3.1129 \(\int \frac{\cos ^4(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=163 \[ \frac{2 \left (-5 a^2 b^2+4 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}-\frac{\cos (c+d x) \left (4 a^2-2 a b \sin (c+d x)-b^2\right )}{b^4 d}-\frac{a x \left (4 a^2-3 b^2\right )}{b^5}+\frac{\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.30615, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2863, 2865, 2735, 2660, 618, 204} \[ \frac{2 \left (-5 a^2 b^2+4 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \sqrt{a^2-b^2}}-\frac{\cos (c+d x) \left (4 a^2-2 a b \sin (c+d x)-b^2\right )}{b^4 d}-\frac{a x \left (4 a^2-3 b^2\right )}{b^5}+\frac{\cos ^3(c+d x) (4 a+b \sin (c+d x))}{3 b^2 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 2.1696, size = 247, normalized size = 1.52 \[ \frac{\frac{48 \left (-5 a^2 b^2+4 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{-24 a^2 b^2 \sin (2 (c+d x))+\left (60 a b^3-96 a^3 b\right ) \cos (c+d x)+72 a^2 b^2 c+72 a^2 b^2 d x-96 a^3 b c \sin (c+d x)-96 a^3 b d x \sin (c+d x)-96 a^4 c-96 a^4 d x+72 a b^3 c \sin (c+d x)+72 a b^3 d x \sin (c+d x)-4 a b^3 \cos (3 (c+d x))+14 b^4 \sin (2 (c+d x))+b^4 \sin (4 (c+d x))}{a+b \sin (c+d x)}}{24 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*(4*a^4 - 5*a^2*b^2 + b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (-96*a^4*c
+ 72*a^2*b^2*c - 96*a^4*d*x + 72*a^2*b^2*d*x + (-96*a^3*b + 60*a*b^3)*Cos[c + d*x] - 4*a*b^3*Cos[3*(c + d*x)]
- 96*a^3*b*c*Sin[c + d*x] + 72*a*b^3*c*Sin[c + d*x] - 96*a^3*b*d*x*Sin[c + d*x] + 72*a*b^3*d*x*Sin[c + d*x] -
24*a^2*b^2*Sin[2*(c + d*x)] + 14*b^4*Sin[2*(c + d*x)] + b^4*Sin[4*(c + d*x)])/(a + b*Sin[c + d*x]))/(24*b^5*d)

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Maple [B]  time = 0.119, size = 627, normalized size = 3.9 \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))^2,x)

[Out]

-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)^4*a^2+4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4-12/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)^2*a^2+4/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*t
an(1/2*d*x+1/2*c)-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2+8/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3-8/d/b^5*arctan(t
an(1/2*d*x+1/2*c))*a^3+6/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a-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)+2/d/b/(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^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/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)
*b+a)*a+8/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))-10/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))+2/d/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., 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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07301, size = 1141, normalized size = 7. \begin{align*} \left [-\frac{4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \,{\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} d x + 3 \,{\left (4 \, a^{3} - a b^{2} +{\left (4 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\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 ) + 6 \,{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) - 2 \,{\left (b^{4} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} d x - 3 \,{\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}}, -\frac{2 \, a b^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} d x + 3 \,{\left (4 \, a^{3} - a b^{2} +{\left (4 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\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 ) + 3 \,{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) -{\left (b^{4} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} d x - 3 \,{\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}}\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))^2,x, algorithm="fricas")

[Out]

[-1/6*(4*a*b^3*cos(d*x + c)^3 + 6*(4*a^4 - 3*a^2*b^2)*d*x + 3*(4*a^3 - a*b^2 + (4*a^2*b - b^3)*sin(d*x + c))*s
qrt(-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^3*b
- 3*a*b^3)*cos(d*x + c) - 2*(b^4*cos(d*x + c)^3 - 3*(4*a^3*b - 3*a*b^3)*d*x - 3*(2*a^2*b^2 - b^4)*cos(d*x + c)
)*sin(d*x + c))/(b^6*d*sin(d*x + c) + a*b^5*d), -1/3*(2*a*b^3*cos(d*x + c)^3 + 3*(4*a^4 - 3*a^2*b^2)*d*x + 3*(
4*a^3 - a*b^2 + (4*a^2*b - b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*co
s(d*x + c))) + 3*(4*a^3*b - 3*a*b^3)*cos(d*x + c) - (b^4*cos(d*x + c)^3 - 3*(4*a^3*b - 3*a*b^3)*d*x - 3*(2*a^2
*b^2 - b^4)*cos(d*x + c))*sin(d*x + c))/(b^6*d*sin(d*x + c) + a*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20063, size = 405, normalized size = 2.48 \begin{align*} -\frac{\frac{3 \,{\left (4 \, a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} - \frac{6 \,{\left (4 \, a^{4} - 5 \, a^{2} b^{2} + 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{6 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} - a b^{2}\right )}}{{\left (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 )} b^{4}} + \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} - 4 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, 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))^2,x, algorithm="giac")

[Out]

-1/3*(3*(4*a^3 - 3*a*b^2)*(d*x + c)/b^5 - 6*(4*a^4 - 5*a^2*b^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^5) + 6*(a^2*b*tan(1/2*d*x + 1/2*c)
- b^3*tan(1/2*d*x + 1/2*c) + a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*b^4) + 2
*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^4 - 6*b^2*tan(1/2*d*x + 1/2*c)^4 + 18*a^2*tan(1/2*
d*x + 1/2*c)^2 - 6*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*a^2 - 4*b^2)/((tan(1/2*d*x + 1/
2*c)^2 + 1)^3*b^4))/d