3.1136 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=284 \[ -\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (-19 a^2 b^2+20 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d \sqrt{a^2-b^2}}+\frac{\left (6 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}-\frac{\left (20 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{6 a^2 b^3 d}+\frac{\left (5 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{a b^4 d}+\frac{a x \left (9-\frac{20 a^2}{b^2}\right )}{2 b^4} \]

[Out]

(a*(9 - (20*a^2)/b^2)*x)/(2*b^4) + ((20*a^4 - 19*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/(b^6*Sqrt[a^2 - b^2]*d) - ((60*a^2 - 17*b^2)*Cos[c + d*x])/(6*b^5*d) + ((5*a^2 - b^2)*Cos[c + d*x]*Sin[c
+ d*x])/(a*b^4*d) - ((20*a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(6*a^2*b^3*d) - ((a^2 - b^2)*Cos[c + d*x]*
Sin[c + d*x]^3)/(2*a*b^2*d*(a + b*Sin[c + d*x])^2) + ((6*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(2*a^2*b^2*d*
(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.738213, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2891, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (-19 a^2 b^2+20 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d \sqrt{a^2-b^2}}+\frac{\left (6 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}-\frac{\left (20 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{6 a^2 b^3 d}+\frac{\left (5 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{a b^4 d}+\frac{a x \left (9-\frac{20 a^2}{b^2}\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(a*(9 - (20*a^2)/b^2)*x)/(2*b^4) + ((20*a^4 - 19*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/(b^6*Sqrt[a^2 - b^2]*d) - ((60*a^2 - 17*b^2)*Cos[c + d*x])/(6*b^5*d) + ((5*a^2 - b^2)*Cos[c + d*x]*Sin[c
+ d*x])/(a*b^4*d) - ((20*a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(6*a^2*b^3*d) - ((a^2 - b^2)*Cos[c + d*x]*
Sin[c + d*x]^3)/(2*a*b^2*d*(a + b*Sin[c + d*x])^2) + ((6*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(2*a^2*b^2*d*
(a + b*Sin[c + d*x]))

Rule 2891

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

Rule 3049

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])))

Rule 3023

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 2735

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 2660

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 618

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 204

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

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\int \frac{\sin ^2(c+d x) \left (15 a^2-2 b^2-a b \sin (c+d x)-\left (20 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 b^2}\\ &=-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\int \frac{\sin (c+d x) \left (-2 a \left (20 a^2-3 b^2\right )+5 a^2 b \sin (c+d x)+12 a \left (5 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 a^2 b^3}\\ &=\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\int \frac{12 a^2 \left (5 a^2-b^2\right )-20 a^3 b \sin (c+d x)-2 a^2 \left (60 a^2-17 b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{12 a^2 b^4}\\ &=-\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\int \frac{12 a^2 b \left (5 a^2-b^2\right )+6 a^3 \left (20 a^2-9 b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{12 a^2 b^5}\\ &=-\frac{a \left (20 a^2-9 b^2\right ) x}{2 b^6}-\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac{\left (6 a^4 \left (20 a^2-9 b^2\right )-12 a^2 b^2 \left (5 a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{12 a^2 b^6}\\ &=-\frac{a \left (20 a^2-9 b^2\right ) x}{2 b^6}-\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac{\left (20 a^4-19 a^2 b^2+2 b^4\right ) \operatorname{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 (20 a^2-9 b^2\right ) x}{2 b^6}-\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{\left (2 \left (20 a^4-19 a^2 b^2+2 b^4\right )\right ) \operatorname{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 (20 a^2-9 b^2\right ) x}{2 b^6}+\frac{\left (20 a^4-19 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^6 \sqrt{a^2-b^2} d}-\frac{\left (60 a^2-17 b^2\right ) \cos (c+d x)}{6 b^5 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{a b^4 d}-\frac{\left (20 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{6 a^2 b^3 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac{\left (6 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.25442, size = 1030, normalized size = 3.63 \[ \frac{-\frac{12 \left (-48 a (c+d x)+\frac{6 \left (16 a^6-40 b^2 a^4+30 b^4 a^2-5 b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-16 b \cos (c+d x)+\frac{a b \left (-40 a^4+72 b^2 a^2-29 b^4\right ) \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac{b \left (8 a^4-8 b^2 a^2+b^4\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}\right )}{b^4}+12 \left (\frac{2 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{b \cos (c+d x) \left (4 a^2+3 b \sin (c+d x) a-b^2\right )}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))^2}\right )+\frac{6 \left (\frac{\cos (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)-b \left (2 a^2+b^2\right )\right )}{(a+b \sin (c+d x))^2}-\frac{6 b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}\right )}{(a-b)^2 (a+b)^2}-\frac{\frac{3840 (c+d x) a^9+3840 b \cos (c+d x) a^8+7680 b (c+d x) \sin (c+d x) a^8-6912 b^2 (c+d x) a^7-1920 b^2 (c+d x) \cos (2 (c+d x)) a^7+2880 b^2 \sin (2 (c+d x)) a^7-7872 b^3 \cos (c+d x) a^6-320 b^3 \cos (3 (c+d x)) a^6-17664 b^3 (c+d x) \sin (c+d x) a^6+1728 b^4 (c+d x) a^5+4416 b^4 (c+d x) \cos (2 (c+d x)) a^5-6304 b^4 \sin (2 (c+d x)) a^5+40 b^4 \sin (4 (c+d x)) a^5+4256 b^5 \cos (c+d x) a^4+696 b^5 \cos (3 (c+d x)) a^4+8 b^5 \cos (5 (c+d x)) a^4+12288 b^5 (c+d x) \sin (c+d x) a^4+1920 b^6 (c+d x) a^3-3072 b^6 (c+d x) \cos (2 (c+d x)) a^3+4022 b^6 \sin (2 (c+d x)) a^3-80 b^6 \sin (4 (c+d x)) a^3-172 b^7 \cos (c+d x) a^2-432 b^7 \cos (3 (c+d x)) a^2-16 b^7 \cos (5 (c+d x)) a^2-2304 b^7 (c+d x) \sin (c+d x) a^2-576 b^8 (c+d x) a+576 b^8 (c+d x) \cos (2 (c+d x)) a-607 b^8 \sin (2 (c+d x)) a+40 b^8 \sin (4 (c+d x)) a-70 b^9 \cos (c+d x)+56 b^9 \cos (3 (c+d x))+8 b^9 \cos (5 (c+d x))}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac{12 \left (640 a^8-1792 b^2 a^6+1680 b^4 a^4-560 b^6 a^2+35 b^8\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}}{b^6}}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

