3.1303 \(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=159 \[ -\frac {2 a \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}+\frac {x \left (8 a^4-12 a^2 b^2+3 b^4\right )}{8 b^5}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d} \]

[Out]

1/8*(8*a^4-12*a^2*b^2+3*b^4)*x/b^5-2*a*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^5/d-
1/12*cos(d*x+c)^3*(4*a-3*b*sin(d*x+c))/b^2/d+1/8*cos(d*x+c)*(8*a*(a^2-b^2)-b*(4*a^2-3*b^2)*sin(d*x+c))/b^4/d

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Rubi [A]  time = 0.32, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2865, 2735, 2660, 618, 204} \[ -\frac {2 a \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}+\frac {x \left (-12 a^2 b^2+8 a^4+3 b^4\right )}{8 b^5}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4 - 12*a^2*b^2 + 3*b^4)*x)/(8*b^5) - (2*a*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/(b^5*d) - (Cos[c + d*x]^3*(4*a - 3*b*Sin[c + d*x]))/(12*b^2*d) + (Cos[c + d*x]*(8*a*(a^2 - b^2) - b*(4*a
^2 - 3*b^2)*Sin[c + d*x]))/(8*b^4*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 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\int \frac {\cos ^2(c+d x) \left (-a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 b^2}\\ &=-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}+\frac {\int \frac {a b \left (4 a^2-5 b^2\right )+\left (8 a^4-12 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{8 b^4}\\ &=\frac {\left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}-\frac {\left (a \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^5}\\ &=\frac {\left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}-\frac {\left (2 a \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^5 d}\\ &=\frac {\left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}+\frac {\left (4 a \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^5 d}\\ &=\frac {\left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {2 a \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^5 d}-\frac {\cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^2 d}+\frac {\cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^4 d}\\ \end {align*}

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Mathematica [A]  time = 1.08, size = 155, normalized size = 0.97 \[ \frac {24 a b \left (4 a^2-5 b^2\right ) \cos (c+d x)-192 a \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+3 \left (\left (8 b^4-8 a^2 b^2\right ) \sin (2 (c+d x))+4 \left (8 a^4-12 a^2 b^2+3 b^4\right ) (c+d x)+b^4 \sin (4 (c+d x))\right )-8 a b^3 \cos (3 (c+d x))}{96 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-192*a*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 24*a*b*(4*a^2 - 5*b^2)*Cos[c + d*
x] - 8*a*b^3*Cos[3*(c + d*x)] + 3*(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)]))/(96*b^5*d)

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fricas [A]  time = 0.71, size = 414, normalized size = 2.60 \[ \left [-\frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} d x + 12 \, {\left (a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}, -\frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} d x - 24 \, {\left (a^{3} - a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(8*a*b^3*cos(d*x + c)^3 - 3*(8*a^4 - 12*a^2*b^2 + 3*b^4)*d*x + 12*(a^3 - a*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)) - 24*(a^3*b - a*b^3)*cos(d*x + c)
- 3*(2*b^4*cos(d*x + c)^3 - (4*a^2*b^2 - 3*b^4)*cos(d*x + c))*sin(d*x + c))/(b^5*d), -1/24*(8*a*b^3*cos(d*x +
c)^3 - 3*(8*a^4 - 12*a^2*b^2 + 3*b^4)*d*x - 24*(a^3 - a*b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqr
t(a^2 - b^2)*cos(d*x + c))) - 24*(a^3*b - a*b^3)*cos(d*x + c) - 3*(2*b^4*cos(d*x + c)^3 - (4*a^2*b^2 - 3*b^4)*
cos(d*x + c))*sin(d*x + c))/(b^5*d)]

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giac [B]  time = 0.17, size = 371, normalized size = 2.33 \[ \frac {\frac {3 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {48 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 48 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 96 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} - 32 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*a^4 - 12*a^2*b^2 + 3*b^4)*(d*x + c)/b^5 - 48*(a^5 - 2*a^3*b^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^5) + 2*(12*a^2*b*tan(1
/2*d*x + 1/2*c)^7 - 15*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^3*tan(1/2*d*x + 1/2*c)^6 - 48*a*b^2*tan(1/2*d*x + 1/2
*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^4 - 96*a*
b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 9*b^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^3*tan(1/2*d
*x + 1/2*c)^2 - 80*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 15*b^3*tan(1/2*d*x + 1/2*c)
+ 24*a^3 - 32*a*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d

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maple [B]  time = 0.25, size = 760, normalized size = 4.78 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {20 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{d \,b^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b}-\frac {2 a^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{5} \sqrt {a^{2}-b^{2}}}+\frac {4 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/b^3/(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/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)^7+2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*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)^4*tan(1/2*d
*x+1/2*c)^6*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a^2+3/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*t
an(1/2*d*x+1/2*c)^5+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*a^3-8/d/b^2/(1+tan(1/2*d*x+1/2*c)^
2)^4*tan(1/2*d*x+1/2*c)^4*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a^2-3/4/d/b/(1+tan(1/2*d*x
+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a^3-20/3/d/b^2/(1+ta
n(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^2+5/4/d/b
/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*a^3-8/3/d/b^2/(1+tan(1/2*d*x
+1/2*c)^2)^4*a+2/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4-3/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+3/4/d/b*arctan(ta
n(1/2*d*x+1/2*c))-2/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))+4/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))-2/d*a/b/(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))

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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)/(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 = 12.95, size = 453, normalized size = 2.85 \[ \frac {3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,b\,d}+\frac {\sin \left (2\,c+2\,d\,x\right )}{4\,b\,d}+\frac {\sin \left (4\,c+4\,d\,x\right )}{32\,b\,d}-\frac {a\,\cos \left (3\,c+3\,d\,x\right )}{12\,b^2\,d}+\frac {a^3\,\cos \left (c+d\,x\right )}{b^4\,d}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d}+\frac {2\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^5\,d}-\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^3\,d}-\frac {5\,a\,\cos \left (c+d\,x\right )}{4\,b^2\,d}-\frac {2\,a\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{b^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*b*d) + sin(2*c + 2*d*x)/(4*b*d) + sin(4*c + 4*d*x)/(32*b*d)
- (a*cos(3*c + 3*d*x))/(12*b^2*d) + (a^3*cos(c + d*x))/(b^4*d) - (3*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/(b^3*d) + (2*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^5*d) - (a^2*sin(2*c + 2*d*x))/(4*b^3*
d) - (5*a*cos(c + d*x))/(4*b^2*d) - (2*a*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^5*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)/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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