3.1301 \(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=282 \[ \frac {a \left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 b^3 d}+\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \sin (c+d x) \cos (c+d x)}{16 b^5 d}-\frac {2 a^3 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^7 d}+\frac {x \left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right )}{16 b^7}+\frac {a \sin ^4(c+d x) \cos (c+d x)}{5 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 b d} \]

[Out]

1/16*(16*a^6-24*a^4*b^2+6*a^2*b^4+b^6)*x/b^7-2*a^3*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(
1/2))/b^7/d+1/15*a*(15*a^4-20*a^2*b^2+3*b^4)*cos(d*x+c)/b^6/d-1/16*(8*a^4-10*a^2*b^2+b^4)*cos(d*x+c)*sin(d*x+c
)/b^5/d+1/15*a*(5*a^2-6*b^2)*cos(d*x+c)*sin(d*x+c)^2/b^4/d-1/24*(6*a^2-7*b^2)*cos(d*x+c)*sin(d*x+c)^3/b^3/d+1/
5*a*cos(d*x+c)*sin(d*x+c)^4/b^2/d-1/6*cos(d*x+c)*sin(d*x+c)^5/b/d

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Rubi [A]  time = 1.00, antiderivative size = 282, normalized size of antiderivative = 1.00, 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 = {2895, 3049, 3023, 2735, 2660, 618, 204} \[ \frac {a \left (-20 a^2 b^2+15 a^4+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {2 a^3 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^7 d}-\frac {\left (6 a^2-7 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 b^3 d}+\frac {a \left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^4 d}-\frac {\left (-10 a^2 b^2+8 a^4+b^4\right ) \sin (c+d x) \cos (c+d x)}{16 b^5 d}+\frac {x \left (-24 a^4 b^2+6 a^2 b^4+16 a^6+b^6\right )}{16 b^7}+\frac {a \sin ^4(c+d x) \cos (c+d x)}{5 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*a^6 - 24*a^4*b^2 + 6*a^2*b^4 + b^6)*x)/(16*b^7) - (2*a^3*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2]
)/Sqrt[a^2 - b^2]])/(b^7*d) + (a*(15*a^4 - 20*a^2*b^2 + 3*b^4)*Cos[c + d*x])/(15*b^6*d) - ((8*a^4 - 10*a^2*b^2
+ b^4)*Cos[c + d*x]*Sin[c + d*x])/(16*b^5*d) + (a*(5*a^2 - 6*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(15*b^4*d) - (
(6*a^2 - 7*b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(24*b^3*d) + (a*Cos[c + d*x]*Sin[c + d*x]^4)/(5*b^2*d) - (Cos[c +
d*x]*Sin[c + d*x]^5)/(6*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 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 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 ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\int \frac {\sin ^3(c+d x) \left (6 \left (4 a^2-5 b^2\right )-a b \sin (c+d x)-5 \left (6 a^2-7 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{30 b^2}\\ &=-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\int \frac {\sin ^2(c+d x) \left (-15 a \left (6 a^2-7 b^2\right )+3 b \left (2 a^2-5 b^2\right ) \sin (c+d x)+24 a \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{120 b^3}\\ &=\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\int \frac {\sin (c+d x) \left (48 a^2 \left (5 a^2-6 b^2\right )-3 a b \left (10 a^2-9 b^2\right ) \sin (c+d x)-45 \left (8 a^4-10 a^2 b^2+b^4\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{360 b^4}\\ &=-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\int \frac {-45 a \left (8 a^4-10 a^2 b^2+b^4\right )+3 b \left (40 a^4-42 a^2 b^2-15 b^4\right ) \sin (c+d x)+48 a \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}{720 b^5}\\ &=\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\int \frac {-45 a b \left (8 a^4-10 a^2 b^2+b^4\right )-45 \left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{720 b^6}\\ &=\frac {\left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right ) x}{16 b^7}+\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\left (a^3 \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^7}\\ &=\frac {\left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right ) x}{16 b^7}+\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\left (2 a^3 \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^7 d}\\ &=\frac {\left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right ) x}{16 b^7}+\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac {\left (4 a^3 \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^7 d}\\ &=\frac {\left (16 a^6-24 a^4 b^2+6 a^2 b^4+b^6\right ) x}{16 b^7}-\frac {2 a^3 \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^7 d}+\frac {a \left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^6 d}-\frac {\left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{16 b^5 d}+\frac {a \left (5 a^2-6 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^4 d}-\frac {\left (6 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 b^3 d}+\frac {a \cos (c+d x) \sin ^4(c+d x)}{5 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 2.21, size = 274, normalized size = 0.97 \[ \frac {960 a^6 c+960 a^6 d x-240 a^4 b^2 \sin (2 (c+d x))-1440 a^4 b^2 c-1440 a^4 b^2 d x+\left (60 a b^5-80 a^3 b^3\right ) \cos (3 (c+d x))+240 a^2 b^4 \sin (2 (c+d x))+30 a^2 b^4 \sin (4 (c+d x))+360 a^2 b^4 c+360 a^2 b^4 d x+120 a b \left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x)-1920 a^3 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+12 a b^5 \cos (5 (c+d x))+15 b^6 \sin (2 (c+d x))-15 b^6 \sin (4 (c+d x))-5 b^6 \sin (6 (c+d x))+60 b^6 c+60 b^6 d x}{960 b^7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*a^6*c - 1440*a^4*b^2*c + 360*a^2*b^4*c + 60*b^6*c + 960*a^6*d*x - 1440*a^4*b^2*d*x + 360*a^2*b^4*d*x + 60
*b^6*d*x - 1920*a^3*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 120*a*b*(8*a^4 - 10*a
^2*b^2 + b^4)*Cos[c + d*x] + (-80*a^3*b^3 + 60*a*b^5)*Cos[3*(c + d*x)] + 12*a*b^5*Cos[5*(c + d*x)] - 240*a^4*b
^2*Sin[2*(c + d*x)] + 240*a^2*b^4*Sin[2*(c + d*x)] + 15*b^6*Sin[2*(c + d*x)] + 30*a^2*b^4*Sin[4*(c + d*x)] - 1
5*b^6*Sin[4*(c + d*x)] - 5*b^6*Sin[6*(c + d*x)])/(960*b^7*d)

