3.432 \(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)
Optimal. Leaf size=109 \[ -\frac {\cos ^3(c+d x)}{a^3 d}+\frac {7 \cos (c+d x)}{a^3 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^3 d}-\frac {19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac {51 x}{8 a^3} \]
[Out]
51/8*x/a^3+7*cos(d*x+c)/a^3/d-cos(d*x+c)^3/a^3/d-19/8*cos(d*x+c)*sin(d*x+c)/a^3/d-1/4*cos(d*x+c)*sin(d*x+c)^3/
a^3/d+4*cos(d*x+c)/a^3/d/(1+sin(d*x+c))
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Rubi [A] time = 0.26, antiderivative size = 109, normalized size of antiderivative = 1.00,
number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used
= {2875, 2872, 2638, 2635, 8, 2633, 2648} \[ -\frac {\cos ^3(c+d x)}{a^3 d}+\frac {7 \cos (c+d x)}{a^3 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^3 d}-\frac {19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac {51 x}{8 a^3} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
[Out]
(51*x)/(8*a^3) + (7*Cos[c + d*x])/(a^3*d) - Cos[c + d*x]^3/(a^3*d) - (19*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d)
- (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^3*d) + (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2633
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Rule 2648
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 2872
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Rule 2875
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \sin (c+d x) (a-a \sin (c+d x))^3 \tan ^2(c+d x) \, dx}{a^6}\\ &=\frac {\int \left (4 a-4 a \sin (c+d x)+4 a \sin ^2(c+d x)-3 a \sin ^3(c+d x)+a \sin ^4(c+d x)-\frac {4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac {4 x}{a^3}+\frac {\int \sin ^4(c+d x) \, dx}{a^3}-\frac {3 \int \sin ^3(c+d x) \, dx}{a^3}-\frac {4 \int \sin (c+d x) \, dx}{a^3}+\frac {4 \int \sin ^2(c+d x) \, dx}{a^3}-\frac {4 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac {4 x}{a^3}+\frac {4 \cos (c+d x)}{a^3 d}-\frac {2 \cos (c+d x) \sin (c+d x)}{a^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {3 \int \sin ^2(c+d x) \, dx}{4 a^3}+\frac {2 \int 1 \, dx}{a^3}+\frac {3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=\frac {6 x}{a^3}+\frac {7 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{a^3 d}-\frac {19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {3 \int 1 \, dx}{8 a^3}\\ &=\frac {51 x}{8 a^3}+\frac {7 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{a^3 d}-\frac {19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.34, size = 195, normalized size = 1.79 \[ \frac {2040 d x \sin \left (c+\frac {d x}{2}\right )+800 \sin \left (2 c+\frac {3 d x}{2}\right )-160 \sin \left (2 c+\frac {5 d x}{2}\right )-35 \sin \left (4 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {9 d x}{2}\right )+997 \cos \left (c+\frac {d x}{2}\right )+800 \cos \left (c+\frac {3 d x}{2}\right )+160 \cos \left (3 c+\frac {5 d x}{2}\right )-35 \cos \left (3 c+\frac {7 d x}{2}\right )-5 \cos \left (5 c+\frac {9 d x}{2}\right )-3563 \sin \left (\frac {d x}{2}\right )+2040 d x \cos \left (\frac {d x}{2}\right )}{320 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
[Out]
(2040*d*x*Cos[(d*x)/2] + 997*Cos[c + (d*x)/2] + 800*Cos[c + (3*d*x)/2] + 160*Cos[3*c + (5*d*x)/2] - 35*Cos[3*c
+ (7*d*x)/2] - 5*Cos[5*c + (9*d*x)/2] - 3563*Sin[(d*x)/2] + 2040*d*x*Sin[c + (d*x)/2] + 800*Sin[2*c + (3*d*x)
/2] - 160*Sin[2*c + (5*d*x)/2] - 35*Sin[4*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*x)/2])/(320*a^3*d*(Cos[c/2] + Sin[
c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
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fricas [A] time = 0.44, size = 144, normalized size = 1.32 \[ -\frac {2 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{4} - 15 \, \cos \left (d x + c\right )^{3} - 51 \, d x - {\left (51 \, d x + 67\right )} \cos \left (d x + c\right ) - 56 \, \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{3} + 51 \, d x - 21 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right ) - 32}{8 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
[Out]
-1/8*(2*cos(d*x + c)^5 + 8*cos(d*x + c)^4 - 15*cos(d*x + c)^3 - 51*d*x - (51*d*x + 67)*cos(d*x + c) - 56*cos(d
*x + c)^2 - (2*cos(d*x + c)^4 - 6*cos(d*x + c)^3 + 51*d*x - 21*cos(d*x + c)^2 + 35*cos(d*x + c) - 32)*sin(d*x
+ c) - 32)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)
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giac [A] time = 0.24, size = 145, normalized size = 1.33 \[ \frac {\frac {51 \, {\left (d x + c\right )}}{a^{3}} + \frac {64}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} + \frac {2 \, {\left (19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 32 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 144 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")
[Out]
1/8*(51*(d*x + c)/a^3 + 64/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(19*tan(1/2*d*x + 1/2*c)^7 + 32*tan(1/2*d*x +
1/2*c)^6 + 27*tan(1/2*d*x + 1/2*c)^5 + 144*tan(1/2*d*x + 1/2*c)^4 - 27*tan(1/2*d*x + 1/2*c)^3 + 160*tan(1/2*d*
x + 1/2*c)^2 - 19*tan(1/2*d*x + 1/2*c) + 48)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d
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maple [B] time = 0.42, size = 300, normalized size = 2.75 \[ \frac {19 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {27 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {36 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {27 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {40 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {12}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {51 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}+\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
[Out]
19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^6+27/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5+36/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*
x+1/2*c)^4-27/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+40/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(
1/2*d*x+1/2*c)^2-19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+12/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4+
51/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))+8/d/a^3/(tan(1/2*d*x+1/2*c)+1)
