\(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1280]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \log (a+b \sin (c+d x))}{b^3 d}-\frac {a \sin (c+d x)}{b^2 d}+\frac {\sin ^2(c+d x)}{2 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \log (a+b \sin (c+d x))}{b^3 d}-\frac {a \sin (c+d x)}{b^2 d}+\frac {\sin ^2(c+d x)}{2 b d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a^2 \log (a+b \sin (c+d x))-2 a b \sin (c+d x)+b^2 \sin ^2(c+d x)}{2 b^3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {\sin ^{2}\left (d x +c \right )}{2 b d}\)	\(54\)
default	\(\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {\sin ^{2}\left (d x +c \right )}{2 b d}\)	\(54\)
parallelrisch	\(\frac {-4 a b \sin \left (d x +c \right )-b^{2} \cos \left (2 d x +2 c \right )-4 a^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+b^{2}}{4 b^{3} d}\)	\(86\)
risch	\(-\frac {i a^{2} x}{b^{3}}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}-\frac {2 i a^{2} c}{b^{3} d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{3} d}\)	\(135\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} d}-\frac {4 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{3} d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}\)	\(178\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, a b \sin \left (d x + c\right )}{2 \, b^{3} d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.58 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\begin {cases} \frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} & \text {for}\: b = 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\\frac {a^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b^{3} d} - \frac {a \sin {\left (c + d x \right )}}{b^{2} d} + \frac {\sin ^{2}{\left (c + d x \right )}}{2 b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*sin(c)**2*cos(c)/a, Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)**3/(3*a*d), Eq(b, 0)), (x*sin(c)**2*cos(c
)/(a + b*sin(c)), Eq(d, 0)), (a**2*log(a/b + sin(c + d*x))/(b**3*d) - a*sin(c + d*x)/(b**2*d) + sin(c + d*x)**
2/(2*b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}} + \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3}} + \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2\,a^2\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )+b^2\,{\sin \left (c+d\,x\right )}^2-2\,a\,b\,\sin \left (c+d\,x\right )}{2\,b^3\,d} \]

[In]

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

[Out]

(2*a^2*log(a + b*sin(c + d*x)) + b^2*sin(c + d*x)^2 - 2*a*b*sin(c + d*x))/(2*b^3*d)