3.1302 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=235 \[ -\frac{\left (-20 a^2 b^2+15 a^4+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{2 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d}-\frac{\left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}+\frac{a \left (4 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac{a x \left (-12 a^2 b^2+8 a^4+3 b^4\right )}{8 b^6}+\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}-\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]

[Out]

-(a*(8*a^4 - 12*a^2*b^2 + 3*b^4)*x)/(8*b^6) + (2*a^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^
2 - b^2]])/(b^6*d) - ((15*a^4 - 20*a^2*b^2 + 3*b^4)*Cos[c + d*x])/(15*b^5*d) + (a*(4*a^2 - 5*b^2)*Cos[c + d*x]
*Sin[c + d*x])/(8*b^4*d) - ((5*a^2 - 6*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(15*b^3*d) + (a*Cos[c + d*x]*Sin[c +
d*x]^3)/(4*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^4)/(5*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.716639, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2895, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac{\left (-20 a^2 b^2+15 a^4+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{2 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d}-\frac{\left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}+\frac{a \left (4 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac{a x \left (-12 a^2 b^2+8 a^4+3 b^4\right )}{8 b^6}+\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}-\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(8*a^4 - 12*a^2*b^2 + 3*b^4)*x)/(8*b^6) + (2*a^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^
2 - b^2]])/(b^6*d) - ((15*a^4 - 20*a^2*b^2 + 3*b^4)*Cos[c + d*x])/(15*b^5*d) + (a*(4*a^2 - 5*b^2)*Cos[c + d*x]
*Sin[c + d*x])/(8*b^4*d) - ((5*a^2 - 6*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(15*b^3*d) + (a*Cos[c + d*x]*Sin[c +
d*x]^3)/(4*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^4)/(5*b*d)

Rule 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 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 ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\int \frac{\sin ^2(c+d x) \left (5 \left (3 a^2-4 b^2\right )-a b \sin (c+d x)-4 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2}\\ &=-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\int \frac{\sin (c+d x) \left (-8 a \left (5 a^2-6 b^2\right )+b \left (5 a^2-12 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3}\\ &=\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\int \frac{15 a^2 \left (4 a^2-5 b^2\right )-a b \left (20 a^2-21 b^2\right ) \sin (c+d x)-8 \left (15 a^4-20 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4}\\ &=-\frac{\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\int \frac{15 a^2 b \left (4 a^2-5 b^2\right )+15 a \left (8 a^4-12 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5}\\ &=-\frac{a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac{\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\left (a^2 \left (a^2-b^2\right )^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=-\frac{a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac{\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\left (2 a^2 \left (a^2-b^2\right )^2\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^6 d}\\ &=-\frac{a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac{\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\left (4 a^2 \left (a^2-b^2\right )^2\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^6 d}\\ &=-\frac{a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac{2 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^6 d}-\frac{\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac{a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac{\left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}+\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 1.98549, size = 186, normalized size = 0.79 \[ \frac{-15 a \left (4 \left (-12 a^2 b^2+8 a^4+3 b^4\right ) (c+d x)+\left (8 b^4-8 a^2 b^2\right ) \sin (2 (c+d x))+b^4 \sin (4 (c+d x))\right )-60 b \left (-10 a^2 b^2+8 a^4+b^4\right ) \cos (c+d x)+10 \left (4 a^2 b^3-3 b^5\right ) \cos (3 (c+d x))+960 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-6 b^5 \cos (5 (c+d x))}{480 b^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*a^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 60*b*(8*a^4 - 10*a^2*b^2 + b^4)*
Cos[c + d*x] + 10*(4*a^2*b^3 - 3*b^5)*Cos[3*(c + d*x)] - 6*b^5*Cos[5*(c + d*x)] - 15*a*(4*(8*a^4 - 12*a^2*b^2
+ 3*b^4)*(c + d*x) + (-8*a^2*b^2 + 8*b^4)*Sin[2*(c + d*x)] + b^4*Sin[4*(c + d*x)]))/(480*b^6*d)

________________________________________________________________________________________

Maple [B]  time = 0.096, size = 941, normalized size = 4. \begin{align*} \text{result too large to display} \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+b*sin(d*x+c)),x)

