\(\int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1302]

Optimal result

Integrand size = 29, antiderivative size = 235 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {2 a^2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d} \]

[Out]

-1/8*a*(8*a^4-12*a^2*b^2+3*b^4)*x/b^6+2*a^2*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b
^6/d-1/15*(15*a^4-20*a^2*b^2+3*b^4)*cos(d*x+c)/b^5/d+1/8*a*(4*a^2-5*b^2)*cos(d*x+c)*sin(d*x+c)/b^4/d-1/15*(5*a
^2-6*b^2)*cos(d*x+c)*sin(d*x+c)^2/b^3/d+1/4*a*cos(d*x+c)*sin(d*x+c)^3/b^2/d-1/5*cos(d*x+c)*sin(d*x+c)^4/b/d

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2974, 3128, 3102, 2814, 2739, 632, 210} \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a^2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {a \left (4 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {a x \left (8 a^4-12 a^2 b^2+3 b^4\right )}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}-\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]

[In]

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

[Out]

-1/8*(a*(8*a^4 - 12*a^2*b^2 + 3*b^4)*x)/b^6 + (2*a^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^
2 - b^2]])/(b^6*d) - ((15*a^4 - 20*a^2*b^2 + 3*b^4)*Cos[c + d*x])/(15*b^5*d) + (a*(4*a^2 - 5*b^2)*Cos[c + d*x]
*Sin[c + d*x])/(8*b^4*d) - ((5*a^2 - 6*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(15*b^3*d) + (a*Cos[c + d*x]*Sin[c +
d*x]^3)/(4*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^4)/(5*b*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 2974

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(b^2*d*f*(m
+ n + 3)*(m + n + 4))), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*((a + b*Sin[e
+ f*x])^(m + 1)/(b*d^2*f*(m + n + 4))), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\int \frac {\sin ^2(c+d x) \left (5 \left (3 a^2-4 b^2\right )-a b \sin (c+d x)-4 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2} \\ & = -\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\int \frac {\sin (c+d x) \left (-8 a \left (5 a^2-6 b^2\right )+b \left (5 a^2-12 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3} \\ & = \frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\int \frac {15 a^2 \left (4 a^2-5 b^2\right )-a b \left (20 a^2-21 b^2\right ) \sin (c+d x)-8 \left (15 a^4-20 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4} \\ & = -\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\int \frac {15 a^2 b \left (4 a^2-5 b^2\right )+15 a \left (8 a^4-12 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5} \\ & = -\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\left (a^2 \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6} \\ & = -\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\left (2 a^2 \left (a^2-b^2\right )^2\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^6 d} \\ & = -\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\left (4 a^2 \left (a^2-b^2\right )^2\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^6 d} \\ & = -\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {2 a^2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {960 a^2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-60 b \left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x)+10 \left (4 a^2 b^3-3 b^5\right ) \cos (3 (c+d x))-6 b^5 \cos (5 (c+d x))-15 a \left (4 \left (8 a^4-12 a^2 b^2+3 b^4\right ) (c+d x)+\left (-8 a^2 b^2+8 b^4\right ) \sin (2 (c+d x))+b^4 \sin (4 (c+d x))\right )}{480 b^6 d} \]

[In]

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

[Out]

(960*a^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 60*b*(8*a^4 - 10*a^2*b^2 + b^4)*
Cos[c + d*x] + 10*(4*a^2*b^3 - 3*b^5)*Cos[3*(c + d*x)] - 6*b^5*Cos[5*(c + d*x)] - 15*a*(4*(8*a^4 - 12*a^2*b^2
+ 3*b^4)*(c + d*x) + (-8*a^2*b^2 + 8*b^4)*Sin[2*(c + d*x)] + b^4*Sin[4*(c + d*x)]))/(480*b^6*d)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\left (\frac {1}{2} a^{3} b^{2}-\frac {5}{8} a \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{4} b -2 a^{2} b^{3}+b^{5}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b^{2}-\frac {1}{4} a \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -6 a^{2} b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 a^{4} b -\frac {22}{3} a^{2} b^{3}+2 b^{5}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3} b^{2}+\frac {1}{4} a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -\frac {14}{3} a^{2} b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{3} b^{2}+\frac {5}{8} a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{4} b -\frac {4 a^{2} b^{3}}{3}+\frac {b^{5}}{5}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \left (8 a^{4}-12 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{6}}+\frac {2 a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{6} \sqrt {a^{2}-b^{2}}}}{d}\)	\(363\)
default	\(\frac {-\frac {2 \left (\frac {\left (\frac {1}{2} a^{3} b^{2}-\frac {5}{8} a \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{4} b -2 a^{2} b^{3}+b^{5}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b^{2}-\frac {1}{4} a \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -6 a^{2} b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 a^{4} b -\frac {22}{3} a^{2} b^{3}+2 b^{5}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{3} b^{2}+\frac {1}{4} a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{4} b -\frac {14}{3} a^{2} b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{3} b^{2}+\frac {5}{8} a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{4} b -\frac {4 a^{2} b^{3}}{3}+\frac {b^{5}}{5}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \left (8 a^{4}-12 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{6}}+\frac {2 a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{6} \sqrt {a^{2}-b^{2}}}}{d}\)	\(363\)
risch	\(-\frac {a^{5} x}{b^{6}}+\frac {3 a^{3} x}{2 b^{4}}-\frac {3 a x}{8 b^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{5} d}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{16 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{5} d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {\cos \left (5 d x +5 c \right )}{80 b d}-\frac {a \sin \left (4 d x +4 c \right )}{32 b^{2} d}+\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 b^{3} d}-\frac {\cos \left (3 d x +3 c \right )}{16 b d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d \,b^{4}}-\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\)	\(463\)

[In]

