3.1127 \(\int \frac{\cos ^4(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=307 \[ -\frac{\left (-25 a^2 b^2+30 a^4+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{6 a^2 \left (-3 a^2 b^2+2 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^7 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\left (3 a^2-2 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a b^3 d}-\frac{\left (10 a^2-7 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{5 b^4 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{4 b^5 d}-\frac{3 a x \left (-8 a^2 b^2+8 a^4+b^4\right )}{4 b^7}-\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b^2 d} \]

[Out]

(-3*a*(8*a^4 - 8*a^2*b^2 + b^4)*x)/(4*b^7) + (6*a^2*(2*a^4 - 3*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/
Sqrt[a^2 - b^2]])/(b^7*Sqrt[a^2 - b^2]*d) - ((30*a^4 - 25*a^2*b^2 + b^4)*Cos[c + d*x])/(5*b^6*d) + (3*a*(4*a^2
- 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(4*b^5*d) - ((10*a^2 - 7*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(5*b^4*d) + ((
3*a^2 - 2*b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(2*a*b^3*d) - (Cos[c + d*x]*Sin[c + d*x]^4)/(5*b^2*d) - ((a^2 - b^
2)*Cos[c + d*x]*Sin[c + d*x]^4)/(a*b^2*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.01584, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{\left (-25 a^2 b^2+30 a^4+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{6 a^2 \left (-3 a^2 b^2+2 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^7 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\left (3 a^2-2 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{2 a b^3 d}-\frac{\left (10 a^2-7 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{5 b^4 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{4 b^5 d}-\frac{3 a x \left (-8 a^2 b^2+8 a^4+b^4\right )}{4 b^7}-\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*(8*a^4 - 8*a^2*b^2 + b^4)*x)/(4*b^7) + (6*a^2*(2*a^4 - 3*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/
Sqrt[a^2 - b^2]])/(b^7*Sqrt[a^2 - b^2]*d) - ((30*a^4 - 25*a^2*b^2 + b^4)*Cos[c + d*x])/(5*b^6*d) + (3*a*(4*a^2
- 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(4*b^5*d) - ((10*a^2 - 7*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(5*b^4*d) + ((
3*a^2 - 2*b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(2*a*b^3*d) - (Cos[c + d*x]*Sin[c + d*x]^4)/(5*b^2*d) - ((a^2 - b^
2)*Cos[c + d*x]*Sin[c + d*x]^4)/(a*b^2*d*(a + b*Sin[c + d*x]))

