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

Optimal. Leaf size=294 \[ \frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) \cos ^2(c+d x)^{\frac {1}{2} (-m-1)} F_1\left (\frac {1}{2};\frac {1}{2} (-m-1),2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}+\frac {a^2 \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} F_1\left (\frac {1}{2};\frac {1-m}{2},2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac {1}{2};-\frac {m}{2},2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2} \]

[Out]

b^2*AppellF1(1/2,-1/2-1/2*m,2,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d*x+c)^(1+m)*(cos(d*x+c)^2)^(-
1/2-1/2*m)*sin(d*x+c)/(a^2-b^2)^2/d+a^2*AppellF1(1/2,1/2-1/2*m,2,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))
*cos(d*x+c)^(-1+m)*(cos(d*x+c)^2)^(1/2-1/2*m)*sin(d*x+c)/(a^2-b^2)^2/d-2*a*b*AppellF1(1/2,-1/2*m,2,3/2,sin(d*x
+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d*x+c)^m*sin(d*x+c)/(a^2-b^2)^2/d/((cos(d*x+c)^2)^(1/2*m))

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Rubi [A]  time = 0.35, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2824, 3189, 429} \[ \frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) \cos ^2(c+d x)^{\frac {1}{2} (-m-1)} F_1\left (\frac {1}{2};\frac {1}{2} (-m-1),2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}+\frac {a^2 \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} F_1\left (\frac {1}{2};\frac {1-m}{2},2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac {1}{2};-\frac {m}{2},2;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*AppellF1[1/2, (-1 - m)/2, 2, 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])/((a^2 - b^2)^2*d) + (a^2*AppellF1[1/2, (1 - m)/2, 2, 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
])/((a^2 - b^2)^2*d) - (2*a*b*AppellF1[1/2, -m/2, 2, 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^2 - b^2)^2*d*(Cos[c + d*x]^2)^(m/2))

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])

Rule 2824

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

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]

Rubi steps

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

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Mathematica [B]  time = 26.46, size = 7214, normalized size = 24.54 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{m}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{m}\left (d x +c \right )}{\left (a +b \cos \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^m}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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