\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx\) [69]

Optimal result

Integrand size = 29, antiderivative size = 138 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=-\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A-11 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \]

[Out]

-1/7*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/35*(4*A-11*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3+1/105*(8*A+13*B)*
sin(d*x+c)/d/(a^2+a^2*cos(d*x+c))^2+1/105*(8*A+13*B)*sin(d*x+c)/d/(a^4+a^4*cos(d*x+c))

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3047, 3098, 2829, 2729, 2727} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^4 \cos (c+d x)+a^4\right )}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac {(4 A-11 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[In]

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

-1/7*((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + ((4*A - 11*B)*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*
x])^3) + ((8*A + 13*B)*Sin[c + d*x])/(105*d*(a^2 + a^2*Cos[c + d*x])^2) + ((8*A + 13*B)*Sin[c + d*x])/(105*d*(
a^4 + a^4*Cos[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx \\ & = -\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {\int \frac {-4 a (A-B)-7 a B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A-11 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+13 B) \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{35 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A-11 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(8 A+13 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{105 a^3} \\ & = -\frac {(A-B) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(4 A-11 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(8 A+13 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (140 (A+2 B) \sin \left (\frac {d x}{2}\right )-35 (4 A+5 B) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )-105 B \sin \left (2 c+\frac {3 d x}{2}\right )+56 A \sin \left (2 c+\frac {5 d x}{2}\right )+91 B \sin \left (2 c+\frac {5 d x}{2}\right )+8 A \sin \left (3 c+\frac {7 d x}{2}\right )+13 B \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \]

[In]

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(140*(A + 2*B)*Sin[(d*x)/2] - 35*(4*A + 5*B)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*d*
x)/2] + 168*B*Sin[c + (3*d*x)/2] - 105*B*Sin[2*c + (3*d*x)/2] + 56*A*Sin[2*c + (5*d*x)/2] + 91*B*Sin[2*c + (5*
d*x)/2] + 8*A*Sin[3*c + (7*d*x)/2] + 13*B*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (A +B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {7 \left (-A +B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 A -7 B \right )}{56 a^{4} d}\)	\(76\)
derivativedivides	\(\frac {\frac {\left (-A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(88\)
default	\(\frac {\frac {\left (-A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(88\)
risch	\(\frac {2 i \left (105 B \,{\mathrm e}^{5 i \left (d x +c \right )}+140 A \,{\mathrm e}^{4 i \left (d x +c \right )}+175 B \,{\mathrm e}^{4 i \left (d x +c \right )}+140 A \,{\mathrm e}^{3 i \left (d x +c \right )}+280 B \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+56 A \,{\mathrm e}^{i \left (d x +c \right )}+91 B \,{\mathrm e}^{i \left (d x +c \right )}+8 A +13 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(138\)
norman	\(\frac {-\frac {\left (A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (7 A +5 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (11 A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (11 A +31 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 a d}-\frac {\left (17 A -3 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{3}}\)	\(167\)

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBOSE)

[Out]

-1/56*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^6+7/5*(A+B)*tan(1/2*d*x+1/2*c)^4+7/3*(-A+B)*tan(1/2*d*x+1/2
*c)^2-7*A-7*B)/a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left ({\left (8 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (8 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 13 \, A + 8 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*((8*A + 13*B)*cos(d*x + c)^3 + 4*(8*A + 13*B)*cos(d*x + c)^2 + 4*(13*A + 8*B)*cos(d*x + c) + 13*A + 8*B)
*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) +
a^4*d)

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.29 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} - \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((-A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40*a**4*d) + A*tan(c/2 + d*x/2)**3/(24*
a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) + B*tan(c/2 + d*x/2)**7/(56*a**4*d) - B*tan(c/2 + d*x/2)**5/(40*a**4*d
) - B*tan(c/2 + d*x/2)**3/(24*a**4*d) + B*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)/(a*
cos(c) + a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.26 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + B*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=-\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 21*A*tan(1/2*d*x + 1/2*c)^5 + 21*B*tan(1/2
*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/2*c)^3 + 35*B*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 105
*B*tan(1/2*d*x + 1/2*c))/(a^4*d)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.61 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B\right )}{40\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}}{d} \]

[In]

int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^4,x)

[Out]

-((tan(c/2 + (d*x)/2)^5*(A + B))/(40*a^4) - (tan(c/2 + (d*x)/2)^3*(A - B))/(24*a^4) + (tan(c/2 + (d*x)/2)^7*(A
- B))/(56*a^4) - (tan(c/2 + (d*x)/2)*(A + B))/(8*a^4))/d