\(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [410]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{16 a}-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d} \]

[Out]

-1/16*x/a-1/3*cos(d*x+c)^3/a/d+1/5*cos(d*x+c)^5/a/d-1/16*cos(d*x+c)*sin(d*x+c)/a/d+1/8*cos(d*x+c)^3*sin(d*x+c)
/a/d+1/6*cos(d*x+c)^3*sin(d*x+c)^3/a/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2918, 2645, 14, 2648, 2715, 8} \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{8 a d}-\frac {\sin (c+d x) \cos (c+d x)}{16 a d}-\frac {x}{16 a} \]

[In]

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

[Out]

-1/16*x/a - Cos[c + d*x]^3/(3*a*d) + Cos[c + d*x]^5/(5*a*d) - (Cos[c + d*x]*Sin[c + d*x])/(16*a*d) + (Cos[c +
d*x]^3*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin[c + d*x]^3)/(6*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2648

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^2(c+d x) \sin ^3(c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{a} \\ & = \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}-\frac {\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{2 a}-\frac {\text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a d} \\ & = \frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}-\frac {\int \cos ^2(c+d x) \, dx}{8 a}-\frac {\text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}-\frac {\int 1 \, dx}{16 a} \\ & = -\frac {x}{16 a}-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(377\) vs. \(2(117)=234\).

Time = 3.61 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.22 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 (3 c-4 d x) \cos \left (\frac {c}{2}\right )-120 \cos \left (\frac {c}{2}+d x\right )-120 \cos \left (\frac {3 c}{2}+d x\right )+15 \cos \left (\frac {3 c}{2}+2 d x\right )-15 \cos \left (\frac {5 c}{2}+2 d x\right )-20 \cos \left (\frac {5 c}{2}+3 d x\right )-20 \cos \left (\frac {7 c}{2}+3 d x\right )+15 \cos \left (\frac {7 c}{2}+4 d x\right )-15 \cos \left (\frac {9 c}{2}+4 d x\right )+12 \cos \left (\frac {9 c}{2}+5 d x\right )+12 \cos \left (\frac {11 c}{2}+5 d x\right )-5 \cos \left (\frac {11 c}{2}+6 d x\right )+5 \cos \left (\frac {13 c}{2}+6 d x\right )-180 \sin \left (\frac {c}{2}\right )+90 c \sin \left (\frac {c}{2}\right )-120 d x \sin \left (\frac {c}{2}\right )+120 \sin \left (\frac {c}{2}+d x\right )-120 \sin \left (\frac {3 c}{2}+d x\right )+15 \sin \left (\frac {3 c}{2}+2 d x\right )+15 \sin \left (\frac {5 c}{2}+2 d x\right )+20 \sin \left (\frac {5 c}{2}+3 d x\right )-20 \sin \left (\frac {7 c}{2}+3 d x\right )+15 \sin \left (\frac {7 c}{2}+4 d x\right )+15 \sin \left (\frac {9 c}{2}+4 d x\right )-12 \sin \left (\frac {9 c}{2}+5 d x\right )+12 \sin \left (\frac {11 c}{2}+5 d x\right )-5 \sin \left (\frac {11 c}{2}+6 d x\right )-5 \sin \left (\frac {13 c}{2}+6 d x\right )}{1920 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

[In]

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

[Out]

(30*(3*c - 4*d*x)*Cos[c/2] - 120*Cos[c/2 + d*x] - 120*Cos[(3*c)/2 + d*x] + 15*Cos[(3*c)/2 + 2*d*x] - 15*Cos[(5
*c)/2 + 2*d*x] - 20*Cos[(5*c)/2 + 3*d*x] - 20*Cos[(7*c)/2 + 3*d*x] + 15*Cos[(7*c)/2 + 4*d*x] - 15*Cos[(9*c)/2
+ 4*d*x] + 12*Cos[(9*c)/2 + 5*d*x] + 12*Cos[(11*c)/2 + 5*d*x] - 5*Cos[(11*c)/2 + 6*d*x] + 5*Cos[(13*c)/2 + 6*d
*x] - 180*Sin[c/2] + 90*c*Sin[c/2] - 120*d*x*Sin[c/2] + 120*Sin[c/2 + d*x] - 120*Sin[(3*c)/2 + d*x] + 15*Sin[(
3*c)/2 + 2*d*x] + 15*Sin[(5*c)/2 + 2*d*x] + 20*Sin[(5*c)/2 + 3*d*x] - 20*Sin[(7*c)/2 + 3*d*x] + 15*Sin[(7*c)/2
+ 4*d*x] + 15*Sin[(9*c)/2 + 4*d*x] - 12*Sin[(9*c)/2 + 5*d*x] + 12*Sin[(11*c)/2 + 5*d*x] - 5*Sin[(11*c)/2 + 6*
d*x] - 5*Sin[(13*c)/2 + 6*d*x])/(1920*a*d*(Cos[c/2] + Sin[c/2]))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {-60 d x -20 \cos \left (3 d x +3 c \right )-5 \sin \left (6 d x +6 c \right )+12 \cos \left (5 d x +5 c \right )+15 \sin \left (4 d x +4 c \right )+15 \sin \left (2 d x +2 c \right )-120 \cos \left (d x +c \right )-128}{960 d a}\)	\(78\)
risch	\(-\frac {x}{16 a}-\frac {\cos \left (d x +c \right )}{8 a d}-\frac {\sin \left (6 d x +6 c \right )}{192 d a}+\frac {\cos \left (5 d x +5 c \right )}{80 a d}+\frac {\sin \left (4 d x +4 c \right )}{64 d a}-\frac {\cos \left (3 d x +3 c \right )}{48 a d}+\frac {\sin \left (2 d x +2 c \right )}{64 d a}\)	\(107\)
derivativedivides	\(\frac {\frac {16 \left (-\frac {\left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {19 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128}-\frac {1}{60}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d a}\)	\(155\)
default	\(\frac {\frac {16 \left (-\frac {\left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {19 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128}-\frac {1}{60}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d a}\)	\(155\)
norman	\(\frac {-\frac {21 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {17}{120 a d}-\frac {7 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {x}{16 a}-\frac {53 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {x \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {3 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 d a}+\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {41 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a}+\frac {7 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {19 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d a}-\frac {55 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {181 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d a}-\frac {7 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {223 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d a}+\frac {8 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(562\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/960*(-60*d*x-20*cos(3*d*x+3*c)-5*sin(6*d*x+6*c)+12*cos(5*d*x+5*c)+15*sin(4*d*x+4*c)+15*sin(2*d*x+2*c)-120*co
s(d*x+c)-128)/d/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {48 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, d x - 5 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 14 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, a d} \]

