3.274 \(\int \frac{\cos ^m(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=144 \[ -\frac{a^2 \cos ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,m+2,m+3,-\frac{a \cos (c+d x)}{b}\right )}{b d (m+2) \left (a^2-b^2\right )}+\frac{\cos ^{m+2}(c+d x) \text{Hypergeometric2F1}(1,m+2,m+3,-\cos (c+d x))}{2 d (m+2) (a-b)}-\frac{\cos ^{m+2}(c+d x) \text{Hypergeometric2F1}(1,m+2,m+3,\cos (c+d x))}{2 d (m+2) (a+b)} \]

[Out]

(Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, -Cos[c + d*x]])/(2*(a - b)*d*(2 + m)) - (Cos[c + d*x]
^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, Cos[c + d*x]])/(2*(a + b)*d*(2 + m)) - (a^2*Cos[c + d*x]^(2 + m)*H
ypergeometric2F1[1, 2 + m, 3 + m, -((a*Cos[c + d*x])/b)])/(b*(a^2 - b^2)*d*(2 + m))

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Rubi [A]  time = 0.431573, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4397, 2837, 961, 64} \[ -\frac{a^2 \cos ^{m+2}(c+d x) \, _2F_1\left (1,m+2;m+3;-\frac{a \cos (c+d x)}{b}\right )}{b d (m+2) \left (a^2-b^2\right )}+\frac{\cos ^{m+2}(c+d x) \, _2F_1(1,m+2;m+3;-\cos (c+d x))}{2 d (m+2) (a-b)}-\frac{\cos ^{m+2}(c+d x) \, _2F_1(1,m+2;m+3;\cos (c+d x))}{2 d (m+2) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, -Cos[c + d*x]])/(2*(a - b)*d*(2 + m)) - (Cos[c + d*x]
^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, Cos[c + d*x]])/(2*(a + b)*d*(2 + m)) - (a^2*Cos[c + d*x]^(2 + m)*H
ypergeometric2F1[1, 2 + m, 3 + m, -((a*Cos[c + d*x])/b)])/(b*(a^2 - b^2)*d*(2 + m))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \frac{\cos ^m(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx &=\int \frac{\cos ^{1+m}(c+d x) \csc (c+d x)}{b+a \cos (c+d x)} \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a}\right )^{1+m}}{(b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac{a \operatorname{Subst}\left (\int \left (\frac{\left (\frac{x}{a}\right )^{1+m}}{2 a (a+b) (a-x)}-\frac{\left (\frac{x}{a}\right )^{1+m}}{2 a (a-b) (a+x)}+\frac{\left (\frac{x}{a}\right )^{1+m}}{(a-b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{x}{a}\right )^{1+m}}{a+x} \, dx,x,a \cos (c+d x)\right )}{2 (a-b) d}-\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{x}{a}\right )^{1+m}}{a-x} \, dx,x,a \cos (c+d x)\right )}{2 (a+b) d}-\frac{a \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a}\right )^{1+m}}{b+x} \, dx,x,a \cos (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=\frac{\cos ^{2+m}(c+d x) \, _2F_1(1,2+m;3+m;-\cos (c+d x))}{2 (a-b) d (2+m)}-\frac{\cos ^{2+m}(c+d x) \, _2F_1(1,2+m;3+m;\cos (c+d x))}{2 (a+b) d (2+m)}-\frac{a^2 \cos ^{2+m}(c+d x) \, _2F_1\left (1,2+m;3+m;-\frac{a \cos (c+d x)}{b}\right )}{b \left (a^2-b^2\right ) d (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.390116, size = 106, normalized size = 0.74 \[ \frac{\cos ^{m+2}(c+d x) \left (-2 a^2 \text{Hypergeometric2F1}\left (1,m+2,m+3,-\frac{a \cos (c+d x)}{b}\right )+b (a+b) \text{Hypergeometric2F1}(1,m+2,m+3,-\cos (c+d x))-b (a-b) \text{Hypergeometric2F1}(1,m+2,m+3,\cos (c+d x))\right )}{2 b d (m+2) (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(2 + m)*(b*(a + b)*Hypergeometric2F1[1, 2 + m, 3 + m, -Cos[c + d*x]] - (a - b)*b*Hypergeometric2
F1[1, 2 + m, 3 + m, Cos[c + d*x]] - 2*a^2*Hypergeometric2F1[1, 2 + m, 3 + m, -((a*Cos[c + d*x])/b)]))/(2*(a -
b)*b*(a + b)*d*(2 + m))

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Maple [F]  time = 0.547, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m}}{a\sin \left ( dx+c \right ) +b\tan \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

int(cos(d*x+c)^m/(a*sin(d*x+c)+b*tan(d*x+c)),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}}{a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^m/(a*sin(d*x + c) + b*tan(d*x + c)), 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 \sin \left (d x + c\right ) + b \tan \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^m/(a*sin(d*x + c) + b*tan(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**m/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**m/(a*sin(c + d*x) + b*tan(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{m}}{a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^m/(a*sin(d*x + c) + b*tan(d*x + c)), x)