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fricas [A]  time = 1.03, size = 526, normalized size = 1.87 \[ \left [\frac {48 \, a b^{5} \cos \left (d x + c\right )^{5} - 80 \, a^{3} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (16 \, a^{6} - 24 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} d x - 120 \, {\left (a^{5} - a^{3} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 240 \, {\left (a^{5} b - a^{3} b^{3}\right )} \cos \left (d x + c\right ) - 5 \, {\left (8 \, b^{6} \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{4} b^{2} - 6 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, b^{7} d}, \frac {48 \, a b^{5} \cos \left (d x + c\right )^{5} - 80 \, a^{3} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (16 \, a^{6} - 24 \, a^{4} b^{2} + 6 \, a^{2} b^{4} + b^{6}\right )} d x + 240 \, {\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 240 \, {\left (a^{5} b - a^{3} b^{3}\right )} \cos \left (d x + c\right ) - 5 \, {\left (8 \, b^{6} \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{4} b^{2} - 6 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, b^{7} d}\right ] \]

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)),x, algorithm="fricas")

[Out]

[1/240*(48*a*b^5*cos(d*x + c)^5 - 80*a^3*b^3*cos(d*x + c)^3 + 15*(16*a^6 - 24*a^4*b^2 + 6*a^2*b^4 + b^6)*d*x -
120*(a^5 - a^3*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)) + 240*(a^5*b - a^3*b^3)*cos(d*x + c) - 5*(8*b^6*cos(d*x + c)^5 - 2*(6*a^2*b^4 + b^6)*cos(d*x + c)^3
+ 3*(8*a^4*b^2 - 6*a^2*b^4 - b^6)*cos(d*x + c))*sin(d*x + c))/(b^7*d), 1/240*(48*a*b^5*cos(d*x + c)^5 - 80*a^3
*b^3*cos(d*x + c)^3 + 15*(16*a^6 - 24*a^4*b^2 + 6*a^2*b^4 + b^6)*d*x + 240*(a^5 - a^3*b^2)*sqrt(a^2 - b^2)*arc
tan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 240*(a^5*b - a^3*b^3)*cos(d*x + c) - 5*(8*b^6*cos(
d*x + c)^5 - 2*(6*a^2*b^4 + b^6)*cos(d*x + c)^3 + 3*(8*a^4*b^2 - 6*a^2*b^4 - b^6)*cos(d*x + c))*sin(d*x + c))/
(b^7*d)]

