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

Optimal. Leaf size=102 \[ -\frac{\cos ^5(c+d x)}{5 a^2 d}+\frac{\cos ^3(c+d x)}{a^2 d}-\frac{2 \cos (c+d x)}{a^2 d}+\frac{\sin ^3(c+d x) \cos (c+d x)}{2 a^2 d}+\frac{3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac{3 x}{4 a^2} \]

[Out]

(-3*x)/(4*a^2) - (2*Cos[c + d*x])/(a^2*d) + Cos[c + d*x]^3/(a^2*d) - Cos[c + d*x]^5/(5*a^2*d) + (3*Cos[c + d*x
]*Sin[c + d*x])/(4*a^2*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(2*a^2*d)

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Rubi [A]  time = 0.199063, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2869, 2757, 2633, 2635, 8} \[ -\frac{\cos ^5(c+d x)}{5 a^2 d}+\frac{\cos ^3(c+d x)}{a^2 d}-\frac{2 \cos (c+d x)}{a^2 d}+\frac{\sin ^3(c+d x) \cos (c+d x)}{2 a^2 d}+\frac{3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac{3 x}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(4*a^2) - (2*Cos[c + d*x])/(a^2*d) + Cos[c + d*x]^3/(a^2*d) - Cos[c + d*x]^5/(5*a^2*d) + (3*Cos[c + d*x
]*Sin[c + d*x])/(4*a^2*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(2*a^2*d)

Rule 2869

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

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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 8

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

Rubi steps

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

Mathematica [B]  time = 1.41871, size = 308, normalized size = 3.02 \[ -\frac{120 d x \sin \left (\frac{c}{2}\right )-110 \sin \left (\frac{c}{2}+d x\right )+110 \sin \left (\frac{3 c}{2}+d x\right )-40 \sin \left (\frac{3 c}{2}+2 d x\right )-40 \sin \left (\frac{5 c}{2}+2 d x\right )+15 \sin \left (\frac{5 c}{2}+3 d x\right )-15 \sin \left (\frac{7 c}{2}+3 d x\right )+5 \sin \left (\frac{7 c}{2}+4 d x\right )+5 \sin \left (\frac{9 c}{2}+4 d x\right )-\sin \left (\frac{9 c}{2}+5 d x\right )+\sin \left (\frac{11 c}{2}+5 d x\right )+5 \cos \left (\frac{c}{2}\right ) (24 d x+1)+110 \cos \left (\frac{c}{2}+d x\right )+110 \cos \left (\frac{3 c}{2}+d x\right )-40 \cos \left (\frac{3 c}{2}+2 d x\right )+40 \cos \left (\frac{5 c}{2}+2 d x\right )-15 \cos \left (\frac{5 c}{2}+3 d x\right )-15 \cos \left (\frac{7 c}{2}+3 d x\right )+5 \cos \left (\frac{7 c}{2}+4 d x\right )-5 \cos \left (\frac{9 c}{2}+4 d x\right )+\cos \left (\frac{9 c}{2}+5 d x\right )+\cos \left (\frac{11 c}{2}+5 d x\right )-5 \sin \left (\frac{c}{2}\right )}{160 a^2 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*(1 + 24*d*x)*Cos[c/2] + 110*Cos[c/2 + d*x] + 110*Cos[(3*c)/2 + d*x] - 40*Cos[(3*c)/2 + 2*d*x] + 40*Cos[(5*
c)/2 + 2*d*x] - 15*Cos[(5*c)/2 + 3*d*x] - 15*Cos[(7*c)/2 + 3*d*x] + 5*Cos[(7*c)/2 + 4*d*x] - 5*Cos[(9*c)/2 + 4
*d*x] + Cos[(9*c)/2 + 5*d*x] + Cos[(11*c)/2 + 5*d*x] - 5*Sin[c/2] + 120*d*x*Sin[c/2] - 110*Sin[c/2 + d*x] + 11
0*Sin[(3*c)/2 + d*x] - 40*Sin[(3*c)/2 + 2*d*x] - 40*Sin[(5*c)/2 + 2*d*x] + 15*Sin[(5*c)/2 + 3*d*x] - 15*Sin[(7
*c)/2 + 3*d*x] + 5*Sin[(7*c)/2 + 4*d*x] + 5*Sin[(9*c)/2 + 4*d*x] - Sin[(9*c)/2 + 5*d*x] + Sin[(11*c)/2 + 5*d*x
])/(160*a^2*d*(Cos[c/2] + Sin[c/2]))

