\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1129]

Optimal result

Integrand size = 27, antiderivative size = 163 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {a \left (4 a^2-3 b^2\right ) x}{b^5}+\frac {2 \left (4 a^4-5 a^2 b^2+b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 \sqrt {a^2-b^2} d}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d} \]

[Out]

-a*(4*a^2-3*b^2)*x/b^5+1/3*cos(d*x+c)^3*(4*a+b*sin(d*x+c))/b^2/d/(a+b*sin(d*x+c))-cos(d*x+c)*(4*a^2-b^2-2*a*b*
sin(d*x+c))/b^4/d+2*(4*a^4-5*a^2*b^2+b^4)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^5/d/(a^2-b^2)^(1/
2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2942, 2944, 2814, 2739, 632, 210} \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {a x \left (4 a^2-3 b^2\right )}{b^5}-\frac {\cos (c+d x) \left (4 a^2-2 a b \sin (c+d x)-b^2\right )}{b^4 d}+\frac {2 \left (4 a^4-5 a^2 b^2+b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d \sqrt {a^2-b^2}}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))} \]

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

-((a*(4*a^2 - 3*b^2)*x)/b^5) + (2*(4*a^4 - 5*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/
(b^5*Sqrt[a^2 - b^2]*d) + (Cos[c + d*x]^3*(4*a + b*Sin[c + d*x]))/(3*b^2*d*(a + b*Sin[c + d*x])) - (Cos[c + d*
x]*(4*a^2 - b^2 - 2*a*b*Sin[c + d*x]))/(b^4*d)

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]

Rule 2944

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

Rubi steps \begin{align*} \text {integral}& = \frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {\cos ^2(c+d x) (-b-4 a \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{b^2} \\ & = \frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d}-\frac {\int \frac {2 b \left (2 a^2-b^2\right )+2 a \left (4 a^2-3 b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^4} \\ & = -\frac {a \left (4 a^2-3 b^2\right ) x}{b^5}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d}+\frac {\left (4 a^4-5 a^2 b^2+b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^5} \\ & = -\frac {a \left (4 a^2-3 b^2\right ) x}{b^5}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d}+\frac {\left (2 \left (4 a^4-5 a^2 b^2+b^4\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^5 d} \\ & = -\frac {a \left (4 a^2-3 b^2\right ) x}{b^5}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d}-\frac {\left (4 \left (4 a^4-5 a^2 b^2+b^4\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^5 d} \\ & = -\frac {a \left (4 a^2-3 b^2\right ) x}{b^5}+\frac {2 \left (4 a^4-5 a^2 b^2+b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 \sqrt {a^2-b^2} d}+\frac {\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x) \left (4 a^2-b^2-2 a b \sin (c+d x)\right )}{b^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {48 \left (4 a^4-5 a^2 b^2+b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {-96 a^4 c+72 a^2 b^2 c-96 a^4 d x+72 a^2 b^2 d x+\left (-96 a^3 b+60 a b^3\right ) \cos (c+d x)-4 a b^3 \cos (3 (c+d x))-96 a^3 b c \sin (c+d x)+72 a b^3 c \sin (c+d x)-96 a^3 b d x \sin (c+d x)+72 a b^3 d x \sin (c+d x)-24 a^2 b^2 \sin (2 (c+d x))+14 b^4 \sin (2 (c+d x))+b^4 \sin (4 (c+d x))}{a+b \sin (c+d x)}}{24 b^5 d} \]

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((48*(4*a^4 - 5*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (-96*a^4*c
+ 72*a^2*b^2*c - 96*a^4*d*x + 72*a^2*b^2*d*x + (-96*a^3*b + 60*a*b^3)*Cos[c + d*x] - 4*a*b^3*Cos[3*(c + d*x)]
- 96*a^3*b*c*Sin[c + d*x] + 72*a*b^3*c*Sin[c + d*x] - 96*a^3*b*d*x*Sin[c + d*x] + 72*a*b^3*d*x*Sin[c + d*x] -
24*a^2*b^2*Sin[2*(c + d*x)] + 14*b^4*Sin[2*(c + d*x)] + b^4*Sin[4*(c + d*x)])/(a + b*Sin[c + d*x]))/(24*b^5*d)