[Out]

-2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7*a^3+5/4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+
1/2*c)^9*a-3/4/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-8/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^2*a^4-
2/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^8-2/d/b^6*arctan(tan(1/2*d*x+1/2*c))*a^5-4/d/b/(1+tan(1/2*
d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4-2/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*a^4+8/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)
^5*a^2+2/d/b^2/(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))*a^2+3/d/b^4*arctan(tan
(1/2*d*x+1/2*c))*a^3+12/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6*a^2+28/3/d/b^3/(1+tan(1/2*d*x+1/
2*c)^2)^5*tan(1/2*d*x+1/2*c)^2*a^2-2/5/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5-2/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1
/2*d*x+1/2*c)^8*a^4+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^8*a^2-8/d/b^5/(1+tan(1/2*d*x+1/2*c)^
2)^5*tan(1/2*d*x+1/2*c)^6*a^4+2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3*a^3-1/2/d/b^2/(1+tan(1/2
*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3*a+2/d*a^6/b^6/(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))-4/d*a^4/b^4/(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))-5/4/d/b^2/
(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)*a+44/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4*a^2
-1/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9*a^3-12/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1
/2*c)^4*a^4+1/2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7*a+1/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan
(1/2*d*x+1/2*c)*a^3

________________________________________________________________________________________

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)^4*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.90645, size = 1056, normalized size = 4.49 \begin{align*} \left [-\frac{24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \,{\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 60 \,{\left (a^{4} - a^{2} b^{2}\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 ) + 120 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, -\frac{24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \,{\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 120 \,{\left (a^{4} - a^{2} b^{2}\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 ) + 120 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} 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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/120*(24*b^5*cos(d*x + c)^5 - 40*a^2*b^3*cos(d*x + c)^3 + 15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*d*x + 60*(a^4 -
a^2*b^2)*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)) +
120*(a^4*b - a^2*b^3)*cos(d*x + c) + 15*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c))*sin(d*x
+ c))/(b^6*d), -1/120*(24*b^5*cos(d*x + c)^5 - 40*a^2*b^3*cos(d*x + c)^3 + 15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*d
*x + 120*(a^4 - a^2*b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 120*(a
^4*b - a^2*b^3)*cos(d*x + c) + 15*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c))*sin(d*x + c))/
(b^6*d)]

________________________________________________________________________________________

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)**4*sin(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.18468, size = 618, normalized size = 2.63 \begin{align*} -\frac{\frac{15 \,{\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )}{\left (d x + c\right )}}{b^{6}} - \frac{240 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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^{6}} + \frac{2 \,{\left (60 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 120 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 240 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 120 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 120 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 30 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 480 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 720 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 720 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 880 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 240 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 480 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 560 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 60 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 120 \, a^{4} - 160 \, a^{2} b^{2} + 24 \, b^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5} b^{5}}}{120 \, 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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*(d*x + c)/b^6 - 240*(a^6 - 2*a^4*b^2 + a^2*b^4)*(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^6) + 2*(60*a^3
*b*tan(1/2*d*x + 1/2*c)^9 - 75*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*a^4*tan(1/2*d*x + 1/2*c)^8 - 240*a^2*b^2*tan
(1/2*d*x + 1/2*c)^8 + 120*b^4*tan(1/2*d*x + 1/2*c)^8 + 120*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 30*a*b^3*tan(1/2*d*x
+ 1/2*c)^7 + 480*a^4*tan(1/2*d*x + 1/2*c)^6 - 720*a^2*b^2*tan(1/2*d*x + 1/2*c)^6 + 720*a^4*tan(1/2*d*x + 1/2*
c)^4 - 880*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 240*b^4*tan(1/2*d*x + 1/2*c)^4 - 120*a^3*b*tan(1/2*d*x + 1/2*c)^3
+ 30*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*a^4*tan(1/2*d*x + 1/2*c)^2 - 560*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 60*a
^3*b*tan(1/2*d*x + 1/2*c) + 75*a*b^3*tan(1/2*d*x + 1/2*c) + 120*a^4 - 160*a^2*b^2 + 24*b^4)/((tan(1/2*d*x + 1/
2*c)^2 + 1)^5*b^5))/d