3.1285 \(\int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

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

[Out]

1/8*a*(8*a^4-4*a^2*b^2-b^4)*x/b^6+1/15*(15*a^4-5*a^2*b^2-2*b^4)*cos(d*x+c)/b^5/d-1/8*a*(4*a^2-b^2)*cos(d*x+c)*
sin(d*x+c)/b^4/d+1/15*(5*a^2-b^2)*cos(d*x+c)*sin(d*x+c)^2/b^3/d-1/4*a*cos(d*x+c)*sin(d*x+c)^3/b^2/d+1/5*cos(d*
x+c)*sin(d*x+c)^4/b/d-2*a^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/b^6/d

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Rubi [A]  time = 0.91, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac {\left (-5 a^2 b^2+15 a^4-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {2 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {a x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^6}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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])))

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac {\sin ^4(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin ^3(c+d x) \left (-4 a+b \sin (c+d x)+5 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b}\\ &=-\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 (15 a^2-a b \sin (c+d x)-4 \left (5 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2}\\ &=\frac {\left (5 a^2-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-b^2\right )+b \left (5 a^2+8 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-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-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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-b^2\right )-a b \left (20 a^2-b^2\right ) \sin (c+d x)-8 \left (15 a^4-5 a^2 b^2-2 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4}\\ &=\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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-b^2\right )+15 a \left (8 a^4-4 a^2 b^2-b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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^4 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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^4 \left (a^2-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^6 d}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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^4 \left (a^2-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^6 d}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}-\frac {2 a^4 \sqrt {a^2-b^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^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-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*}

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Mathematica [A]  time = 1.77, size = 177, normalized size = 0.75 \[ \frac {-10 \left (4 a^2 b^3+b^5\right ) \cos (3 (c+d x))-960 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+15 a \left (-8 a^2 b^2 \sin (2 (c+d x))+4 \left (8 a^4-4 a^2 b^2-b^4\right ) (c+d x)+b^4 \sin (4 (c+d x))\right )-60 b \left (-8 a^4+2 a^2 b^2+b^4\right ) \cos (c+d x)+6 b^5 \cos (5 (c+d x))}{480 b^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 427, normalized size = 1.82 \[ \left [\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 60 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, \frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 120 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 467, normalized size = 1.99 \[ \frac {\frac {15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{6} - a^{4} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \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} - 15 \, 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} - 120 \, a^{2} b^{2} \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} + 90 \, 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} - 240 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, b^{4} \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} - 160 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, 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} - 90 \, 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} - 80 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, b^{4} \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 ) + 15 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, a^{4} - 40 \, a^{2} b^{2} - 16 \, b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} b^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(8*a^5 - 4*a^3*b^2 - a*b^4)*(d*x + c)/b^6 - 240*(a^6 - a^4*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sg
n(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 - 15*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*a^4*tan(1/2*d*x + 1/2*c)^8 - 120*a^2*b^2*tan(1/2*d*x + 1/2*c
)^8 + 120*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 90*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*a^4*tan(1/2*d*x + 1/2*c)^6 - 24
0*a^2*b^2*tan(1/2*d*x + 1/2*c)^6 - 240*b^4*tan(1/2*d*x + 1/2*c)^6 + 720*a^4*tan(1/2*d*x + 1/2*c)^4 - 160*a^2*b
^2*tan(1/2*d*x + 1/2*c)^4 + 80*b^4*tan(1/2*d*x + 1/2*c)^4 - 120*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 90*a*b^3*tan(1/
2*d*x + 1/2*c)^3 + 480*a^4*tan(1/2*d*x + 1/2*c)^2 - 80*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 80*b^4*tan(1/2*d*x + 1
/2*c)^2 - 60*a^3*b*tan(1/2*d*x + 1/2*c) + 15*a*b^3*tan(1/2*d*x + 1/2*c) + 120*a^4 - 40*a^2*b^2 - 16*b^4)/((tan
(1/2*d*x + 1/2*c)^2 + 1)^5*b^5))/d

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maple [B]  time = 0.32, size = 871, normalized size = 3.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))+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^4-2/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+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-
4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6+4/3/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4-
4/3/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^2+2/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*a^4-2/3/d/b^3/(1+ta
n(1/2*d*x+1/2*c)^2)^5*a^2-1/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^3+2/d/b^6*arctan(tan(1/2*d*x+1/2*c))*a^5-1/4/d/
b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9*a-4/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^
2*a^2-1/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)*a^3+1/4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d
*x+1/2*c)*a+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^3-4/15/d/b/(1+tan(1/2*d*x+1/2*c)^2)^5-2/
d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3*a^3-3/2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2
*c)^3*a-8/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4*a^2-4/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+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*a^4*(a^2-b^2)^(1/2)/b^6*ar
ctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+3/2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)
^7*a+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+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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 13.39, size = 376, normalized size = 1.60 \[ \frac {a^4\,\cos \left (c+d\,x\right )}{b^5\,d}-\frac {\frac {\cos \left (c+d\,x\right )}{8}+\frac {\cos \left (3\,c+3\,d\,x\right )}{48}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80}}{b\,d}-\frac {\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32}}{b^2\,d}-\frac {\frac {a^2\,\cos \left (c+d\,x\right )}{4}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{12}}{b^3\,d}-\frac {a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^4\,d}+\frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^6\,d}-\frac {2\,a^4\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}}{b^6\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*cos(c + d*x))/(b^5*d) - (cos(c + d*x)/8 + cos(3*c + 3*d*x)/48 - cos(5*c + 5*d*x)/80)/(b*d) - ((a*atan(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - (a*sin(4*c + 4*d*x))/32)/(b^2*d) - ((a^2*cos(c + d*x))/4 + (a^2*cos(3
*c + 3*d*x))/12)/(b^3*d) - (a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (a^3*sin(2*c + 2*d*x))/4)/(b^4*d
) + (2*a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^6*d) - (2*a^4*atanh((2*b^2*sin(c/2 + (d*x)/2)*(b^2
- a^2)^(1/2) - a^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + a*b*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(a^3*cos(c
/2 + (d*x)/2) - 2*b^3*sin(c/2 + (d*x)/2) - a*b^2*cos(c/2 + (d*x)/2) + 2*a^2*b*sin(c/2 + (d*x)/2)))*(b^2 - a^2)
^(1/2))/(b^6*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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