3.1137 \(\int \frac{\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=173 \[ -\frac{3 a \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}+\frac{3 \cos (c+d x) \left (4 a^2+2 a b \sin (c+d x)-b^2\right )}{2 b^4 d (a+b \sin (c+d x))}+\frac{3 x \left (4 a^2-b^2\right )}{2 b^5}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.274943, antiderivative size = 173, 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 = {2863, 2735, 2660, 618, 204} \[ -\frac{3 a \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}+\frac{3 \cos (c+d x) \left (4 a^2+2 a b \sin (c+d x)-b^2\right )}{2 b^4 d (a+b \sin (c+d x))}+\frac{3 x \left (4 a^2-b^2\right )}{2 b^5}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2863

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 + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[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))^3} \, dx &=\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}-\frac{3 \int \frac{\cos ^2(c+d x) (-2 b-4 a \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx}{4 b^2}\\ &=\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x) \left (4 a^2-b^2+2 a b \sin (c+d x)\right )}{2 b^4 d (a+b \sin (c+d x))}+\frac{3 \int \frac{4 a b+2 \left (4 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{4 b^4}\\ &=\frac{3 \left (4 a^2-b^2\right ) x}{2 b^5}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x) \left (4 a^2-b^2+2 a b \sin (c+d x)\right )}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (3 \left (-4 a b^2+2 a \left (4 a^2-b^2\right )\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{4 b^5}\\ &=\frac{3 \left (4 a^2-b^2\right ) x}{2 b^5}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x) \left (4 a^2-b^2+2 a b \sin (c+d x)\right )}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (3 a \left (4 a^2-3 b^2\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^5 d}\\ &=\frac{3 \left (4 a^2-b^2\right ) x}{2 b^5}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x) \left (4 a^2-b^2+2 a b \sin (c+d x)\right )}{2 b^4 d (a+b \sin (c+d x))}+\frac{\left (6 a \left (4 a^2-3 b^2\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^5 d}\\ &=\frac{3 \left (4 a^2-b^2\right ) x}{2 b^5}-\frac{3 a \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^5 \sqrt{a^2-b^2} d}+\frac{\cos ^3(c+d x) (2 a+b \sin (c+d x))}{2 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x) \left (4 a^2-b^2+2 a b \sin (c+d x)\right )}{2 b^4 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.47278, size = 274, normalized size = 1.58 \[ \frac{\frac{72 a^2 b^2 \sin (2 (c+d x))+12 b^2 \left (b^2-4 a^2\right ) (c+d x) \cos (2 (c+d x))+24 a^2 b^2 c+24 a^2 b^2 d x+192 a^3 b c \sin (c+d x)+192 a^3 b d x \sin (c+d x)+96 a^3 b \cos (c+d x)+96 a^4 c+96 a^4 d x-48 a b^3 c \sin (c+d x)-48 a b^3 d x \sin (c+d x)-8 a b^3 \cos (3 (c+d x))-10 b^4 \sin (2 (c+d x))+b^4 \sin (4 (c+d x))-12 b^4 c-12 b^4 d x}{(a+b \sin (c+d x))^2}-\frac{48 a \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{16 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-48*a*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (96*a^4*c + 24*a^2
*b^2*c - 12*b^4*c + 96*a^4*d*x + 24*a^2*b^2*d*x - 12*b^4*d*x + 96*a^3*b*Cos[c + d*x] + 12*b^2*(-4*a^2 + b^2)*(
c + d*x)*Cos[2*(c + d*x)] - 8*a*b^3*Cos[3*(c + d*x)] + 192*a^3*b*c*Sin[c + d*x] - 48*a*b^3*c*Sin[c + d*x] + 19
2*a^3*b*d*x*Sin[c + d*x] - 48*a*b^3*d*x*Sin[c + d*x] + 72*a^2*b^2*Sin[2*(c + d*x)] - 10*b^4*Sin[2*(c + d*x)] +
b^4*Sin[4*(c + d*x)])/(a + b*Sin[c + d*x])^2)/(16*b^5*d)

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Maple [B]  time = 0.139, size = 639, normalized size = 3.7 \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))^3,x)

[Out]

