\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [61]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 A+3 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/5*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(2*A+3*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/15*(2*A+3*B)*sin(
d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2829, 2729, 2727} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(2 A+3 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(2 A+3 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac {(A-B) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[In]

Int[(A + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

((A - B)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((2*A + 3*B)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^
2) + ((2*A + 3*B)*Sin[c + d*x])/(15*d*(a^3 + a^3*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)]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B) \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{5 a} \\ & = \frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 A+3 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{15 a^2} \\ & = \frac {(A-B) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(2 A+3 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (7 A+3 B+(6 A+9 B) \cos (c+d x)+(2 A+3 B) \cos ^2(c+d x)\right ) \sin (c+d x)}{15 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[(A + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

((7*A + 3*B + (6*A + 9*B)*Cos[c + d*x] + (2*A + 3*B)*Cos[c + d*x]^2)*Sin[c + d*x])/(15*a^3*d*(1 + Cos[c + d*x]
)^3)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 A +5 B \right )}{20 a^{3} d}\)	\(56\)
derivativedivides	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(64\)
default	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(64\)
risch	\(\frac {2 i \left (15 B \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+10 A \,{\mathrm e}^{i \left (d x +c \right )}+15 B \,{\mathrm e}^{i \left (d x +c \right )}+2 A +3 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(90\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (5 A +3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (13 A -3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}\)	\(117\)

[In]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/20*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^4+10/3*A*tan(1/2*d*x+1/2*c)^2+5*A+5*B)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left ({\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 3 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((2*A + 3*B)*cos(d*x + c)^2 + 3*(2*A + 3*B)*cos(d*x + c) + 7*A + 3*B)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**5/(20*a**3*d) + A*tan(c/2 + d*x/2)**3/(6*a**3*d) + A*tan(c/2 + d*x/2)/(4*a**3*d
) - B*tan(c/2 + d*x/2)**5/(20*a**3*d) + B*tan(c/2 + d*x/2)/(4*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))/(a*cos(c)
+ a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(A*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 + 3*B*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]

[In]

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 10*A*tan(1/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x
+ 1/2*c) + 15*B*tan(1/2*d*x + 1/2*c))/(a^3*d)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B+10\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]

[In]

int((A + B*cos(c + d*x))/(a + a*cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(15*A + 15*B + 10*A*tan(c/2 + (d*x)/2)^2 + 3*A*tan(c/2 + (d*x)/2)^4 - 3*B*tan(c/2 + (d*x)/
2)^4))/(60*a^3*d)