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

Optimal. Leaf size=250 \[ -\frac{a \left (a^2 (m+2)+3 b^2 (m+1)\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a b^2 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))}{d (m+3)} \]

[Out]

(a*b^2*(7 + 2*m)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)) + (b^2*Cos[c + d*x]^(1 + m)*(a + b*Cos
[c + d*x])*Sin[c + d*x])/(d*(3 + m)) - (a*(3*b^2*(1 + m) + a^2*(2 + 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]) - (b*(b^2*(
2 + m) + 3*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])

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Rubi [A]  time = 0.316659, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2793, 3023, 2748, 2643} \[ -\frac{a \left (a^2 (m+2)+3 b^2 (m+1)\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt{\sin ^2(c+d x)}}+\frac{a b^2 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*(7 + 2*m)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)) + (b^2*Cos[c + d*x]^(1 + m)*(a + b*Cos
[c + d*x])*Sin[c + d*x])/(d*(3 + m)) - (a*(3*b^2*(1 + m) + a^2*(2 + 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]) - (b*(b^2*(
2 + m) + 3*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 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 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+b \cos (c+d x))^3 \, dx &=\frac{b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{d (3+m)}+\frac{\int \cos ^m(c+d x) \left (a \left (b^2 (1+m)+a^2 (3+m)\right )+b \left (b^2 (2+m)+3 a^2 (3+m)\right ) \cos (c+d x)+a b^2 (7+2 m) \cos ^2(c+d x)\right ) \, dx}{3+m}\\ &=\frac{a b^2 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{d (3+m)}+\frac{\int \cos ^m(c+d x) \left (a (3+m) \left (3 b^2 (1+m)+a^2 (2+m)\right )+b (2+m) \left (b^2 (2+m)+3 a^2 (3+m)\right ) \cos (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=\frac{a b^2 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{d (3+m)}+\left (a \left (a^2+\frac{3 b^2 (1+m)}{2+m}\right )\right ) \int \cos ^m(c+d x) \, dx+\left (b \left (3 a^2+\frac{b^2 (2+m)}{3+m}\right )\right ) \int \cos ^{1+m}(c+d x) \, dx\\ &=\frac{a b^2 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac{b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{d (3+m)}-\frac{a \left (a^2+\frac{3 b^2 (1+m)}{2+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) \sqrt{\sin ^2(c+d x)}}-\frac{b \left (3 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*}

Mathematica [A]  time = 0.807916, size = 197, normalized size = 0.79 \[ \frac{\sqrt{\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (b \cos (c+d x) \left (b \cos (c+d x) \left (-\frac{3 a \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(c+d x)\right )}{m+3}-\frac{b \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+4}{2};\frac{m+6}{2};\cos ^2(c+d x)\right )}{m+4}\right )-\frac{3 a^2 \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{m+2}\right )-\frac{a^3 \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{m+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((a^3*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2])/(1 +
m)) + b*Cos[c + d*x]*((-3*a^2*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m) + b*Cos[c
+ d*x]*((-3*a*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2])/(3 + m) - (b*Cos[c + d*x]*Hypergeo
metric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[c + d*x]^2])/(4 + m))))*Sqrt[Sin[c + d*x]^2])/d

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Maple [F]  time = 1.426, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a+b\cos \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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