\(\int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\) [1013]

Optimal result

Integrand size = 31, antiderivative size = 66 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {2 (A-B) \log (1+\sin (c+d x))}{a^2 d}-\frac {(A-B) \sin (c+d x)}{a^2 d}-\frac {B (a-a \sin (c+d x))^2}{2 a^4 d} \]

[Out]

2*(A-B)*ln(1+sin(d*x+c))/a^2/d-(A-B)*sin(d*x+c)/a^2/d-1/2*B*(a-a*sin(d*x+c))^2/a^4/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 78} \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {B (a-a \sin (c+d x))^2}{2 a^4 d}-\frac {(A-B) \sin (c+d x)}{a^2 d}+\frac {2 (A-B) \log (\sin (c+d x)+1)}{a^2 d} \]

[In]

Int[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*(A - B)*Log[1 + Sin[c + d*x]])/(a^2*d) - ((A - B)*Sin[c + d*x])/(a^2*d) - (B*(a - a*Sin[c + d*x])^2)/(2*a^4
*d)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x) \left (A+\frac {B x}{a}\right )}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (-A+B+\frac {B (a-x)}{a}+\frac {2 a (A-B)}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {2 (A-B) \log (1+\sin (c+d x))}{a^2 d}-\frac {(A-B) \sin (c+d x)}{a^2 d}-\frac {B (a-a \sin (c+d x))^2}{2 a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {B-4 (A-B) \log (1+\sin (c+d x))+2 (A-2 B) \sin (c+d x)+B \sin ^2(c+d x)}{2 a^2 d} \]

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/2*(B - 4*(A - B)*Log[1 + Sin[c + d*x]] + 2*(A - 2*B)*Sin[c + d*x] + B*Sin[c + d*x]^2)/(a^2*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {\frac {B \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )-2 B \sin \left (d x +c \right )+\left (2 B -2 A \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{a^{2} d}\)	\(55\)
default	\(-\frac {\frac {B \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )-2 B \sin \left (d x +c \right )+\left (2 B -2 A \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{a^{2} d}\)	\(55\)
parallelrisch	\(\frac {8 \left (-A +B \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+B \cos \left (2 d x +2 c \right )+4 \left (2 B -A \right ) \sin \left (d x +c \right )-B}{4 a^{2} d}\)	\(77\)
risch	\(-\frac {2 i x A}{a^{2}}+\frac {2 i x B}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a^{2} d}-\frac {4 i A c}{a^{2} d}+\frac {4 i B c}{a^{2} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{2} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}+\frac {\cos \left (2 d x +2 c \right ) B}{4 a^{2} d}\)	\(178\)
norman	\(\frac {-\frac {\left (12 A -18 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (12 A -18 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (12 A -17 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (12 A -17 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (10 A -15 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (10 A -15 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (6 A -10 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (6 A -10 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (2 A -4 B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {2 \left (A -B \right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(345\)

[In]

int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/a^2/d*(1/2*B*sin(d*x+c)^2+A*sin(d*x+c)-2*B*sin(d*x+c)+(2*B-2*A)*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B \cos \left (d x + c\right )^{2} + 4 \, {\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{2 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(B*cos(d*x + c)^2 + 4*(A - B)*log(sin(d*x + c) + 1) - 2*(A - 2*B)*sin(d*x + c))/(a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (56) = 112\).

Time = 7.08 (sec) , antiderivative size = 1096, normalized size of antiderivative = 16.61 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((4*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 +
d*x/2)**2 + a**2*d) + 8*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d
*tan(c/2 + d*x/2)**2 + a**2*d) + 4*A*log(tan(c/2 + d*x/2) + 1)/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2
+ d*x/2)**2 + a**2*d) - 2*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a
**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 4*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d
*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*log(tan(c/2 + d*x/2)**2 + 1)/(a**2*d*tan(c/2 + d*x/2)*
*4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*tan(c/2 + d*x/2)**3/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*t
an(c/2 + d*x/2)**2 + a**2*d) - 2*A*tan(c/2 + d*x/2)/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2
+ a**2*d) - 4*B*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2
+ d*x/2)**2 + a**2*d) - 8*B*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2
*d*tan(c/2 + d*x/2)**2 + a**2*d) - 4*B*log(tan(c/2 + d*x/2) + 1)/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/
2 + d*x/2)**2 + a**2*d) + 2*B*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2
*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) + 4*B*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 +
d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) + 2*B*log(tan(c/2 + d*x/2)**2 + 1)/(a**2*d*tan(c/2 + d*x/2
)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) + 4*B*tan(c/2 + d*x/2)**3/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d
*tan(c/2 + d*x/2)**2 + a**2*d) - 2*B*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/
2)**2 + a**2*d) + 4*B*tan(c/2 + d*x/2)/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d), N
e(d, 0)), (x*(A + B*sin(c))*cos(c)**3/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {B \sin \left (d x + c\right )^{2} + 2 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{a^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(4*(A - B)*log(sin(d*x + c) + 1)/a^2 - (B*sin(d*x + c)^2 + 2*(A - 2*B)*sin(d*x + c))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (A - B\right )} \log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a^{2}} + \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (B + \frac {2 \, {\left (A a^{2} - 3 \, B a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )} a}\right )}}{a^{4}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*(A - B)*log(abs(a*sin(d*x + c) + a)/((a*sin(d*x + c) + a)^2*abs(a)))/a^2 + (a*sin(d*x + c) + a)^2*(B +
2*(A*a^2 - 3*B*a^2)/((a*sin(d*x + c) + a)*a))/a^4)/d

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {2\,A\,\sin \left (c+d\,x\right )-4\,B\,\sin \left (c+d\,x\right )+B\,{\sin \left (c+d\,x\right )}^2-4\,A\,\ln \left (\sin \left (c+d\,x\right )+1\right )+4\,B\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,a^2\,d} \]

[In]

int((cos(c + d*x)^3*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x))^2,x)

[Out]

-(2*A*sin(c + d*x) - 4*B*sin(c + d*x) + B*sin(c + d*x)^2 - 4*A*log(sin(c + d*x) + 1) + 4*B*log(sin(c + d*x) +
1))/(2*a^2*d)