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.69

method	result	size
derivativedivides	\(\frac {\frac {\frac {4 \left (-\frac {b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3} b}{2}+\frac {a \,b^{3}}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (4 a^{4}-5 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{5}}-\frac {4 \left (\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{2}+\left (\frac {3}{2} a^{2} b -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{2} b -b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}}{2}+\frac {3 a^{2} b}{2}-\frac {2 b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (4 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}}{d}\)	\(276\)
default	\(\frac {\frac {\frac {4 \left (-\frac {b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3} b}{2}+\frac {a \,b^{3}}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (4 a^{4}-5 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{5}}-\frac {4 \left (\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{2}+\left (\frac {3}{2} a^{2} b -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{2} b -b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}}{2}+\frac {3 a^{2} b}{2}-\frac {2 b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (4 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}}{d}\)	\(276\)
risch	\(-\frac {4 a^{3} x}{b^{5}}+\frac {3 a x}{b^{3}}-\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{3} d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{4} d}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{2} d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{4} d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}+\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{3} d}+\frac {2 i a \left (-a^{2}+b^{2}\right ) \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{5} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {4 i \sqrt {a^{2}-b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{5}}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{3}}+\frac {4 i \sqrt {a^{2}-b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{5}}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{3}}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,b^{2}}\)	\(421\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.11 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \, {\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} d x + 3 \, {\left (4 \, a^{3} - a b^{2} + {\left (4 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 6 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) - 2 \, {\left (b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} d x - 3 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}}, -\frac {2 \, a b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} d x + 3 \, {\left (4 \, a^{3} - a b^{2} + {\left (4 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 3 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) - {\left (b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} d x - 3 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(4*a*b^3*cos(d*x + c)^3 + 6*(4*a^4 - 3*a^2*b^2)*d*x + 3*(4*a^3 - a*b^2 + (4*a^2*b - b^3)*sin(d*x + c))*s
qrt(-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)) + 6*(4*a^3*b
- 3*a*b^3)*cos(d*x + c) - 2*(b^4*cos(d*x + c)^3 - 3*(4*a^3*b - 3*a*b^3)*d*x - 3*(2*a^2*b^2 - b^4)*cos(d*x + c)
)*sin(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d), -1/3*(2*a*b^3*cos(d*x + c)^3 + 3*(4*a^4 - 3*a^2*b^2)*d*x + 3*(
4*a^3 - a*b^2 + (4*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)*co
s(d*x + c))) + 3*(4*a^3*b - 3*a*b^3)*cos(d*x + c) - (b^4*cos(d*x + c)^3 - 3*(4*a^3*b - 3*a*b^3)*d*x - 3*(2*a^2
*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^4*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 = 0.33 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {6 \, {\left (4 \, a^{4} - 5 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {6 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} - 4 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(3*(4*a^3 - 3*a*b^2)*(d*x + c)/b^5 - 6*(4*a^4 - 5*a^2*b^2 + b^4)*(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)))/(sqrt(a^2 - b^2)*b^5) + 6*(a^2*b*tan(1/2*d*x + 1/2*c)
- b^3*tan(1/2*d*x + 1/2*c) + a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*b^4) + 2
*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^4 - 6*b^2*tan(1/2*d*x + 1/2*c)^4 + 18*a^2*tan(1/2*
d*x + 1/2*c)^2 - 6*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*a^2 - 4*b^2)/((tan(1/2*d*x + 1/
2*c)^2 + 1)^3*b^4))/d

Mupad [B] (verification not implemented)

Time = 12.60 (sec) , antiderivative size = 964, normalized size of antiderivative = 5.91 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int((cos(c + d*x)^4*sin(c + d*x))/(a + b*sin(c + d*x))^2,x)

[Out]

((2*(7*a*b^2 - 12*a^3))/(3*b^4) + (2*tan(c/2 + (d*x)/2)^6*(a*b^2 - 4*a^3))/b^4 + (2*tan(c/2 + (d*x)/2)^4*(7*a*
b^2 - 12*a^3))/b^4 + (2*tan(c/2 + (d*x)/2)^2*(25*a*b^2 - 36*a^3))/(3*b^4) - (2*tan(c/2 + (d*x)/2)*(18*a^2 - 11
*b^2))/(3*b^3) - (14*tan(c/2 + (d*x)/2)^3*(2*a^2 - b^2))/b^3 - (2*tan(c/2 + (d*x)/2)^7*(2*a^2 - b^2))/b^3 - (2
*tan(c/2 + (d*x)/2)^5*(10*a^2 - 7*b^2))/b^3)/(d*(a + 2*b*tan(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*t
an(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + (d*x)/2)^6 + a*tan(c/2 + (d*x)/2)^8 + 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c
/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*x)/2)^7)) - (2*atanh((64*a^2*(b^2 - a^2)^(1/2))/(64*a^2*b - (320*a^4)/b + (
256*a^6)/b^3 - 640*a^3*tan(c/2 + (d*x)/2) + 128*a*b^2*tan(c/2 + (d*x)/2) + (512*a^5*tan(c/2 + (d*x)/2))/b^2) -
(256*a^4*(b^2 - a^2)^(1/2))/(64*a^2*b^3 - 320*a^4*b + (256*a^6)/b + 512*a^5*tan(c/2 + (d*x)/2) + 128*a*b^4*ta
n(c/2 + (d*x)/2) - 640*a^3*b^2*tan(c/2 + (d*x)/2)) + (128*a*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(64*a^2 - (3
20*a^4)/b^2 + (256*a^6)/b^4 - (640*a^3*tan(c/2 + (d*x)/2))/b + (512*a^5*tan(c/2 + (d*x)/2))/b^3 + 128*a*b*tan(
c/2 + (d*x)/2)) - (576*a^3*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(64*a^2*b^2 - 320*a^4 + (256*a^6)/b^2 + 128*a
*b^3*tan(c/2 + (d*x)/2) - 640*a^3*b*tan(c/2 + (d*x)/2) + (512*a^5*tan(c/2 + (d*x)/2))/b) + (256*a^5*tan(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2))/(256*a^6 + 64*a^2*b^4 - 320*a^4*b^2 + 128*a*b^5*tan(c/2 + (d*x)/2) + 512*a^5*b*tan
(c/2 + (d*x)/2) - 640*a^3*b^3*tan(c/2 + (d*x)/2)))*(4*a^2*(b^2 - a^2)^(1/2) - b^2*(b^2 - a^2)^(1/2)))/(b^5*d)
- (2*a*atan((192*a^2*tan(c/2 + (d*x)/2))/(192*a^2 - (448*a^4)/b^2 + (256*a^6)/b^4) - (448*a^4*tan(c/2 + (d*x)/
2))/(192*a^2*b^2 - 448*a^4 + (256*a^6)/b^2) + (256*a^6*tan(c/2 + (d*x)/2))/(256*a^6 + 192*a^2*b^4 - 448*a^4*b^
2))*(4*a^2 - 3*b^2))/(b^5*d)