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maxima [B] time = 0.43, size = 398, normalized size = 3.65 \[ \frac {\frac {\frac {29 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {269 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {133 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {309 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {171 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {187 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {51 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {51 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 80}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {51 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
[Out]
1/4*((29*sin(d*x + c)/(cos(d*x + c) + 1) + 269*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 133*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 309*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 171*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 187*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 51*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 51*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 + 80)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*a^3*sin(d*x + c)^5/(cos(d*x + c)
+ 1)^5 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x
+ c)^8/(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 51*arctan(sin(d*x + c)/(cos(d*x + c)
+ 1))/a^3)/d
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mupad [B] time = 12.35, size = 146, normalized size = 1.34 \[ \frac {51\,x}{8\,a^3}+\frac {\frac {51\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+\frac {51\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {187\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+\frac {171\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {309\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {133\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {269\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {29\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+20}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)
[Out]
(51*x)/(8*a^3) + ((29*tan(c/2 + (d*x)/2))/4 + (269*tan(c/2 + (d*x)/2)^2)/4 + (133*tan(c/2 + (d*x)/2)^3)/4 + (3
09*tan(c/2 + (d*x)/2)^4)/4 + (171*tan(c/2 + (d*x)/2)^5)/4 + (187*tan(c/2 + (d*x)/2)^6)/4 + (51*tan(c/2 + (d*x)
/2)^7)/4 + (51*tan(c/2 + (d*x)/2)^8)/4 + 20)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^4)
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sympy [A] time = 147.46, size = 3578, normalized size = 32.83 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
[Out]
Piecewise((51*d*x*tan(c/2 + d*x/2)**9/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d
*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x
/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d)
+ 51*d*x*tan(c/2 + d*x/2)**8/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2
+ d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 +
32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 204*d
*x*tan(c/2 + d*x/2)**7/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/
2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**
3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 204*d*x*tan(
c/2 + d*x/2)**6/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 +
32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan
(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 306*d*x*tan(c/2 + d
*x/2)**5/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**
3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 +
d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 306*d*x*tan(c/2 + d*x/2)**
4/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan
(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)*
*3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 204*d*x*tan(c/2 + d*x/2)**3/(8*a*
*3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 +
d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32
*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 204*d*x*tan(c/2 + d*x/2)**2/(8*a**3*d*ta
n(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)*
*6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d
*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 51*d*x*tan(c/2 + d*x/2)/(8*a**3*d*tan(c/2 + d*x
/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**
3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 +
d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 51*d*x/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 +
d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48
*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2
+ d*x/2) + 8*a**3*d) + 102*tan(c/2 + d*x/2)**8/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 +
32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan
(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) +
8*a**3*d) + 102*tan(c/2 + d*x/2)**7/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*
tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/
2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d)
+ 374*tan(c/2 + d*x/2)**6/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d
*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*
a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 342*tan(c
/2 + d*x/2)**5/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 +
32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(
c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 618*tan(c/2 + d*x/2)
**4/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*t
an(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2
)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 266*tan(c/2 + d*x/2)**3/(8*a**3
*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*
x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a
**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 538*tan(c/2 + d*x/2)**2/(8*a**3*d*tan(c/2
+ d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 4
8*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c
/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 58*tan(c/2 + d*x/2)/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8
*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/
2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2
+ 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 160/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 3
2*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c
/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8
*a**3*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**4/(a*sin(c) + a)**3, True))
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