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giac [B]  time = 0.21, size = 726, normalized size = 2.57 \[ \text {result too large to display} \]

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)),x, algorithm="giac")

[Out]

1/240*(15*(16*a^6 - 24*a^4*b^2 + 6*a^2*b^4 + b^6)*(d*x + c)/b^7 - 480*(a^7 - 2*a^5*b^2 + a^3*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) + 2
*(120*a^4*b*tan(1/2*d*x + 1/2*c)^11 - 150*a^2*b^3*tan(1/2*d*x + 1/2*c)^11 + 15*b^5*tan(1/2*d*x + 1/2*c)^11 + 2
40*a^5*tan(1/2*d*x + 1/2*c)^10 - 480*a^3*b^2*tan(1/2*d*x + 1/2*c)^10 + 240*a*b^4*tan(1/2*d*x + 1/2*c)^10 + 360
*a^4*b*tan(1/2*d*x + 1/2*c)^9 - 210*a^2*b^3*tan(1/2*d*x + 1/2*c)^9 - 235*b^5*tan(1/2*d*x + 1/2*c)^9 + 1200*a^5
*tan(1/2*d*x + 1/2*c)^8 - 1920*a^3*b^2*tan(1/2*d*x + 1/2*c)^8 + 240*a*b^4*tan(1/2*d*x + 1/2*c)^8 + 240*a^4*b*t
an(1/2*d*x + 1/2*c)^7 - 60*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 + 390*b^5*tan(1/2*d*x + 1/2*c)^7 + 2400*a^5*tan(1/2*
d*x + 1/2*c)^6 - 3200*a^3*b^2*tan(1/2*d*x + 1/2*c)^6 + 480*a*b^4*tan(1/2*d*x + 1/2*c)^6 - 240*a^4*b*tan(1/2*d*
x + 1/2*c)^5 + 60*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 390*b^5*tan(1/2*d*x + 1/2*c)^5 + 2400*a^5*tan(1/2*d*x + 1/2
*c)^4 - 2880*a^3*b^2*tan(1/2*d*x + 1/2*c)^4 + 480*a*b^4*tan(1/2*d*x + 1/2*c)^4 - 360*a^4*b*tan(1/2*d*x + 1/2*c
)^3 + 210*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 235*b^5*tan(1/2*d*x + 1/2*c)^3 + 1200*a^5*tan(1/2*d*x + 1/2*c)^2 -
1440*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + 48*a*b^4*tan(1/2*d*x + 1/2*c)^2 - 120*a^4*b*tan(1/2*d*x + 1/2*c) + 150*a
^2*b^3*tan(1/2*d*x + 1/2*c) - 15*b^5*tan(1/2*d*x + 1/2*c) + 240*a^5 - 320*a^3*b^2 + 48*a*b^4)/((tan(1/2*d*x +
1/2*c)^2 + 1)^6*b^6))/d

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maple [B]  time = 0.31, size = 1501, normalized size = 5.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

