3.398 \(\int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx\)

Optimal. Leaf size=232 \[ -\frac{a^3 (4 m+5) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{a^3 (4 m+11) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)} \]

[Out]

(a^3*(7 + 2*m)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)) + (Cos[c + d*x]^(1 + m)*(a^3 + a^3*Cos[c
+ d*x])*Sin[c + d*x])/(d*(3 + m)) - (a^3*(5 + 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)*Sqrt[Sin[c + d*x]^2]) - (a^3*(11 + 4*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 = 0.307741, antiderivative size = 232, normalized size of antiderivative = 1., 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 = {2763, 2968, 3023, 2748, 2643} \[ -\frac{a^3 (4 m+5) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{a^3 (4 m+11) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(a + a*Cos[c + d*x])^3,x]

[Out]

(a^3*(7 + 2*m)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)) + (Cos[c + d*x]^(1 + m)*(a^3 + a^3*Cos[c
+ d*x])*Sin[c + d*x])/(d*(3 + m)) - (a^3*(5 + 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)*Sqrt[Sin[c + d*x]^2]) - (a^3*(11 + 4*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 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))^3 \, dx &=\frac{\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}+\frac{\int \cos ^m(c+d x) (a+a \cos (c+d x)) \left (2 a^2 (2+m)+a^2 (7+2 m) \cos (c+d x)\right ) \, dx}{3+m}\\ &=\frac{\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}+\frac{\int \cos ^m(c+d x) \left (2 a^3 (2+m)+\left (2 a^3 (2+m)+a^3 (7+2 m)\right ) \cos (c+d x)+a^3 (7+2 m) \cos ^2(c+d x)\right ) \, dx}{3+m}\\ &=\frac{a^3 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}+\frac{\int \cos ^m(c+d x) \left (a^3 (3+m) (5+4 m)+a^3 (2+m) (11+4 m) \cos (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=\frac{a^3 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}+\frac{\left (a^3 (5+4 m)\right ) \int \cos ^m(c+d x) \, dx}{2+m}+\frac{\left (a^3 (11+4 m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{3+m}\\ &=\frac{a^3 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}-\frac{a^3 (5+4 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{a^3 (11+4 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 = 1.35356, size = 0, normalized size = 0. \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cos[c + d*x]^m*(a + a*Cos[c + d*x])^3,x]

[Out]

Integrate[Cos[c + d*x]^m*(a + a*Cos[c + d*x])^3, x]

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Maple [F]  time = 2.316, 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 ) ^{3}\, 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)^3,x)

[Out]

int(cos(d*x+c)^m*(a+cos(d*x+c)*a)^3,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 )}^{3} \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))^3,x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^3*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^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}\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))^3,x, algorithm="fricas")

[Out]

integral((a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3)*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))**3,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 )}^{3} \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))^3,x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^3*cos(d*x + c)^m, x)