1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^
2*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*a+12/d/b^5*arctan
(tan(1/2*d*x+1/2*c))*a^2-3/d/b^3*arctan(tan(1/2*d*x+1/2*c))+5/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)^2*tan(1/2*d*x+1/2*c)^3+6/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*tan(1/2*d*x
+1/2*c)^2+11/d*a/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-2/d/(tan(1/2*d*x
+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^2+19/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)^2*tan(1/2*d*x+1/2*c)-4/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2
*c)+6/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)^2-1/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*
d*x+1/2*c)*b+a)^2*a-12/d*a^3/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))+9/d*
a/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))

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35628, size = 1800, normalized size = 10.4 \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)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(6*(4*a^4*b^2 - 5*a^2*b^4 + b^6)*d*x*cos(d*x + c)^2 + 8*(a^3*b^3 - a*b^5)*cos(d*x + c)^3 - 6*(4*a^6 - a^4
*b^2 - 4*a^2*b^4 + b^6)*d*x - 3*(4*a^5 + a^3*b^2 - 3*a*b^4 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(4*a^4*b
- 3*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)) - 6*(4*a^5*b - 3*a^3*b^3 - a*b^5)*cos(d*x + c) - 2*((a^2*b^4 - b^6)*cos(d*x + c)^3 + 6*(4*a^5*
b - 5*a^3*b^3 + a*b^5)*d*x + 3*(6*a^4*b^2 - 7*a^2*b^4 + b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^7 - b^9)*d*co
s(d*x + c)^2 - 2*(a^3*b^6 - a*b^8)*d*sin(d*x + c) - (a^4*b^5 - b^9)*d), 1/2*(3*(4*a^4*b^2 - 5*a^2*b^4 + b^6)*d
*x*cos(d*x + c)^2 + 4*(a^3*b^3 - a*b^5)*cos(d*x + c)^3 - 3*(4*a^6 - a^4*b^2 - 4*a^2*b^4 + b^6)*d*x - 3*(4*a^5
+ a^3*b^2 - 3*a*b^4 - (4*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^2 + 2*(4*a^4*b - 3*a^2*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*(4*a^5*b - 3*a^3*b^3 - a*b^5)*cos(d*x +
c) - ((a^2*b^4 - b^6)*cos(d*x + c)^3 + 6*(4*a^5*b - 5*a^3*b^3 + a*b^5)*d*x + 3*(6*a^4*b^2 - 7*a^2*b^4 + b^6)*c
os(d*x + c))*sin(d*x + c))/((a^2*b^7 - b^9)*d*cos(d*x + c)^2 - 2*(a^3*b^6 - a*b^8)*d*sin(d*x + c) - (a^4*b^5 -
b^9)*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.24212, size = 579, normalized size = 3.35 \begin{align*} \frac{\frac{3 \,{\left (4 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{b^{5}} - \frac{6 \,{\left (4 \, a^{3} - 3 \, a b^{2}\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 (6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 15 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 54 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 45 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 90 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 29 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 42 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, a^{4} - a^{2} b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, 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 )}^{2} a b^{4}}}{2 \, 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))^3,x, algorithm="giac")

[Out]

1/2*(3*(4*a^2 - b^2)*(d*x + c)/b^5 - 6*(4*a^3 - 3*a*b^2)*(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*(6*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 12*a^4
*tan(1/2*d*x + 1/2*c)^6 + 15*a^2*b^2*tan(1/2*d*x + 1/2*c)^6 - 2*b^4*tan(1/2*d*x + 1/2*c)^6 + 54*a^3*b*tan(1/2*
d*x + 1/2*c)^5 + 36*a^4*tan(1/2*d*x + 1/2*c)^4 + 45*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 - 4*b^4*tan(1/2*d*x + 1/2*c
)^4 + 90*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*a^4*tan(1/2*d*x + 1/2*c)^2 + 29*a
^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 2*b^4*tan(1/2*d*x + 1/2*c)^2 + 42*a^3*b*tan(1/2*d*x + 1/2*c) - 4*a*b^3*tan(1/2
*d*x + 1/2*c) + 12*a^4 - a^2*b^2)/((a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x +
1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a*b^4))/d