3.771 \(\int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=179 \[ -\frac{\left (a^2 (m+2)+b^2 (m+1)\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) \sqrt{\sin ^2(c+d x)}}+\frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)} \]

[Out]

(b^2*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) - ((b^2*(1 + m) + a^2*(2 + m))*Cos[c + d*x]^(1 + m)*Hyperg
eometric2F1[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])
- (2*a*b*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)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.127536, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2789, 2643, 3014} \[ -\frac{\left (a^2 (m+2)+b^2 (m+1)\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{2 a b \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) \sqrt{\sin ^2(c+d x)}}+\frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) - ((b^2*(1 + m) + a^2*(2 + m))*Cos[c + d*x]^(1 + m)*Hyperg
eometric2F1[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])
- (2*a*b*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)*Sqrt[Sin[c + d*x]^2])

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), 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]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin{align*} \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^{1+m}(c+d x) \, dx+\int \cos ^m(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac{2 a b \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)}}+\left (a^2+\frac{b^2 (1+m)}{2+m}\right ) \int \cos ^m(c+d x) \, dx\\ &=\frac{b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac{\left (a^2+\frac{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{2 a b \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 = 0.316214, size = 168, normalized size = 0.94 \[ -\frac{\sqrt{\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (a^2 \left (m^2+5 m+6\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )+b (m+1) \cos (c+d x) \left (2 a (m+3) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )+b (m+2) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(c+d x)\right )\right )\right )}{d (m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.218, 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 ) ^{2}\, 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))^2,x)

[Out]

int(cos(d*x+c)^m*(a+b*cos(d*x+c))^2,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 )}^{2} \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))^2,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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