\(\int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) [1546]

Optimal result

Integrand size = 31, antiderivative size = 111 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\left (a^2-b^2\right ) (A b-a B) \log (a+b \sin (c+d x))}{b^4 d}+\frac {\left (a A b-a^2 B+b^2 B\right ) \sin (c+d x)}{b^3 d}-\frac {(A b-a B) \sin ^2(c+d x)}{2 b^2 d}-\frac {B \sin ^3(c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 111, 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 = {2916, 786} \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\left (a^2-b^2\right ) (A b-a B) \log (a+b \sin (c+d x))}{b^4 d}+\frac {\left (a^2 (-B)+a A b+b^2 B\right ) \sin (c+d x)}{b^3 d}-\frac {(A b-a B) \sin ^2(c+d x)}{2 b^2 d}-\frac {B \sin ^3(c+d x)}{3 b d} \]

[In]

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

[Out]

-(((a^2 - b^2)*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/(b^4*d)) + ((a*A*b - a^2*B + b^2*B)*Sin[c + d*x])/(b^3*d)
- ((A*b - a*B)*Sin[c + d*x]^2)/(2*b^2*d) - (B*Sin[c + d*x]^3)/(3*b*d)

Rule 786

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\left (A-\frac {a B}{b}\right ) \left (\left (-a^2+b^2\right ) \log (a+b \sin (c+d x))+a b \sin (c+d x)-\frac {1}{2} b^2 \sin ^2(c+d x)\right )+\frac {1}{12} b^2 B (9 \sin (c+d x)+\sin (3 (c+d x)))}{b^3 d} \]

[In]

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

[Out]

((A - (a*B)/b)*((-a^2 + b^2)*Log[a + b*Sin[c + d*x]] + a*b*Sin[c + d*x] - (b^2*Sin[c + d*x]^2)/2) + (b^2*B*(9*
Sin[c + d*x] + Sin[3*(c + d*x)]))/12)/(b^3*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(-\frac {-\frac {-\frac {B \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {A \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}}{2}+\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) a b}{2}+A \sin \left (d x +c \right ) a b -B \sin \left (d x +c \right ) a^{2}+B \sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(127\)
default	\(-\frac {-\frac {-\frac {B \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {A \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}}{2}+\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) a b}{2}+A \sin \left (d x +c \right ) a b -B \sin \left (d x +c \right ) a^{2}+B \sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(127\)
parallelrisch	\(\frac {-12 \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+12 \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,b^{3}-3 B a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right ) B \,b^{3}+\left (12 A a \,b^{2}-12 B \,a^{2} b +9 B \,b^{3}\right ) \sin \left (d x +c \right )-3 A \,b^{3}+3 B a \,b^{2}}{12 b^{4} d}\)	\(164\)
norman	\(\frac {-\frac {2 \left (2 A b -2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}+\frac {2 \left (9 A a b -9 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{3}}+\frac {2 \left (9 A a b -9 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{3}}+\frac {2 \left (A a b -B \,a^{2}+B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}+\frac {2 \left (A a b -B \,a^{2}+B \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {2 \left (A b -B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {2 \left (A b -B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {\left (A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{4} d}\)	\(350\)
risch	\(\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{8 b d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B \,a^{3}}{d \,b^{4}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B a}{d \,b^{2}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 b d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A \,a^{2}}{d \,b^{3}}+\frac {2 i B a c}{d \,b^{2}}+\frac {2 i A \,a^{2} c}{b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A a}{2 b^{2} d}-\frac {i x A}{b}-\frac {2 i A c}{d b}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 b^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{2 b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A}{d b}+\frac {\sin \left (3 d x +3 c \right ) B}{12 b d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A a}{2 b^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B a}{8 b^{2} d}+\frac {i x A \,a^{2}}{b^{3}}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} B a}{8 b^{2} d}+\frac {i x B a}{b^{2}}-\frac {i x B \,a^{3}}{b^{4}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B}{8 b d}-\frac {2 i B \,a^{3} c}{b^{4} d}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} A}{8 b d}\)	\(459\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {3 \, {\left (B a b^{2} - A b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (B b^{3} \cos \left (d x + c\right )^{2} - 3 \, B a^{2} b + 3 \, A a b^{2} + 2 \, B b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \]

[In]

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

[Out]

-1/6*(3*(B*a*b^2 - A*b^3)*cos(d*x + c)^2 - 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(b*sin(d*x + c) + a) - 2*(
B*b^3*cos(d*x + c)^2 - 3*B*a^2*b + 3*A*a*b^2 + 2*B*b^3)*sin(d*x + c))/(b^4*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, B b^{2} \sin \left (d x + c\right )^{3} - 3 \, {\left (B a b - A b^{2}\right )} \sin \left (d x + c\right )^{2} + 6 \, {\left (B a^{2} - A a b - B b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{6 \, d} \]

[In]

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

[Out]

-1/6*((2*B*b^2*sin(d*x + c)^3 - 3*(B*a*b - A*b^2)*sin(d*x + c)^2 + 6*(B*a^2 - A*a*b - B*b^2)*sin(d*x + c))/b^3
- 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(b*sin(d*x + c) + a)/b^4)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, B b^{2} \sin \left (d x + c\right )^{3} - 3 \, B a b \sin \left (d x + c\right )^{2} + 3 \, A b^{2} \sin \left (d x + c\right )^{2} + 6 \, B a^{2} \sin \left (d x + c\right ) - 6 \, A a b \sin \left (d x + c\right ) - 6 \, B b^{2} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}}}{6 \, d} \]

[In]

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

[Out]

-1/6*((2*B*b^2*sin(d*x + c)^3 - 3*B*a*b*sin(d*x + c)^2 + 3*A*b^2*sin(d*x + c)^2 + 6*B*a^2*sin(d*x + c) - 6*A*a
*b*sin(d*x + c) - 6*B*b^2*sin(d*x + c))/b^3 - 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(abs(b*sin(d*x + c) + a
))/b^4)/d

Mupad [B] (verification not implemented)

Time = 12.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A}{2\,b}-\frac {B\,a}{2\,b^2}\right )}{d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (B\,a^3-A\,a^2\,b-B\,a\,b^2+A\,b^3\right )}{b^4\,d}-\frac {B\,{\sin \left (c+d\,x\right )}^3}{3\,b\,d} \]

[In]

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

[Out]

(sin(c + d*x)*(B/b + (a*(A/b - (B*a)/b^2))/b))/d - (sin(c + d*x)^2*(A/(2*b) - (B*a)/(2*b^2)))/d + (log(a + b*s
in(c + d*x))*(A*b^3 + B*a^3 - A*a^2*b - B*a*b^2))/(b^4*d) - (B*sin(c + d*x)^3)/(3*b*d)