3.1128 \(\int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=267 \[ -\frac {\left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 a b^3 d}-\frac {2 a \left (5 a^4-7 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d \sqrt {a^2-b^2}}+\frac {x \left (40 a^4-36 a^2 b^2+3 b^4\right )}{8 b^6}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b^2 d} \]

[Out]

1/8*(40*a^4-36*a^2*b^2+3*b^4)*x/b^6+1/3*a*(15*a^2-11*b^2)*cos(d*x+c)/b^5/d-1/8*(20*a^2-13*b^2)*cos(d*x+c)*sin(
d*x+c)/b^4/d+1/3*(5*a^2-3*b^2)*cos(d*x+c)*sin(d*x+c)^2/a/b^3/d-1/4*cos(d*x+c)*sin(d*x+c)^3/b^2/d-(a^2-b^2)*cos
(d*x+c)*sin(d*x+c)^3/a/b^2/d/(a+b*sin(d*x+c))-2*a*(5*a^4-7*a^2*b^2+2*b^4)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2
-b^2)^(1/2))/b^6/d/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.76, antiderivative size = 267, normalized size of antiderivative = 1.00, 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 = {2892, 3049, 3023, 2735, 2660, 618, 204} \[ \frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {2 a \left (-7 a^2 b^2+5 a^4+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d \sqrt {a^2-b^2}}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\left (5 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{3 a b^3 d}-\frac {\left (20 a^2-13 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {x \left (-36 a^2 b^2+40 a^4+3 b^4\right )}{8 b^6}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((40*a^4 - 36*a^2*b^2 + 3*b^4)*x)/(8*b^6) - (2*a*(5*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]])/(b^6*Sqrt[a^2 - b^2]*d) + (a*(15*a^2 - 11*b^2)*Cos[c + d*x])/(3*b^5*d) - ((20*a^2 - 13*b^2)*C
os[c + d*x]*Sin[c + d*x])/(8*b^4*d) + ((5*a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(3*a*b^3*d) - (Cos[c + d*x
]*Sin[c + d*x]^3)/(4*b^2*d) - ((a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(a*b^2*d*(a + b*Sin[c + d*x]))

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 2892

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[((a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*b^2*d*
f*(m + 1)), x] + (-Dist[1/(a*b^2*(m + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Sim
p[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m
+ n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e +
f*x])^(n + 1))/(b^2*d*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2
*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && NeQ[m + n + 4, 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 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])))

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\int \frac {\sin ^2(c+d x) \left (15 a^2-8 b^2-a b \sin (c+d x)-4 \left (5 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a b^2}\\ &=\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\int \frac {\sin (c+d x) \left (-8 a \left (5 a^2-3 b^2\right )+5 a^2 b \sin (c+d x)+3 a \left (20 a^2-13 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{12 a b^3}\\ &=-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\int \frac {3 a^2 \left (20 a^2-13 b^2\right )-a b \left (20 a^2-9 b^2\right ) \sin (c+d x)-8 a^2 \left (15 a^2-11 b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{24 a b^4}\\ &=\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\int \frac {3 a^2 b \left (20 a^2-13 b^2\right )+3 a \left (40 a^4-36 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{24 a b^5}\\ &=\frac {\left (40 a^4-36 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\left (a \left (5 a^4-7 a^2 b^2+2 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=\frac {\left (40 a^4-36 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\left (2 a \left (5 a^4-7 a^2 b^2+2 b^4\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 {\left (40 a^4-36 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {\left (4 a \left (5 a^4-7 a^2 b^2+2 b^4\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 {\left (40 a^4-36 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {2 a \left (5 a^4-7 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 \sqrt {a^2-b^2} d}+\frac {a \left (15 a^2-11 b^2\right ) \cos (c+d x)}{3 b^5 d}-\frac {\left (20 a^2-13 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{3 a b^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{a b^2 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 3.67, size = 325, normalized size = 1.22 \[ \frac {\frac {960 a^5 c+960 a^5 d x+960 a^4 b c \sin (c+d x)+960 a^4 b d x \sin (c+d x)+240 a^3 b^2 \sin (2 (c+d x))-864 a^3 b^2 c-864 a^3 b^2 d x-864 a^2 b^3 c \sin (c+d x)-864 a^2 b^3 d x \sin (c+d x)+\left (40 a^2 b^3-21 b^5\right ) \cos (3 (c+d x))+24 b \left (40 a^4-31 a^2 b^2+b^4\right ) \cos (c+d x)-176 a b^4 \sin (2 (c+d x))-10 a b^4 \sin (4 (c+d x))+72 a b^4 c+72 a b^4 d x+72 b^5 c \sin (c+d x)+72 b^5 d x \sin (c+d x)-3 b^5 \cos (5 (c+d x))}{a+b \sin (c+d x)}-\frac {384 a \left (5 a^4-7 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{192 b^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-384*a*(5*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (960*
a^5*c - 864*a^3*b^2*c + 72*a*b^4*c + 960*a^5*d*x - 864*a^3*b^2*d*x + 72*a*b^4*d*x + 24*b*(40*a^4 - 31*a^2*b^2
+ b^4)*Cos[c + d*x] + (40*a^2*b^3 - 21*b^5)*Cos[3*(c + d*x)] - 3*b^5*Cos[5*(c + d*x)] + 960*a^4*b*c*Sin[c + d*
x] - 864*a^2*b^3*c*Sin[c + d*x] + 72*b^5*c*Sin[c + d*x] + 960*a^4*b*d*x*Sin[c + d*x] - 864*a^2*b^3*d*x*Sin[c +
d*x] + 72*b^5*d*x*Sin[c + d*x] + 240*a^3*b^2*Sin[2*(c + d*x)] - 176*a*b^4*Sin[2*(c + d*x)] - 10*a*b^4*Sin[4*(
c + d*x)])/(a + b*Sin[c + d*x]))/(192*b^6*d)

