3.34 \(\int (b \cos (c+d x))^m (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=117 \[ \frac{C \sin (c+d x) (b \cos (c+d x))^{m+1}}{b d (m+2)}-\frac{(A (m+2)+C (m+1)) \sin (c+d x) (b \cos (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(C*(b*Cos[c + d*x])^(1 + m)*Sin[c + d*x])/(b*d*(2 + m)) - ((C*(1 + m) + A*(2 + m))*(b*Cos[c + d*x])^(1 + m)*Hy
pergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(1 + m)*(2 + m)*Sqrt[Sin[c + d*x
]^2])

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Rubi [A]  time = 0.0721987, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3014, 2643} \[ \frac{C \sin (c+d x) (b \cos (c+d x))^{m+1}}{b d (m+2)}-\frac{(A (m+2)+C (m+1)) \sin (c+d x) (b \cos (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b d (m+1) (m+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(b*Cos[c + d*x])^(1 + m)*Sin[c + d*x])/(b*d*(2 + m)) - ((C*(1 + m) + A*(2 + m))*(b*Cos[c + d*x])^(1 + m)*Hy
pergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(1 + m)*(2 + m)*Sqrt[Sin[c + d*x
]^2])

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]

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 (b \cos (c+d x))^m \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C (b \cos (c+d x))^{1+m} \sin (c+d x)}{b d (2+m)}+\left (A+\frac{C (1+m)}{2+m}\right ) \int (b \cos (c+d x))^m \, dx\\ &=\frac{C (b \cos (c+d x))^{1+m} \sin (c+d x)}{b d (2+m)}-\frac{\left (A+\frac{C (1+m)}{2+m}\right ) (b \cos (c+d x))^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.173454, size = 114, normalized size = 0.97 \[ -\frac{\sqrt{\sin ^2(c+d x)} \cot (c+d x) (b \cos (c+d x))^m \left (A (m+3) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )+C (m+1) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(c+d x)\right )\right )}{d (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*Cos[c + d*x])^m*Cot[c + d*x]*(A*(3 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2] + C
*(1 + m)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d
*(1 + m)*(3 + m)))

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Maple [F]  time = 1.754, size = 0, normalized size = 0. \begin{align*} \int \left ( b\cos \left ( dx+c \right ) \right ) ^{m} \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^m*(A+C*cos(d*x+c)^2),x)

[Out]

int((b*cos(d*x+c))^m*(A+C*cos(d*x+c)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^m*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*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 (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^m*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*(b*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((b*cos(d*x+c))**m*(A+C*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 (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^m*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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