((-12*(-48*a*(c + d*x) + (6*(16*a^6 - 40*a^4*b^2 + 30*a^2*b^4 - 5*b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^
2 - b^2]])/(a^2 - b^2)^(5/2) - 16*b*Cos[c + d*x] + (b*(8*a^4 - 8*a^2*b^2 + b^4)*Cos[c + d*x])/((a - b)*(a + b)
*(a + b*Sin[c + d*x])^2) + (a*b*(-40*a^4 + 72*a^2*b^2 - 29*b^4)*Cos[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[
c + d*x]))))/b^4 + 12*((2*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) +
(b*Cos[c + d*x]*(4*a^2 - b^2 + 3*a*b*Sin[c + d*x]))/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])^2)) + (6*((-6*b^
2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (Cos[c + d*x]*(-(b*(2*a^2 + b^2)) + a*(2
*a^2 - 5*b^2)*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2))/((a - b)^2*(a + b)^2) - ((-12*(640*a^8 - 1792*a^6*b^2 +
1680*a^4*b^4 - 560*a^2*b^6 + 35*b^8)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (38
40*a^9*(c + d*x) - 6912*a^7*b^2*(c + d*x) + 1728*a^5*b^4*(c + d*x) + 1920*a^3*b^6*(c + d*x) - 576*a*b^8*(c + d
*x) + 3840*a^8*b*Cos[c + d*x] - 7872*a^6*b^3*Cos[c + d*x] + 4256*a^4*b^5*Cos[c + d*x] - 172*a^2*b^7*Cos[c + d*
x] - 70*b^9*Cos[c + d*x] - 1920*a^7*b^2*(c + d*x)*Cos[2*(c + d*x)] + 4416*a^5*b^4*(c + d*x)*Cos[2*(c + d*x)] -
3072*a^3*b^6*(c + d*x)*Cos[2*(c + d*x)] + 576*a*b^8*(c + d*x)*Cos[2*(c + d*x)] - 320*a^6*b^3*Cos[3*(c + d*x)]
+ 696*a^4*b^5*Cos[3*(c + d*x)] - 432*a^2*b^7*Cos[3*(c + d*x)] + 56*b^9*Cos[3*(c + d*x)] + 8*a^4*b^5*Cos[5*(c
+ d*x)] - 16*a^2*b^7*Cos[5*(c + d*x)] + 8*b^9*Cos[5*(c + d*x)] + 7680*a^8*b*(c + d*x)*Sin[c + d*x] - 17664*a^6
*b^3*(c + d*x)*Sin[c + d*x] + 12288*a^4*b^5*(c + d*x)*Sin[c + d*x] - 2304*a^2*b^7*(c + d*x)*Sin[c + d*x] + 288
0*a^7*b^2*Sin[2*(c + d*x)] - 6304*a^5*b^4*Sin[2*(c + d*x)] + 4022*a^3*b^6*Sin[2*(c + d*x)] - 607*a*b^8*Sin[2*(
c + d*x)] + 40*a^5*b^4*Sin[4*(c + d*x)] - 80*a^3*b^6*Sin[4*(c + d*x)] + 40*a*b^8*Sin[4*(c + d*x)])/((a^2 - b^2
)^2*(a + b*Sin[c + d*x])^2))/b^6)/(384*d)

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Maple [B]  time = 0.152, size = 880, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x)

[Out]

