3.433 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)
Optimal. Leaf size=87 \[ \frac{\cos ^3(c+d x)}{3 a^3 d}-\frac{5 \cos (c+d x)}{a^3 d}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac{11 x}{2 a^3} \]
[Out]
(-11*x)/(2*a^3) - (5*Cos[c + d*x])/(a^3*d) + Cos[c + d*x]^3/(3*a^3*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d
) - (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))
________________________________________________________________________________________
Rubi [A] time = 0.226797, antiderivative size = 87, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used =
{2875, 2709, 2638, 2635, 8, 2633, 2648} \[ \frac{\cos ^3(c+d x)}{3 a^3 d}-\frac{5 \cos (c+d x)}{a^3 d}+\frac{3 \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac{11 x}{2 a^3} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
[Out]
(-11*x)/(2*a^3) - (5*Cos[c + d*x])/(a^3*d) + Cos[c + d*x]^3/(3*a^3*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d
) - (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))
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]
Rule 2709
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Rule 2638
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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 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 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]
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int (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)-3 a \sin ^2(c+d x)+a \sin ^3(c+d x)+\frac{4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=-\frac{4 x}{a^3}+\frac{\int \sin ^3(c+d x) \, dx}{a^3}-\frac{3 \int \sin ^2(c+d x) \, dx}{a^3}+\frac{4 \int \sin (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{3 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac{3 \int 1 \, dx}{2 a^3}-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=-\frac{11 x}{2 a^3}-\frac{5 \cos (c+d x)}{a^3 d}+\frac{\cos ^3(c+d x)}{3 a^3 d}+\frac{3 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}\\ \end{align*}
Mathematica [B] time = 1.11834, size = 181, normalized size = 2.08 \[ \frac{-660 d x \sin \left (c+\frac{d x}{2}\right )+\sin \left (c+\frac{d x}{2}\right )-240 \sin \left (2 c+\frac{3 d x}{2}\right )+40 \sin \left (2 c+\frac{5 d x}{2}\right )+5 \sin \left (4 c+\frac{7 d x}{2}\right )-286 \cos \left (c+\frac{d x}{2}\right )-240 \cos \left (c+\frac{3 d x}{2}\right )-40 \cos \left (3 c+\frac{5 d x}{2}\right )+5 \cos \left (3 c+\frac{7 d x}{2}\right )+1244 \sin \left (\frac{d x}{2}\right )+(1-660 d x) \cos \left (\frac{d x}{2}\right )}{120 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
[Out]
((1 - 660*d*x)*Cos[(d*x)/2] - 286*Cos[c + (d*x)/2] - 240*Cos[c + (3*d*x)/2] - 40*Cos[3*c + (5*d*x)/2] + 5*Cos[
3*c + (7*d*x)/2] + 1244*Sin[(d*x)/2] + Sin[c + (d*x)/2] - 660*d*x*Sin[c + (d*x)/2] - 240*Sin[2*c + (3*d*x)/2]
+ 40*Sin[2*c + (5*d*x)/2] + 5*Sin[4*c + (7*d*x)/2])/(120*a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]))
________________________________________________________________________________________
Maple [B] time = 0.106, size = 198, normalized size = 2.3 \begin{align*} -3\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-8\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-20\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+3\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{28}{3\,d{a}^{3}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-11\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-8\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \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+a*sin(d*x+c))^3,x)
[Out]
-3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5-8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)
^4-20/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2
*c)-28/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3-11/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)
________________________________________________________________________________________
Maxima [B] time = 1.6924, size = 421, normalized size = 4.84 \begin{align*} -\frac{\frac{\frac{19 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{123 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{96 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{33 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 52}{a^{3} + \frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac{33 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, 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+a*sin(d*x+c))^3,x, algorithm="maxima")
[Out]
-1/3*((19*sin(d*x + c)/(cos(d*x + c) + 1) + 123*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 96*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 33*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 33*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 52)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 3*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*a^3*sin
(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7) + 33*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
