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

Optimal. Leaf size=514 \[ \frac{\sin (c+d x) \left (a^2 b^2 (A (-m)+A+C (m+2))+a^4 (-C) (m+1)+A b^4 m\right ) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \left (a^2 b^2 (A (-m)+A+C (m+2))+a^4 (-C) (m+1)+A b^4 m\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \left (a^2 C (m+1)-b^2 (C-A m)\right ) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

((A*b^4*m - a^4*C*(1 + m) + a^2*b^2*(A - A*m + C*(2 + m)))*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)^2*d) - ((A*b^4*m - a^4*C*(1 + m) + a^2*b^2*(A - A*m + C*(2 + m)))*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])/(a*b*(a^2 - b^2)^2*d*(Cos[c + d*x]^2)
^(m/2)) + ((A*b^2 + a^2*C)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) - ((a^2*C
*(1 + m) - b^2*(C - A*m))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Si
n[c + d*x])/(b^2*(a^2 - b^2)*d*(1 + m)*Sqrt[Sin[c + d*x]^2]) + ((A*b^2 + a^2*C)*(1 + m)*Cos[c + d*x]^(2 + m)*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a*b*(a^2 - b^2)*d*(2 + m)*Sqrt[Sin[
c + d*x]^2])

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Rubi [A]  time = 0.880797, antiderivative size = 514, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3056, 3063, 2643, 2823, 3189, 429} \[ \frac{\sin (c+d x) \left (a^2 b^2 (A (-m)+A+C (m+2))+a^4 (-C) (m+1)+A b^4 m\right ) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \left (a^2 b^2 (A (-m)+A+C (m+2))+a^4 (-C) (m+1)+A b^4 m\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \left (a^2 C (m+1)-b^2 (C-A m)\right ) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b^4*m - a^4*C*(1 + m) + a^2*b^2*(A - A*m + C*(2 + m)))*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)^2*d) - ((A*b^4*m - a^4*C*(1 + m) + a^2*b^2*(A - A*m + C*(2 + m)))*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])/(a*b*(a^2 - b^2)^2*d*(Cos[c + d*x]^2)
^(m/2)) + ((A*b^2 + a^2*C)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) - ((a^2*C
*(1 + m) - b^2*(C - A*m))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Si
n[c + d*x])/(b^2*(a^2 - b^2)*d*(1 + m)*Sqrt[Sin[c + d*x]^2]) + ((A*b^2 + a^2*C)*(1 + m)*Cos[c + d*x]^(2 + m)*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a*b*(a^2 - b^2)*d*(2 + m)*Sqrt[Sin[
c + d*x]^2])

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

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

Mathematica [B]  time = 34.2314, size = 17999, normalized size = 35.02 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.523, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) }{ \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

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

[Out]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*cos(d*x + c)^m/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2), 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+C*cos(d*x+c)**2)/(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 \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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