3.1158 \(\int \frac{\cos ^m(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=372 \[ \frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b d \left (a^2-b^2\right )}-\frac{(b B-a C) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{b d (m+2) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(a*(A*b^2 - a*(b*B - a*C))*AppellF1[1/2, (1 - m)/2, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2)
)]*Cos[c + d*x]^(-1 + m)*(Cos[c + d*x]^2)^((1 - m)/2)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a
*C))*AppellF1[1/2, -m/2, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^m*Sin[c + d
*x])/(b*(a^2 - b^2)*d*(Cos[c + d*x]^2)^(m/2)) - ((b*B - a*C)*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b^2*d*(1 + m)*Sqrt[Sin[c + d*x]^2]) - (C*Cos[c + d*x]^(2 + m)*
Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(2 + m)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.420838, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3063, 2643, 2823, 3189, 429} \[ \frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b d \left (a^2-b^2\right )}-\frac{(b B-a C) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{b d (m+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*b^2 - a*(b*B - a*C))*AppellF1[1/2, (1 - m)/2, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2)
)]*Cos[c + d*x]^(-1 + m)*(Cos[c + d*x]^2)^((1 - m)/2)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d) - ((A*b^2 - a*(b*B - a
*C))*AppellF1[1/2, -m/2, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^m*Sin[c + d
*x])/(b*(a^2 - b^2)*d*(Cos[c + d*x]^2)^(m/2)) - ((b*B - a*C)*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b^2*d*(1 + m)*Sqrt[Sin[c + d*x]^2]) - (C*Cos[c + d*x]^(2 + m)*
Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(2 + m)*Sqrt[Sin[c + d*x]^2])

Rule 3063

Int[(((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x
_)]^2))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(b*B - a*C)/b^2, Int[(d*Sin[e + f*x])^n, x],
x] + (Dist[(A*b^2 - a*b*B + a^2*C)/b^2, Int[(d*Sin[e + f*x])^n/(a + b*Sin[e + f*x]), x], x] + Dist[C/(b*d), I
nt[(d*Sin[e + f*x])^(n + 1), x], x]) /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0]

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 2823

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

Rule 3189

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, -Dist[(ff*d^(2*IntPart[(m - 1)/2] + 1)*(d*Sin[e + f*x])^(2*FracPart[(m - 1)/
2]))/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x]
, x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac{C \int \cos ^{1+m}(c+d x) \, dx}{b}+\frac{(b B-a C) \int \cos ^m(c+d x) \, dx}{b^2}+\left (A-\frac{a (b B-a C)}{b^2}\right ) \int \frac{\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx\\ &=-\frac{(b B-a C) \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)}{b^2 d (1+m) \sqrt{\sin ^2(c+d x)}}-\frac{C \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)}{b d (2+m) \sqrt{\sin ^2(c+d x)}}+\left (a \left (A-\frac{a (b B-a C)}{b^2}\right )\right ) \int \frac{\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx+\left (b \left (-A+\frac{a (b B-a C)}{b^2}\right )\right ) \int \frac{\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx\\ &=-\frac{(b B-a C) \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)}{b^2 d (1+m) \sqrt{\sin ^2(c+d x)}}-\frac{C \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)}{b d (2+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a \left (A-\frac{a (b B-a C)}{b^2}\right ) \cos ^{2 \left (-\frac{1}{2}+\frac{m}{2}\right )}(c+d x) \cos ^2(c+d x)^{\frac{1}{2}-\frac{m}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1}{2} (-1+m)}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac{\left (b \left (-A+\frac{a (b B-a C)}{b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{a \left (A-\frac{a (b B-a C)}{b^2}\right ) F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} \sin (c+d x)}{\left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{b \left (a^2-b^2\right ) d}-\frac{(b B-a C) \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)}{b^2 d (1+m) \sqrt{\sin ^2(c+d x)}}-\frac{C \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)}{b d (2+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [B]  time = 30.1837, size = 15557, normalized size = 41.82 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.754, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) }{a+b\cos \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+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

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

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

[Out]

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

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

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^m/(b*cos(d*x + c) + a), 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)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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

[Out]

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