\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1080]

Optimal result

Integrand size = 27, antiderivative size = 106 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 a x}{b^3}-\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2942, 2814, 2739, 632, 210} \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}+\frac {2 a x}{b^3}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))} \]

[In]

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

[Out]

(2*a*x)/b^3 - (2*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^3*Sqrt[a^2 - b^2]*d) + (Co
s[c + d*x]*(2*a + b*Sin[c + d*x]))/(b^2*d*(a + b*Sin[c + d*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2942

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(m + 1)*(m + p + 1
))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {4 a^2 c+4 a^2 d x+4 a b \cos (c+d x)+4 a b (c+d x) \sin (c+d x)+b^2 \sin (2 (c+d x))}{a+b \sin (c+d x)}}{2 b^3 d} \]

[In]

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

[Out]

((-4*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (4*a^2*c + 4*a^2*d*x +
4*a*b*Cos[c + d*x] + 4*a*b*(c + d*x)*Sin[c + d*x] + b^2*Sin[2*(c + d*x)])/(a + b*Sin[c + d*x]))/(2*b^3*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {\frac {\frac {4 b}{2+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {4 \left (\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}-\frac {a b}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3}}}{d}\)	\(151\)
default	\(\frac {\frac {\frac {4 b}{2+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {4 \left (\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}-\frac {a b}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3}}}{d}\)	\(151\)
risch	\(\frac {2 a x}{b^{3}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {2 i a \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d b}\)	\(384\)

[In]

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

[Out]

1/d*(4/b^3*(1/2*b/(1+tan(1/2*d*x+1/2*c)^2)+a*arctan(tan(1/2*d*x+1/2*c)))-4/b^3*((-1/2*tan(1/2*d*x+1/2*c)*b^2-1
/2*a*b)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+1/2*(2*a^2-b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(
1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 479, normalized size of antiderivative = 4.52 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {4 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 4 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + 2 \, {\left (2 \, {\left (a^{3} b - a b^{3}\right )} d x + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d\right )}}, \frac {2 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + {\left (2 \, {\left (a^{3} b - a b^{3}\right )} d x + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d}\right ] \]

[In]

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

[Out]

[1/2*(4*(a^4 - a^2*b^2)*d*x + (2*a^3 - a*b^2 + (2*a^2*b - b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^
2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^
2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 4*(a^3*b - a*b^3)*cos(d*x + c) + 2*(2*(a^3*
b - a*b^3)*d*x + (a^2*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*sin(d*x + c) + (a^3*b^3 - a*b^
5)*d), (2*(a^4 - a^2*b^2)*d*x + (2*a^3 - a*b^2 + (2*a^2*b - b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(
d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 2*(a^3*b - a*b^3)*cos(d*x + c) + (2*(a^3*b - a*b^3)*d*x + (a^2
*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*sin(d*x + c) + (a^3*b^3 - a*b^5)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )} a}{b^{3}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} - b^{2}\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{2}}\right )}}{d} \]

[In]

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

[Out]

2*((d*x + c)*a/b^3 - (pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 -
b^2)))*(2*a^2 - b^2)/(sqrt(a^2 - b^2)*b^3) + (b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 3*b*tan
(1/2*d*x + 1/2*c) + 2*a)/((a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2
+ 2*b*tan(1/2*d*x + 1/2*c) + a)*b^2))/d

Mupad [B] (verification not implemented)

Time = 10.53 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.54 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b}+\frac {4\,a}{b^2}+\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}+\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b^2}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {2\,a\,x}{b^3}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-b^2\right )}{d\,\left (b^5-a^2\,b^3\right )}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (2\,a^2\,\sqrt {b^2-a^2}-b^2\,\sqrt {b^2-a^2}\right )}{b^3\,d\,\left (a^2-b^2\right )} \]

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3)/b + (4*a)/b^2 + (6*tan(c/2 + (d*x)/2))/b + (4*a*tan(c/2 + (d*x)/2)^2)/b^2)/(d*(a + 2
*b*tan(c/2 + (d*x)/2) + 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4 + 2*b*tan(c/2 + (d*x)/2)^3)) + (2*a*
x)/b^3 - (log(b + a*tan(c/2 + (d*x)/2) - (b^2 - a^2)^(1/2))*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2))/(d*(b^5 -
a^2*b^3)) - (log(b + a*tan(c/2 + (d*x)/2) + (b^2 - a^2)^(1/2))*(2*a^2*(b^2 - a^2)^(1/2) - b^2*(b^2 - a^2)^(1/2
)))/(b^3*d*(a^2 - b^2))