________________________________________________________________________________________
Fricas [A] time = 1.10882, size = 328, normalized size = 3.77 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{3} - 33 \, d x - 3 \,{\left (11 \, d x + 15\right )} \cos \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )^{2} +{\left (2 \, \cos \left (d x + c\right )^{3} - 33 \, d x + 9 \, \cos \left (d x + c\right )^{2} - 21 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right ) - 24}{6 \,{\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \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+a*sin(d*x+c))^3,x, algorithm="fricas")
[Out]
1/6*(2*cos(d*x + c)^4 - 7*cos(d*x + c)^3 - 33*d*x - 3*(11*d*x + 15)*cos(d*x + c) - 30*cos(d*x + c)^2 + (2*cos(
d*x + c)^3 - 33*d*x + 9*cos(d*x + c)^2 - 21*cos(d*x + c) + 24)*sin(d*x + c) - 24)/(a^3*d*cos(d*x + c) + a^3*d*
sin(d*x + c) + a^3*d)
________________________________________________________________________________________
Sympy [A] time = 130.34, size = 2412, normalized size = 27.72 \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+a*sin(d*x+c))**3,x)
[Out]
Piecewise((-1320*d*x*tan(c/2 + d*x/2)**7/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 72
0*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*ta
n(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 1320*d*x*tan(c/2 + d*x/2)**6/(240*a**3*d*tan(c
/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)
**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3
*d) - 3960*d*x*tan(c/2 + d*x/2)**5/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3
*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2
+ d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 3960*d*x*tan(c/2 + d*x/2)**4/(240*a**3*d*tan(c/2 + d
*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 +
720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) -
3960*d*x*tan(c/2 + d*x/2)**3/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan
(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/
2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 3960*d*x*tan(c/2 + d*x/2)**2/(240*a**3*d*tan(c/2 + d*x/2)*
*7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a*
*3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 1320*d
*x*tan(c/2 + d*x/2)/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*
x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 2
40*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 1320*d*x/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x
/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 72
0*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) + 1269*tan(c/2 + d*x/2)**7/(240*a**3*
d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 +
d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 2
40*a**3*d) - 1371*tan(c/2 + d*x/2)**6/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a
**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c
/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) + 1167*tan(c/2 + d*x/2)**5/(240*a**3*d*tan(c/2 + d*
x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 7
20*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 3
873*tan(c/2 + d*x/2)**4/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2
+ d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2
+ 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 993*tan(c/2 + d*x/2)**3/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a
**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c
/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 6033*tan(c/2 + d
*x/2)**2/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 7
20*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*t
an(c/2 + d*x/2) + 240*a**3*d) - 251*tan(c/2 + d*x/2)/(240*a**3*d*tan(c/2 + d*x/2)**7 + 240*a**3*d*tan(c/2 + d*
x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a**3*d*tan(c/2 + d*x/2)**3 + 7
20*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d) - 2891/(240*a**3*d*tan(c/2 + d*x/2)*
*7 + 240*a**3*d*tan(c/2 + d*x/2)**6 + 720*a**3*d*tan(c/2 + d*x/2)**5 + 720*a**3*d*tan(c/2 + d*x/2)**4 + 720*a*
*3*d*tan(c/2 + d*x/2)**3 + 720*a**3*d*tan(c/2 + d*x/2)**2 + 240*a**3*d*tan(c/2 + d*x/2) + 240*a**3*d), Ne(d, 0
)), (x*sin(c)**2*cos(c)**4/(a*sin(c) + a)**3, True))
________________________________________________________________________________________
Giac [A] time = 1.32942, size = 143, normalized size = 1.64 \begin{align*} -\frac{\frac{33 \,{\left (d x + c\right )}}{a^{3}} + \frac{48}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 60 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 28\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{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+a*sin(d*x+c))^3,x, algorithm="giac")
[Out]
-1/6*(33*(d*x + c)/a^3 + 48/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(9*tan(1/2*d*x + 1/2*c)^5 + 24*tan(1/2*d*x +
1/2*c)^4 + 60*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 28)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d