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fricas [A]  time = 0.94, size = 604, normalized size = 2.26 \[ \left [-\frac {6 \, b^{5} \cos \left (d x + c\right )^{5} - {\left (20 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (40 \, a^{5} - 36 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 12 \, {\left (5 \, a^{4} - 2 \, a^{2} b^{2} + {\left (5 \, a^{3} b - 2 \, a 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 ) - 3 \, {\left (40 \, a^{4} b - 36 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right ) + {\left (10 \, a b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (40 \, a^{4} b - 36 \, a^{2} b^{3} + 3 \, b^{5}\right )} d x - 3 \, {\left (20 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}}, -\frac {6 \, b^{5} \cos \left (d x + c\right )^{5} - {\left (20 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (40 \, a^{5} - 36 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x - 24 \, {\left (5 \, a^{4} - 2 \, a^{2} b^{2} + {\left (5 \, a^{3} b - 2 \, a 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 (40 \, a^{4} b - 36 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right ) + {\left (10 \, a b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (40 \, a^{4} b - 36 \, a^{2} b^{3} + 3 \, b^{5}\right )} d x - 3 \, {\left (20 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}}\right ] \]

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))^2,x, algorithm="fricas")

[Out]

[-1/24*(6*b^5*cos(d*x + c)^5 - (20*a^2*b^3 - 3*b^5)*cos(d*x + c)^3 - 3*(40*a^5 - 36*a^3*b^2 + 3*a*b^4)*d*x + 1
2*(5*a^4 - 2*a^2*b^2 + (5*a^3*b - 2*a*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)) - 3*(40*a^4*b - 36*a^2*b^3 + 3*b^5)*cos(d*x + c) + (10*a*b^4*cos
(d*x + c)^3 - 3*(40*a^4*b - 36*a^2*b^3 + 3*b^5)*d*x - 3*(20*a^3*b^2 - 13*a*b^4)*cos(d*x + c))*sin(d*x + c))/(b
^7*d*sin(d*x + c) + a*b^6*d), -1/24*(6*b^5*cos(d*x + c)^5 - (20*a^2*b^3 - 3*b^5)*cos(d*x + c)^3 - 3*(40*a^5 -
36*a^3*b^2 + 3*a*b^4)*d*x - 24*(5*a^4 - 2*a^2*b^2 + (5*a^3*b - 2*a*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*(40*a^4*b - 36*a^2*b^3 + 3*b^5)*cos(d*x + c) + (10*a*
b^4*cos(d*x + c)^3 - 3*(40*a^4*b - 36*a^2*b^3 + 3*b^5)*d*x - 3*(20*a^3*b^2 - 13*a*b^4)*cos(d*x + c))*sin(d*x +
c))/(b^7*d*sin(d*x + c) + a*b^6*d)]

