\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [310]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 47 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 x}{a^2}+\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2936, 2718} \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {2 x}{a^2} \]

[In]

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

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2936

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
 /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(117\) vs. \(2(47)=94\).

Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.49 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {12 d x \cos \left (\frac {d x}{2}\right )+2 \cos \left (c+\frac {d x}{2}\right )+3 \cos \left (c+\frac {3 d x}{2}\right )-28 \sin \left (\frac {d x}{2}\right )+12 d x \sin \left (c+\frac {d x}{2}\right )+3 \sin \left (2 c+\frac {3 d x}{2}\right )}{6 a^2 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(12*d*x*Cos[(d*x)/2] + 2*Cos[c + (d*x)/2] + 3*Cos[c + (3*d*x)/2] - 28*Sin[(d*x)/2] + 12*d*x*Sin[c + (d*x)/2] +
 3*Sin[2*c + (3*d*x)/2])/(6*a^2*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06

method result size
derivativedivides \(\frac {\frac {4}{2+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\) \(50\)
default \(\frac {\frac {4}{2+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\) \(50\)
parallelrisch \(\frac {4 d x \cos \left (d x +c \right )-2 \cos \left (d x +c \right )-4 \sin \left (d x +c \right )+\cos \left (2 d x +2 c \right )+5}{2 d \,a^{2} \cos \left (d x +c \right )}\) \(54\)
risch \(\frac {2 x}{a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {4}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\) \(64\)
norman \(\frac {\frac {6}{a d}+\frac {4 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 x}{a}+\frac {6 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+\frac {12 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {20 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {24 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {24 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {20 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {12 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {6 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {10 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {26 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {22 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {38 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {34 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {38 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) \(349\)

[In]

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

[Out]

4/d/a^2*(1/2/(1+tan(1/2*d*x+1/2*c)^2)+arctan(tan(1/2*d*x+1/2*c))+1/(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.64 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, d x + {\left (2 \, d x + 3\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + {\left (2 \, d x + \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 2}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]

[In]

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

[Out]

(2*d*x + (2*d*x + 3)*cos(d*x + c) + cos(d*x + c)^2 + (2*d*x + cos(d*x + c) - 2)*sin(d*x + c) + 2)/(a^2*d*cos(d
*x + c) + a^2*d*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. 479 vs. \(2 (41) = 82\).

Time = 3.80 (sec) , antiderivative size = 479, normalized size of antiderivative = 10.19 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {2 d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {2 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {2 d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {2 d x}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {4 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} + \frac {6}{a^{2} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos ^{2}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*d*x*tan(c/2 + d*x/2)**3/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)**2 + a**2*d*tan(c/2
 + d*x/2) + a**2*d) + 2*d*x*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)**2 + a**
2*d*tan(c/2 + d*x/2) + a**2*d) + 2*d*x*tan(c/2 + d*x/2)/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)*
*2 + a**2*d*tan(c/2 + d*x/2) + a**2*d) + 2*d*x/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)**2 + a**2
*d*tan(c/2 + d*x/2) + a**2*d) + 4*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)**2
 + a**2*d*tan(c/2 + d*x/2) + a**2*d) + 2*tan(c/2 + d*x/2)/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2
)**2 + a**2*d*tan(c/2 + d*x/2) + a**2*d) + 6/(a**2*d*tan(c/2 + d*x/2)**3 + a**2*d*tan(c/2 + d*x/2)**2 + a**2*d
*tan(c/2 + d*x/2) + a**2*d), Ne(d, 0)), (x*sin(c)*cos(c)**2/(a*sin(c) + a)**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (47) = 94\).

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.96 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 3}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{d} \]

[In]

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

[Out]

2*((sin(d*x + c)/(cos(d*x + c) + 1) + 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3)/(a^2 + a^2*sin(d*x + c)/(cos(
d*x + c) + 1) + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + 2*arctan(
sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {d x + c}{a^{2}} + \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} a^{2}}\right )}}{d} \]

[In]

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

[Out]

2*((d*x + c)/a^2 + (2*tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 3)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*
x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 1)*a^2))/d

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.45 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2\,x}{a^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(2*x)/a^2 + (2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^2 + 6)/(a^2*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*
x)/2)^2 + 1))