3.402 \(\int \frac{\cos ^m(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=229 \[ \frac{(1-2 m) m \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{3 a^2 d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{2 (1-m) (m+1) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{3 a^2 d (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{2 (1-m) \sin (c+d x) \cos ^{m+1}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{m+1}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.304709, antiderivative size = 229, 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 = {2766, 2978, 2748, 2643} \[ \frac{(1-2 m) m \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{3 a^2 d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{2 (1-m) (m+1) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{3 a^2 d (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{2 (1-m) \sin (c+d x) \cos ^{m+1}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{m+1}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

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 \frac{\cos ^m(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{\cos ^{1+m}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^m(c+d x) (a (2-m)+a m \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{2 (1-m) \cos ^{1+m}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{1+m}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos ^m(c+d x) \left (-a^2 (1-2 m) m+2 a^2 (1-m) (1+m) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{2 (1-m) \cos ^{1+m}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{1+m}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{((1-2 m) m) \int \cos ^m(c+d x) \, dx}{3 a^2}+\frac{(2 (1-m) (1+m)) \int \cos ^{1+m}(c+d x) \, dx}{3 a^2}\\ &=-\frac{2 (1-m) \cos ^{1+m}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{1+m}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(1-2 m) 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)}{3 a^2 d (1+m) \sqrt{\sin ^2(c+d x)}}-\frac{2 (1-m) (1+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)}{3 a^2 d (2+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [F]  time = 1.05838, size = 0, normalized size = 0. \[ \int \frac{\cos ^m(c+d x)}{(a+a \cos (c+d x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.492, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m}}{ \left ( a+\cos \left ( dx+c \right ) a \right ) ^{2}}}\, 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)^2,x)

[Out]

int(cos(d*x+c)^m/(a+cos(d*x+c)*a)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{m}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

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

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

[Out]

integral(cos(d*x + c)^m/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{m}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

Integral(cos(c + d*x)**m/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x)/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{m}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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