^3 \left (25+23 m+4 m^2\right )+a^3 \left (55+29 m+4 m^2\right ) \cos (c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac{\int \cos ^m(c+d x) \left (a^4 \left (25+23 m+4 m^2\right )+\left (a^4 \left (25+23 m+4 m^2\right )+a^4 \left (55+29 m+4 m^2\right )\right ) \cos (c+d x)+a^4 \left (55+29 m+4 m^2\right ) \cos ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac{a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac{\int \cos ^m(c+d x) \left (a^4 (3+m) \left (35+40 m+8 m^2\right )+4 a^4 (2+m) (4+m) (5+2 m) \cos (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=\frac{a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac{\left (4 a^4 (5+2 m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{3+m}+\frac{\left (a^4 \left (35+40 m+8 m^2\right )\right ) \int \cos ^m(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac{a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac{2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}-\frac{a^4 \left (35+40 m+8 m^2\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^4 (5+2 m) \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*}

Mathematica [F]  time = 3.23122, size = 0, normalized size = 0. \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 2.855, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a+\cos \left ( dx+c \right ) a \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \cos \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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