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

[Out]

1/d*(-2/b^6*(((1/2*a^3*b^2-5/8*a*b^4)*tan(1/2*d*x+1/2*c)^9+(a^4*b-2*a^2*b^3+b^5)*tan(1/2*d*x+1/2*c)^8+(a^3*b^2
-1/4*a*b^4)*tan(1/2*d*x+1/2*c)^7+(4*a^4*b-6*a^2*b^3)*tan(1/2*d*x+1/2*c)^6+(6*a^4*b-22/3*a^2*b^3+2*b^5)*tan(1/2
*d*x+1/2*c)^4+(-a^3*b^2+1/4*a*b^4)*tan(1/2*d*x+1/2*c)^3+(4*a^4*b-14/3*a^2*b^3)*tan(1/2*d*x+1/2*c)^2+(-1/2*a^3*
b^2+5/8*a*b^4)*tan(1/2*d*x+1/2*c)+a^4*b-4/3*a^2*b^3+1/5*b^5)/(1+tan(1/2*d*x+1/2*c)^2)^5+1/8*a*(8*a^4-12*a^2*b^
2+3*b^4)*arctan(tan(1/2*d*x+1/2*c)))+2*a^2*(a^4-2*a^2*b^2+b^4)/b^6/(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 = 0.44 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.94 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 60 \, {\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 120 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, -\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 120 \, {\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 120 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\right ] \]

[In]

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

[Out]

[-1/120*(24*b^5*cos(d*x + c)^5 - 40*a^2*b^3*cos(d*x + c)^3 + 15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*d*x + 60*(a^4 -
a^2*b^2)*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)) +
120*(a^4*b - a^2*b^3)*cos(d*x + c) + 15*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c))*sin(d*x
+ c))/(b^6*d), -1/120*(24*b^5*cos(d*x + c)^5 - 40*a^2*b^3*cos(d*x + c)^3 + 15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*d
*x + 120*(a^4 - a^2*b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 120*(a
^4*b - a^2*b^3)*cos(d*x + c) + 15*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c))*sin(d*x + c))/
(b^6*d)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c)),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (218) = 436\).

Time = 0.37 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.95 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{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^{6}} + \frac {2 \, {\left (60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 240 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 720 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 720 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 880 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 560 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, a^{4} - 160 \, a^{2} b^{2} + 24 \, b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} b^{5}}}{120 \, d} \]

[In]

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

[Out]

-1/120*(15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*(d*x + c)/b^6 - 240*(a^6 - 2*a^4*b^2 + a^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^6) + 2*(60*a^3
*b*tan(1/2*d*x + 1/2*c)^9 - 75*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*a^4*tan(1/2*d*x + 1/2*c)^8 - 240*a^2*b^2*tan
(1/2*d*x + 1/2*c)^8 + 120*b^4*tan(1/2*d*x + 1/2*c)^8 + 120*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 30*a*b^3*tan(1/2*d*x
+ 1/2*c)^7 + 480*a^4*tan(1/2*d*x + 1/2*c)^6 - 720*a^2*b^2*tan(1/2*d*x + 1/2*c)^6 + 720*a^4*tan(1/2*d*x + 1/2*
c)^4 - 880*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 240*b^4*tan(1/2*d*x + 1/2*c)^4 - 120*a^3*b*tan(1/2*d*x + 1/2*c)^3
+ 30*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*a^4*tan(1/2*d*x + 1/2*c)^2 - 560*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 60*a
^3*b*tan(1/2*d*x + 1/2*c) + 75*a*b^3*tan(1/2*d*x + 1/2*c) + 120*a^4 - 160*a^2*b^2 + 24*b^4)/((tan(1/2*d*x + 1/
2*c)^2 + 1)^5*b^5))/d

Mupad [B] (verification not implemented)

Time = 13.19 (sec) , antiderivative size = 511, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {5\,a^2\,\cos \left (c+d\,x\right )}{4\,b^3\,d}-\frac {\cos \left (3\,c+3\,d\,x\right )}{16\,b\,d}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80\,b\,d}-\frac {\cos \left (c+d\,x\right )}{8\,b\,d}-\frac {a^4\,\cos \left (c+d\,x\right )}{b^5\,d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^2\,d}-\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32\,b^2\,d}+\frac {3\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}-\frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^6\,d}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{12\,b^3\,d}+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^4\,d}-\frac {3\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,b^2\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{b^6\,d} \]

[In]

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

[Out]

(5*a^2*cos(c + d*x))/(4*b^3*d) - cos(3*c + 3*d*x)/(16*b*d) - cos(5*c + 5*d*x)/(80*b*d) - cos(c + d*x)/(8*b*d)
- (a^4*cos(c + d*x))/(b^5*d) - (a*sin(2*c + 2*d*x))/(4*b^2*d) - (a*sin(4*c + 4*d*x))/(32*b^2*d) + (3*a^3*atan(
sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^4*d) - (2*a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^6*d)
+ (a^2*cos(3*c + 3*d*x))/(12*b^3*d) + (a^3*sin(2*c + 2*d*x))/(4*b^4*d) - (3*a*atan(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2)))/(4*b^2*d) + (2*a^2*atanh((2*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - a^
2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4
+ 3*a^4*b^2)^(1/2))/(a^5*cos(c/2 + (d*x)/2) + 2*b^5*sin(c/2 + (d*x)/2) + a*b^4*cos(c/2 + (d*x)/2) + 2*a^4*b*s
in(c/2 + (d*x)/2) - 2*a^3*b^2*cos(c/2 + (d*x)/2) - 4*a^2*b^3*sin(c/2 + (d*x)/2)))*(b^6 - a^6 - 3*a^2*b^4 + 3*a
^4*b^2)^(1/2))/(b^6*d)