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giac [A]  time = 0.21, size = 449, normalized size = 1.68 \[ \frac {\frac {3 \, {\left (40 \, a^{4} - 36 \, a^{2} b^{2} + 3 \, b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {48 \, {\left (5 \, a^{5} - 7 \, a^{3} b^{2} + 2 \, a b^{4}\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 {48 \, {\left (a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4} - a^{2} b^{2}\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^{5}} + \frac {2 \, {\left (36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 96 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 192 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 288 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 160 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} - 64 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{5}}}{24 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/24*(3*(40*a^4 - 36*a^2*b^2 + 3*b^4)*(d*x + c)/b^6 - 48*(5*a^5 - 7*a^3*b^2 + 2*a*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) + 48*(a^3*b*ta
n(1/2*d*x + 1/2*c) - a*b^3*tan(1/2*d*x + 1/2*c) + a^4 - a^2*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x
+ 1/2*c) + a)*b^5) + 2*(36*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 15*b^3*tan(1/2*d*x + 1/2*c)^7 + 96*a^3*tan(1/2*d*x +
1/2*c)^6 - 96*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 36*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^3*tan(1/2*d*x + 1/2*c)^5 +
288*a^3*tan(1/2*d*x + 1/2*c)^4 - 192*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 9*b^3*t
an(1/2*d*x + 1/2*c)^3 + 288*a^3*tan(1/2*d*x + 1/2*c)^2 - 160*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 36*a^2*b*tan(1/2*d
*x + 1/2*c) + 15*b^3*tan(1/2*d*x + 1/2*c) + 96*a^3 - 64*a*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^5))/d

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maple [B]  time = 0.50, size = 938, normalized size = 3.51 \[ \text {result too large to display} \]

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))^2,x)

[Out]

-10/d/b^6*a^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))+14/d/b^4*a^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))-4/d/b^2*a/(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))+3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*a^2-5/4/d/b^2/(1+t
an(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+3/4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-3/4/d/b^
2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+5/4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+8/d/
b^5/(1+tan(1/2*d*x+1/2*c)^2)^4*a^3-16/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*a+10/d/b^6*arctan(tan(1/2*d*x+1/2*c))
*a^4-9/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^2+2/d/b^5/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^4-2/d/
b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^2+3/4/d/b^2*arctan(tan(1/2*d*x+1/2*c))+8/d/b^5/(1+tan(
1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6*a^3-8/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6*a+3/d/b^4
/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a^2+24/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*
a^3-16/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*a-3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+
1/2*c)^3*a^2+24/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a^3-40/3/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^
4*tan(1/2*d*x+1/2*c)^2*a-3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^2+2/d/b^4/(tan(1/2*d*x+1/2*c)
^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^3*tan(1/2*d*x+1/2*c)-2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a
)*a*tan(1/2*d*x+1/2*c)

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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)^4*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,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 = 11.79, size = 1003, normalized size = 3.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(60*a^4*b + (3*b^5*cos(c + d*x))/2 + 120*a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - 44*a^2*b^3 - (21*b^
5*cos(3*c + 3*d*x))/16 - (3*b^5*cos(5*c + 5*d*x))/16 + 9*a*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (
93*a^2*b^3*cos(c + d*x))/2 + 9*b^5*sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - 11*a*b^4*sin(2*c
+ 2*d*x) - (5*a*b^4*sin(4*c + 4*d*x))/8 + 60*a^3*b^2*sin(c + d*x) - 108*a^3*b^2*atan(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2)) + (5*a^2*b^3*cos(3*c + 3*d*x))/2 - 120*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) + 15*a^3
*b^2*sin(2*c + 2*d*x) + 60*a^4*b*cos(c + d*x) - 44*a*b^4*sin(c + d*x) - 108*a^2*b^3*sin(c + d*x)*atan(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)) + 48*a^2*b^2*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) + 120*a^4*b*sin(c + d*
x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 48*a*b^3*sin(c + d*x)*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) - 120*a^3*b*sin(c + d*x)*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))/(12*b^6*d*(a + b*sin(c + d*x)))

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

[Out]

Timed out

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