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 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 ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{\sin ^3(c+d x) \left (3 \left (8 a^2-5 b^2\right )-a b \sin (c+d x)-10 \left (3 a^2-2 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 a b^2}\\ &=\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{\sin ^2(c+d x) \left (-30 a \left (3 a^2-2 b^2\right )+6 a^2 b \sin (c+d x)+12 a \left (10 a^2-7 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 a b^3}\\ &=-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{\sin (c+d x) \left (24 a^2 \left (10 a^2-7 b^2\right )-6 a b \left (5 a^2-2 b^2\right ) \sin (c+d x)-90 a^2 \left (4 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 a b^4}\\ &=\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{-90 a^3 \left (4 a^2-3 b^2\right )+6 a^2 b \left (20 a^2-11 b^2\right ) \sin (c+d x)+24 a \left (30 a^4-25 a^2 b^2+b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 a b^5}\\ &=-\frac{\left (30 a^4-25 a^2 b^2+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\int \frac{-90 a^3 b \left (4 a^2-3 b^2\right )-90 a^2 \left (8 a^4-8 a^2 b^2+b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 a b^6}\\ &=-\frac{3 a \left (8 a^4-8 a^2 b^2+b^4\right ) x}{4 b^7}-\frac{\left (30 a^4-25 a^2 b^2+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\left (3 a^2 \left (2 a^4-3 a^2 b^2+b^4\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^7}\\ &=-\frac{3 a \left (8 a^4-8 a^2 b^2+b^4\right ) x}{4 b^7}-\frac{\left (30 a^4-25 a^2 b^2+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac{\left (6 a^2 \left (2 a^4-3 a^2 b^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^7 d}\\ &=-\frac{3 a \left (8 a^4-8 a^2 b^2+b^4\right ) x}{4 b^7}-\frac{\left (30 a^4-25 a^2 b^2+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac{\left (12 a^2 \left (2 a^4-3 a^2 b^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^7 d}\\ &=-\frac{3 a \left (8 a^4-8 a^2 b^2+b^4\right ) x}{4 b^7}+\frac{6 a^2 \left (2 a^4-3 a^2 b^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^7 \sqrt{a^2-b^2} d}-\frac{\left (30 a^4-25 a^2 b^2+b^4\right ) \cos (c+d x)}{5 b^6 d}+\frac{3 a \left (4 a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 b^5 d}-\frac{\left (10 a^2-7 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{5 b^4 d}+\frac{\left (3 a^2-2 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{2 a b^3 d}-\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac{\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{a b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.11241, size = 378, normalized size = 1.23 \[ \frac{\frac{960 a^2 \left (-3 a^2 b^2+2 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{240 a^4 b^2 \sin (2 (c+d x))-960 a^3 b^3 c \sin (c+d x)-960 a^3 b^3 d x \sin (c+d x)-200 a^2 b^4 \sin (2 (c+d x))-10 a^2 b^4 \sin (4 (c+d x))+60 a b \left (-14 a^2 b^2+16 a^4+b^4\right ) \cos (c+d x)+5 \left (8 a^3 b^3-5 a b^5\right ) \cos (3 (c+d x))-960 a^4 b^2 c+120 a^2 b^4 c-960 a^4 b^2 d x+120 a^2 b^4 d x+960 a^5 b c \sin (c+d x)+960 a^5 b d x \sin (c+d x)+960 a^6 c+960 a^6 d x+120 a b^5 c \sin (c+d x)+120 a b^5 d x \sin (c+d x)-3 a b^5 \cos (5 (c+d x))+5 b^6 \sin (2 (c+d x))+4 b^6 \sin (4 (c+d x))+b^6 \sin (6 (c+d x))}{a+b \sin (c+d x)}}{160 b^7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((960*a^2*(2*a^4 - 3*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (960*a
^6*c - 960*a^4*b^2*c + 120*a^2*b^4*c + 960*a^6*d*x - 960*a^4*b^2*d*x + 120*a^2*b^4*d*x + 60*a*b*(16*a^4 - 14*a
^2*b^2 + b^4)*Cos[c + d*x] + 5*(8*a^3*b^3 - 5*a*b^5)*Cos[3*(c + d*x)] - 3*a*b^5*Cos[5*(c + d*x)] + 960*a^5*b*c
*Sin[c + d*x] - 960*a^3*b^3*c*Sin[c + d*x] + 120*a*b^5*c*Sin[c + d*x] + 960*a^5*b*d*x*Sin[c + d*x] - 960*a^3*b
^3*d*x*Sin[c + d*x] + 120*a*b^5*d*x*Sin[c + d*x] + 240*a^4*b^2*Sin[2*(c + d*x)] - 200*a^2*b^4*Sin[2*(c + d*x)]
+ 5*b^6*Sin[2*(c + d*x)] - 10*a^2*b^4*Sin[4*(c + d*x)] + 4*b^6*Sin[4*(c + d*x)] + b^6*Sin[6*(c + d*x)])/(a +
b*Sin[c + d*x]))/(160*b^7*d)

________________________________________________________________________________________

Maple [B]  time = 0.139, size = 1119, normalized size = 3.6 \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)^3/(a+b*sin(d*x+c))^2,x)

[Out]