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Maple [B]  time = 0.099, size = 279, normalized size = 2.7 \begin{align*} -{\frac{3}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-7\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}-20\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}+7\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}-12\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}+{\frac{3}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-{\frac{12}{5\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-{\frac{3}{2\,d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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))^2,x)

[Out]

-3/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9-7/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*
c)^7-4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6-20/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1
/2*c)^4+7/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3-12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*
x+1/2*c)^2+3/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)-12/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5-3/2/d
/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.72114, size = 392, normalized size = 3.84 \begin{align*} \frac{\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{200 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{40 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{70 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 24}{a^{2} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{10 \, d} \end{align*}

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))^2,x, algorithm="maxima")

[Out]

1/10*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 70*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 200*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 40*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 70*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 24)/(a^2 + 5*a^2*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a
^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) - 15*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.12741, size = 181, normalized size = 1.77 \begin{align*} -\frac{4 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 15 \, d x + 5 \,{\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 40 \, \cos \left (d x + c\right )}{20 \, a^{2} d} \end{align*}

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))^2,x, algorithm="fricas")

[Out]

-1/20*(4*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 15*d*x + 5*(2*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 40
*cos(d*x + c))/(a^2*d)

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Sympy [A]  time = 127.725, size = 1836, normalized size = 18. \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)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-315*d*x*tan(c/2 + d*x/2)**10/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 +
4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2
*d) - 1575*d*x*tan(c/2 + d*x/2)**8/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a
**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) -
3150*d*x*tan(c/2 + d*x/2)**6/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*
tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 3150*d
*x*tan(c/2 + d*x/2)**4/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/
2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 1575*d*x*tan
(c/2 + d*x/2)**2/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*
x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 315*d*x/(420*a**2*
d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c
/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) + 355*tan(c/2 + d*x/2)**10/(420*a**2*d*tan(c/2
+ d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2
)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 630*tan(c/2 + d*x/2)**9/(420*a**2*d*tan(c/2 + d*x/2)**1
0 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100
*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) + 1775*tan(c/2 + d*x/2)**8/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a
**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*ta
n(c/2 + d*x/2)**2 + 420*a**2*d) - 2940*tan(c/2 + d*x/2)**7/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(
c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*
x/2)**2 + 420*a**2*d) + 1870*tan(c/2 + d*x/2)**6/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/
2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 +
420*a**2*d) - 4850*tan(c/2 + d*x/2)**4/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 42
00*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d
) + 2940*tan(c/2 + d*x/2)**3/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*
tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 3265*t
an(c/2 + d*x/2)**2/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 +
d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) + 630*tan(c/2 + d*
x/2)/(420*a**2*d*tan(c/2 + d*x/2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 42
00*a**2*d*tan(c/2 + d*x/2)**4 + 2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d) - 653/(420*a**2*d*tan(c/2 + d*x/
2)**10 + 2100*a**2*d*tan(c/2 + d*x/2)**8 + 4200*a**2*d*tan(c/2 + d*x/2)**6 + 4200*a**2*d*tan(c/2 + d*x/2)**4 +
2100*a**2*d*tan(c/2 + d*x/2)**2 + 420*a**2*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**4/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.35405, size = 171, normalized size = 1.68 \begin{align*} -\frac{\frac{15 \,{\left (d x + c\right )}}{a^{2}} + \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 70 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 200 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 70 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, \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 ) + 24\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{20 \, d} \end{align*}

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))^2,x, algorithm="giac")

[Out]

-1/20*(15*(d*x + c)/a^2 + 2*(15*tan(1/2*d*x + 1/2*c)^9 + 70*tan(1/2*d*x + 1/2*c)^7 + 40*tan(1/2*d*x + 1/2*c)^6
+ 200*tan(1/2*d*x + 1/2*c)^4 - 70*tan(1/2*d*x + 1/2*c)^3 + 120*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*
c) + 24)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^2))/d