3.1078 \(\int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=188 \[ \frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}+\frac{a x \left (4 a^2-b^2\right )}{b^5}-\frac{2 a \sin (c+d x) \cos (c+d x)}{b^3 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac{4 \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d} \]

[Out]

(a*(4*a^2 - b^2)*x)/b^5 - (2*a^2*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^5*Sqrt[a
^2 - b^2]*d) + ((12*a^2 - b^2)*Cos[c + d*x])/(3*b^4*d) - (2*a*Cos[c + d*x]*Sin[c + d*x])/(b^3*d) + (4*Cos[c +
d*x]*Sin[c + d*x]^2)/(3*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(b*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.741585, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2889, 3048, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}+\frac{a x \left (4 a^2-b^2\right )}{b^5}-\frac{2 a \sin (c+d x) \cos (c+d x)}{b^3 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac{4 \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*a^2 - b^2)*x)/b^5 - (2*a^2*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^5*Sqrt[a
^2 - b^2]*d) + ((12*a^2 - b^2)*Cos[c + d*x])/(3*b^4*d) - (2*a*Cos[c + d*x]*Sin[c + d*x])/(b^3*d) + (4*Cos[c +
d*x]*Sin[c + d*x]^2)/(3*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(b*d*(a + b*Sin[c + d*x]))

Rule 2889

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

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 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 ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac{\sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{\sin ^2(c+d x) \left (-3 \left (a^2-b^2\right )+4 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{\sin (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (a^2-b^2\right ) \sin (c+d x)-12 a \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{-12 a^2 \left (a^2-b^2\right )+4 a b \left (a^2-b^2\right ) \sin (c+d x)+2 \left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\int \frac{-12 a^2 b \left (a^2-b^2\right )-6 a \left (a^2-b^2\right ) \left (4 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=\frac{a \left (4 a^2-b^2\right ) x}{b^5}+\frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (a^2 \left (4 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^5}\\ &=\frac{a \left (4 a^2-b^2\right ) x}{b^5}+\frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (2 a^2 \left (4 a^2-3 b^2\right )\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^5 d}\\ &=\frac{a \left (4 a^2-b^2\right ) x}{b^5}+\frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}+\frac{\left (4 a^2 \left (4 a^2-3 b^2\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^5 d}\\ &=\frac{a \left (4 a^2-b^2\right ) x}{b^5}-\frac{2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^5 \sqrt{a^2-b^2} d}+\frac{\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac{2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac{4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.47963, size = 246, normalized size = 1.31 \[ \frac{\frac{24 a^2 b^2 \sin (2 (c+d x))+12 a b \left (8 a^2-b^2\right ) \cos (c+d x)-24 a^2 b^2 c-24 a^2 b^2 d x+96 a^3 b c \sin (c+d x)+96 a^3 b d x \sin (c+d x)+96 a^4 c+96 a^4 d x-24 a b^3 c \sin (c+d x)-24 a b^3 d x \sin (c+d x)+4 a b^3 \cos (3 (c+d x))-2 b^4 \sin (2 (c+d x))-b^4 \sin (4 (c+d x))}{a+b \sin (c+d x)}-\frac{48 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{24 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-48*a^2*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (96*a^4*c - 24*a
^2*b^2*c + 96*a^4*d*x - 24*a^2*b^2*d*x + 12*a*b*(8*a^2 - b^2)*Cos[c + d*x] + 4*a*b^3*Cos[3*(c + d*x)] + 96*a^3
*b*c*Sin[c + d*x] - 24*a*b^3*c*Sin[c + d*x] + 96*a^3*b*d*x*Sin[c + d*x] - 24*a*b^3*d*x*Sin[c + d*x] + 24*a^2*b
^2*Sin[2*(c + d*x)] - 2*b^4*Sin[2*(c + d*x)] - b^4*Sin[4*(c + d*x)])/(a + b*Sin[c + d*x]))/(24*b^5*d)

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Maple [B]  time = 0.125, size = 460, normalized size = 2.5 \begin{align*} 2\,{\frac{a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}{a}^{2}}{d{b}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+12\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{b}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{{a}^{2}}{d{b}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{2}{3\,d{b}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+8\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{3}}{d{b}^{5}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{3}}}+2\,{\frac{{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{{a}^{3}}{d{b}^{4} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-8\,{\frac{{a}^{4}}{d{b}^{5}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+6\,{\frac{{a}^{2}}{d{b}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^4*a^2-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+12/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2*tan(1/
2*d*x+1/2*c)^2-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-
2/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3+8/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^3-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))
*a+2/d*a^2/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d*a^3/b^4/(tan(1/2*d*x+1
/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-8/d*a^4/b^5/(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))+6/d*a^2/b^3/(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)^2*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86673, size = 1381, normalized size = 7.35 \begin{align*} \left [\frac{4 \,{\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \,{\left (4 \, a^{5} - 3 \, a^{3} b^{2} +{\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 6 \,{\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - 2 \,{\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \,{\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}, \frac{2 \,{\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \,{\left (4 \, a^{5} - 3 \, a^{3} b^{2} +{\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 3 \,{\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) -{\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \,{\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(4*(a^3*b^3 - a*b^5)*cos(d*x + c)^3 + 6*(4*a^6 - 5*a^4*b^2 + a^2*b^4)*d*x + 3*(4*a^5 - 3*a^3*b^2 + (4*a^4
*b - 3*a^2*b^3)*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*(4*a^5*b - 5*a^3*b^3 + a*b^5)*cos(d*x + c) - 2*((a^2*b^4 - b^6)*cos(d*x + c)^3 - 3*(4*a^
5*b - 5*a^3*b^3 + a*b^5)*d*x - 6*(a^4*b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^6 - b^8)*d*sin(d*x +
c) + (a^3*b^5 - a*b^7)*d), 1/3*(2*(a^3*b^3 - a*b^5)*cos(d*x + c)^3 + 3*(4*a^6 - 5*a^4*b^2 + a^2*b^4)*d*x + 3*(
4*a^5 - 3*a^3*b^2 + (4*a^4*b - 3*a^2*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2
- b^2)*cos(d*x + c))) + 3*(4*a^5*b - 5*a^3*b^3 + a*b^5)*cos(d*x + c) - ((a^2*b^4 - b^6)*cos(d*x + c)^3 - 3*(4
*a^5*b - 5*a^3*b^3 + a*b^5)*d*x - 6*(a^4*b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^6 - b^8)*d*sin(d*x
+ c) + (a^3*b^5 - a*b^7)*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)**2*sin(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.31987, size = 352, normalized size = 1.87 \begin{align*} \frac{\frac{3 \,{\left (4 \, a^{3} - a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} - \frac{6 \,{\left (4 \, a^{4} - 3 \, a^{2} b^{2}\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^{5}} + \frac{6 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}\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 )} b^{4}} + \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} - b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(4*a^3 - a*b^2)*(d*x + c)/b^5 - 6*(4*a^4 - 3*a^2*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan
((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^5) + 6*(a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(
(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*b^4) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1
/2*d*x + 1/2*c)^4 - 3*b^2*tan(1/2*d*x + 1/2*c)^4 + 18*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c)
+ 9*a^2 - b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4))/d