12/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^3-3/2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a-2/5/d/b^2/(1+tan(1/2*d*x+1/2*c)
^2)^5-2/d*a^5/b^6/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+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)-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^8-4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^5
*tan(1/2*d*x+1/2*c)^4-10/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^5*a^4+8/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*a^2-12/d/b^7*
arctan(tan(1/2*d*x+1/2*c))*a^5+12/d*a^6/b^7/(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))-18/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))+5/2/d/b^3/(1+tan(
1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9*a-10/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^8*a^4+12/d/b
^4/(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)^7
*a^3+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))+1/d/b^3/(1+tan(1/2*d
*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7*a-40/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6*a^4+36/d/b^4/(1
+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6*a^2-60/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4*a^4
+44/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4*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)^3*a^3-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3*a-40/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(
1/2*d*x+1/2*c)^2*a^4-4/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9*a^3+28/d/b^4/(1+tan(1/2*d*x+1/2*c
)^2)^5*tan(1/2*d*x+1/2*c)^2*a^2+4/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)*a^3-5/2/d/b^3/(1+tan(1/2
*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)*a-2/d*a^4/b^5/(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^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)

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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)^4*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 = 2.34225, size = 1451, normalized size = 4.73 \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+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/20*(6*a*b^5*cos(d*x + c)^5 - 5*(4*a^3*b^3 - a*b^5)*cos(d*x + c)^3 - 15*(8*a^6 - 8*a^4*b^2 + a^2*b^4)*d*x -
30*(2*a^5 - a^3*b^2 + (2*a^4*b - 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)) - 15*(8*a^5*b - 8*a^3*b^3 + a*b^5)*cos(d*x + c) - (4*b^6*cos(d*x +
c)^5 - 10*a^2*b^4*cos(d*x + c)^3 + 15*(8*a^5*b - 8*a^3*b^3 + a*b^5)*d*x + 15*(4*a^4*b^2 - 3*a^2*b^4)*cos(d*x
+ c))*sin(d*x + c))/(b^8*d*sin(d*x + c) + a*b^7*d), 1/20*(6*a*b^5*cos(d*x + c)^5 - 5*(4*a^3*b^3 - a*b^5)*cos(d
*x + c)^3 - 15*(8*a^6 - 8*a^4*b^2 + a^2*b^4)*d*x - 60*(2*a^5 - a^3*b^2 + (2*a^4*b - a^2*b^3)*sin(d*x + c))*sqr
t(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 15*(8*a^5*b - 8*a^3*b^3 + a*b^5)*c
os(d*x + c) - (4*b^6*cos(d*x + c)^5 - 10*a^2*b^4*cos(d*x + c)^3 + 15*(8*a^5*b - 8*a^3*b^3 + a*b^5)*d*x + 15*(4
*a^4*b^2 - 3*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(b^8*d*sin(d*x + c) + 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)**4*sin(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.23183, size = 724, normalized size = 2.36 \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+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/20*(15*(8*a^5 - 8*a^3*b^2 + a*b^4)*(d*x + c)/b^7 - 120*(2*a^6 - 3*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^7) + 40*(a^4*b*t
an(1/2*d*x + 1/2*c) - a^2*b^3*tan(1/2*d*x + 1/2*c) + a^5 - a^3*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d
*x + 1/2*c) + a)*b^6) + 2*(40*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 25*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 100*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 + 20*b^4*tan(1/2*d*x + 1/2*c)^8 + 80*a^3*b*tan(1/2*d*x +
1/2*c)^7 - 10*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 400*a^4*tan(1/2*d*x + 1/2*c)^6 - 360*a^2*b^2*tan(1/2*d*x + 1/2*c)
^6 + 600*a^4*tan(1/2*d*x + 1/2*c)^4 - 440*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 40*b^4*tan(1/2*d*x + 1/2*c)^4 - 80*
a^3*b*tan(1/2*d*x + 1/2*c)^3 + 10*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 400*a^4*tan(1/2*d*x + 1/2*c)^2 - 280*a^2*b^2*
tan(1/2*d*x + 1/2*c)^2 - 40*a^3*b*tan(1/2*d*x + 1/2*c) + 25*a*b^3*tan(1/2*d*x + 1/2*c) + 100*a^4 - 80*a^2*b^2
+ 4*b^4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*b^6))/d