[In]

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

[Out]

1/240*(48*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 15*d*x - 5*(8*cos(d*x + c)^5 - 14*cos(d*x + c)^3 + 3*cos(d*x +
c))*sin(d*x + c))/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2067 vs. \(2 (92) = 184\).

Time = 19.66 (sec) , antiderivative size = 2067, normalized size of antiderivative = 17.67 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-15*d*x*tan(c/2 + d*x/2)**12/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a
*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/
2)**2 + 240*a*d) - 90*d*x*tan(c/2 + d*x/2)**10/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 +
3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2
+ d*x/2)**2 + 240*a*d) - 225*d*x*tan(c/2 + d*x/2)**8/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2
)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*
tan(c/2 + d*x/2)**2 + 240*a*d) - 300*d*x*tan(c/2 + d*x/2)**6/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2
+ d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 14
40*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 225*d*x*tan(c/2 + d*x/2)**4/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*t
an(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)*
*4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 90*d*x*tan(c/2 + d*x/2)**2/(240*a*d*tan(c/2 + d*x/2)**12 + 1440
*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d
*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 15*d*x/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2 +
d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 144
0*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 30*tan(c/2 + d*x/2)**11/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan(c/
2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 +
1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 170*tan(c/2 + d*x/2)**9/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*tan
(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4
+ 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 960*tan(c/2 + d*x/2)**8/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*
tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)
**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) + 1140*tan(c/2 + d*x/2)**7/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*
a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*
x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 640*tan(c/2 + d*x/2)**6/(240*a*d*tan(c/2 + d*x/2)**12 + 14
40*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 +
d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 1140*tan(c/2 + d*x/2)**5/(240*a*d*tan(c/2 + d*x/2)**12
+ 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c
/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) + 170*tan(c/2 + d*x/2)**3/(240*a*d*tan(c/2 + d*x/2)**
12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*ta
n(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 384*tan(c/2 + d*x/2)**2/(240*a*d*tan(c/2 + d*x/2
)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d
*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) + 30*tan(c/2 + d*x/2)/(240*a*d*tan(c/2 + d*x/2)
**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*
tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 64/(240*a*d*tan(c/2 + d*x/2)**12 + 1440*a*d*ta
n(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**
4 + 1440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**4/(a*sin(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (105) = 210\).

Time = 0.32 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.90 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {192 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {85 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {570 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {320 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {570 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {480 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {85 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {15 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - 32}{a + \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{120 \, d} \]

[In]

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

[Out]

1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 192*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 85*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 570*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 320*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 570*sin(
d*x + c)^7/(cos(d*x + c) + 1)^7 - 480*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 85*sin(d*x + c)^9/(cos(d*x + c) +
1)^9 - 15*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 32)/(a + 6*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a*sin(
d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 + 6*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 15*arctan(sin
(d*x + c)/(cos(d*x + c) + 1))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 570 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 570 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 192 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 32\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a}}{240 \, d} \]

[In]

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

[Out]

-1/240*(15*(d*x + c)/a + 2*(15*tan(1/2*d*x + 1/2*c)^11 + 85*tan(1/2*d*x + 1/2*c)^9 + 480*tan(1/2*d*x + 1/2*c)^
8 - 570*tan(1/2*d*x + 1/2*c)^7 + 320*tan(1/2*d*x + 1/2*c)^6 + 570*tan(1/2*d*x + 1/2*c)^5 - 85*tan(1/2*d*x + 1/
2*c)^3 + 192*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 32)/((tan(1/2*d*x + 1/2*c)^2 + 1)^6*a))/d

Mupad [B] (verification not implemented)

Time = 12.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{16\,a}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {4}{15}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

[In]

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

[Out]

- x/(16*a) - ((8*tan(c/2 + (d*x)/2)^2)/5 - tan(c/2 + (d*x)/2)/8 - (17*tan(c/2 + (d*x)/2)^3)/24 + (19*tan(c/2 +
(d*x)/2)^5)/4 + (8*tan(c/2 + (d*x)/2)^6)/3 - (19*tan(c/2 + (d*x)/2)^7)/4 + 4*tan(c/2 + (d*x)/2)^8 + (17*tan(c
/2 + (d*x)/2)^9)/24 + tan(c/2 + (d*x)/2)^11/8 + 4/15)/(a*d*(tan(c/2 + (d*x)/2)^2 + 1)^6)