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

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

[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)

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Rubi [A]  time = 0.324701, antiderivative size = 159, normalized size of antiderivative = 1., 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 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]

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 (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*}

Mathematica [A]  time = 1.0446, size = 155, normalized size = 0.97 \[ \frac{3 \left (4 \left (-12 a^2 b^2+8 a^4+3 b^4\right ) (c+d x)+\left (8 b^4-8 a^2 b^2\right ) \sin (2 (c+d x))+b^4 \sin (4 (c+d x))\right )+24 a b \left (4 a^2-5 b^2\right ) \cos (c+d x)-192 a \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-8 a b^3 \cos (3 (c+d x))}{96 b^5 d} \]

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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Maple [B]  time = 0.087, size = 760, normalized size = 4.8 \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)/(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/b*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))

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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)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64871, size = 945, normalized size = 5.94 \begin{align*} \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 ] \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.20264, size = 501, normalized size = 3.15 \begin{align*} \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}\left (a\right ) + \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} \end{align*}

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