3/4/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+1/2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5*a^2-3/d/b^5
/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*a^4+2/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^10*
a^5+1/8/d/b*arctan(tan(1/2*d*x+1/2*c))+5/4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*a^2-4/d/b^4/(1+
tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^10*a^3+20/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^4*a^5
-24/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^4*a^3-80/3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*
x+1/2*c)^6*a^3+4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^6*a-16/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^6*t
an(1/2*d*x+1/2*c)^8*a^3+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^8*a-2/d*a^3/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))-3/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4+2/d/b^7*arctan
(tan(1/2*d*x+1/2*c))*a^6+2/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^6*a^5+1/8/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x
+1/2*c)^11-47/24/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9+13/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1
/2*d*x+1/2*c)^7-13/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5+47/24/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*
tan(1/2*d*x+1/2*c)^3-1/8/d/b/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)-8/3/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^
6*a^3+2/5/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*a+7/4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*a^2+10/
d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^2*a^5-2/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c
)^5*a^4+3/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9*a^4-7/4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1
/2*d*x+1/2*c)^9*a^2+2/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7*a^4-1/2/d/b^3/(1+tan(1/2*d*x+1/2*c
)^2)^6*tan(1/2*d*x+1/2*c)^7*a^2+20/d/b^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^6*a^5-12/d/b^4/(1+tan(1
/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^2*a^3+2/5/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^2*a+10/d/b
^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^8*a^5+4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^4
*a+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^10*a-5/4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x
+1/2*c)^11*a^2-1/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*a^4-2/d*a^7/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))+4/d*a^5/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))+1/d/b^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11*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)^4*sin(d*x+c)^3/(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 = 14.64, size = 600, normalized size = 2.13 \[ \frac {\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 )}{8\,b\,d}+\frac {\sin \left (2\,c+2\,d\,x\right )}{64\,b\,d}-\frac {\sin \left (4\,c+4\,d\,x\right )}{64\,b\,d}-\frac {\sin \left (6\,c+6\,d\,x\right )}{192\,b\,d}+\frac {a\,\cos \left (3\,c+3\,d\,x\right )}{16\,b^2\,d}+\frac {a\,\cos \left (5\,c+5\,d\,x\right )}{80\,b^2\,d}-\frac {5\,a^3\,\cos \left (c+d\,x\right )}{4\,b^4\,d}+\frac {a^5\,\cos \left (c+d\,x\right )}{b^6\,d}+\frac {3\,a^2\,\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\,b^3\,d}-\frac {3\,a^4\,\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^5\,d}+\frac {2\,a^6\,\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^7\,d}-\frac {a^3\,\cos \left (3\,c+3\,d\,x\right )}{12\,b^4\,d}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^3\,d}+\frac {a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,b^3\,d}-\frac {a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^5\,d}+\frac {a\,\cos \left (c+d\,x\right )}{8\,b^2\,d}-\frac {2\,a^3\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{b^7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + b*sin(c + d*x)),x)

[Out]

atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(8*b*d) + sin(2*c + 2*d*x)/(64*b*d) - sin(4*c + 4*d*x)/(64*b*d) -
sin(6*c + 6*d*x)/(192*b*d) + (a*cos(3*c + 3*d*x))/(16*b^2*d) + (a*cos(5*c + 5*d*x))/(80*b^2*d) - (5*a^3*cos(c
+ d*x))/(4*b^4*d) + (a^5*cos(c + d*x))/(b^6*d) + (3*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*b^3*d)
- (3*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^5*d) + (2*a^6*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2)))/(b^7*d) - (a^3*cos(3*c + 3*d*x))/(12*b^4*d) + (a^2*sin(2*c + 2*d*x))/(4*b^3*d) + (a^2*sin(4*c + 4*d*x))
/(32*b^3*d) - (a^4*sin(2*c + 2*d*x))/(4*b^5*d) + (a*cos(c + d*x))/(8*b^2*d) - (2*a^3*atanh((2*b^2*sin(c/2 + (d
*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - a^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^
(1/2) + a*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^5*cos(c/2 + (d*x)/2) + 2*b^5*sin(
c/2 + (d*x)/2) + a*b^4*cos(c/2 + (d*x)/2) + 2*a^4*b*sin(c/2 + (d*x)/2) - 2*a^3*b^2*cos(c/2 + (d*x)/2) - 4*a^2*
b^3*sin(c/2 + (d*x)/2)))*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(b^7*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)**4*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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