-3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5-12/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2
*c)^4*a^2+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4-24/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*
d*x+1/2*c)^2*a^2+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a*
tan(1/2*d*x+1/2*c)-12/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2+8/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3-20/d/b^6*arcta
n(tan(1/2*d*x+1/2*c))*a^3+9/d/b^4*a*arctan(tan(1/2*d*x+1/2*c))-7/d/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1
/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3*a^3+2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+
1/2*c)^3*a-8/d/b^5/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2*a^4-13/d/b^3/(tan(
1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2*a^2+6/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/
2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-25/d/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*
d*x+1/2*c)*a^3+10/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)*a-8/d/b^5/(tan(
1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^4+3/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2
*a^2+20/d/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))*a^4-19/d/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))*a^2+2/d/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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.55714, size = 2152, normalized size = 7.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/12*(4*(a^2*b^5 - b^7)*cos(d*x + c)^5 - 6*(20*a^5*b^2 - 29*a^3*b^4 + 9*a*b^6)*d*x*cos(d*x + c)^2 - 8*(5*a^4*
b^3 - 6*a^2*b^5 + b^7)*cos(d*x + c)^3 + 6*(20*a^7 - 9*a^5*b^2 - 20*a^3*b^4 + 9*a*b^6)*d*x + 3*(20*a^6 + a^4*b^
2 - 17*a^2*b^4 + 2*b^6 - (20*a^4*b^2 - 19*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(20*a^5*b - 19*a^3*b^3 + 2*a*b^5
)*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)) + 6*(20*a^6*b - 19*a^4*b^3 - 3*a^2*b^5 + 2*b^7)*cos(d*x + c) + 2*(5*(a^3*b^4 - a*b^6)*cos(d*x + c)^3 + 6*(
20*a^6*b - 29*a^4*b^3 + 9*a^2*b^5)*d*x + 3*(30*a^5*b^2 - 41*a^3*b^4 + 11*a*b^6)*cos(d*x + c))*sin(d*x + c))/((
a^2*b^8 - b^10)*d*cos(d*x + c)^2 - 2*(a^3*b^7 - a*b^9)*d*sin(d*x + c) - (a^4*b^6 - b^10)*d), 1/6*(2*(a^2*b^5 -
b^7)*cos(d*x + c)^5 - 3*(20*a^5*b^2 - 29*a^3*b^4 + 9*a*b^6)*d*x*cos(d*x + c)^2 - 4*(5*a^4*b^3 - 6*a^2*b^5 + b
^7)*cos(d*x + c)^3 + 3*(20*a^7 - 9*a^5*b^2 - 20*a^3*b^4 + 9*a*b^6)*d*x + 3*(20*a^6 + a^4*b^2 - 17*a^2*b^4 + 2*
b^6 - (20*a^4*b^2 - 19*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(20*a^5*b - 19*a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqr
t(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*(20*a^6*b - 19*a^4*b^3 - 3*a^2*b
^5 + 2*b^7)*cos(d*x + c) + (5*(a^3*b^4 - a*b^6)*cos(d*x + c)^3 + 6*(20*a^6*b - 29*a^4*b^3 + 9*a^2*b^5)*d*x + 3
*(30*a^5*b^2 - 41*a^3*b^4 + 11*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^8 - b^10)*d*cos(d*x + c)^2 - 2*(a^3*
b^7 - a*b^9)*d*sin(d*x + c) - (a^4*b^6 - b^10)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25778, size = 531, normalized size = 1.87 \begin{align*} -\frac{\frac{3 \,{\left (20 \, a^{3} - 9 \, a b^{2}\right )}{\left (d x + c\right )}}{b^{6}} - \frac{6 \,{\left (20 \, a^{4} - 19 \, a^{2} b^{2} + 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{6 \,{\left (7 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 13 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 25 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 10 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a^{4} - 3 \, a^{2} 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 )}^{2} b^{5}} + \frac{2 \,{\left (9 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 72 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, a^{2} - 8 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{5}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(3*(20*a^3 - 9*a*b^2)*(d*x + c)/b^6 - 6*(20*a^4 - 19*a^2*b^2 + 2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*s
gn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^6) + 6*(7*a^3*b*tan(1/2*d*x +
1/2*c)^3 - 2*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*a^4*tan(1/2*d*x + 1/2*c)^2 + 13*a^2*b^2*tan(1/2*d*x + 1/2*c)^2
- 6*b^4*tan(1/2*d*x + 1/2*c)^2 + 25*a^3*b*tan(1/2*d*x + 1/2*c) - 10*a*b^3*tan(1/2*d*x + 1/2*c) + 8*a^4 - 3*a^2
*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*b^5) + 2*(9*a*b*tan(1/2*d*x + 1/2*c)^5 + 36
*a^2*tan(1/2*d*x + 1/2*c)^4 - 12*b^2*tan(1/2*d*x + 1/2*c)^4 + 72*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d
*x + 1/2*c)^2 - 9*a*b*tan(1/2*d*x + 1/2*c) + 36*a^2 - 8*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^5))/d