3.1220 \(\int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=120 \[ \frac{\left (a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac{\left (2 a^2-b^2\right ) \csc (c+d x)}{d}-\frac{a^2 \csc ^3(c+d x)}{3 d}+\frac{a b \sin ^2(c+d x)}{d}-\frac{a b \csc ^2(c+d x)}{d}-\frac{4 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.140823, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 948} \[ \frac{\left (a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac{\left (2 a^2-b^2\right ) \csc (c+d x)}{d}-\frac{a^2 \csc ^3(c+d x)}{3 d}+\frac{a b \sin ^2(c+d x)}{d}-\frac{a b \csc ^2(c+d x)}{d}-\frac{4 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 0.274211, size = 103, normalized size = 0.86 \[ \frac{3 \left (a^2-2 b^2\right ) \sin (c+d x)+\left (6 a^2-3 b^2\right ) \csc (c+d x)-a^2 \csc ^3(c+d x)+3 a b \sin ^2(c+d x)-3 a b \csc ^2(c+d x)-12 a b \log (\sin (c+d x))+b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2 - 3*b^2)*Csc[c + d*x] - 3*a*b*Csc[c + d*x]^2 - a^2*Csc[c + d*x]^3 - 12*a*b*Log[Sin[c + d*x]] + 3*(a^2
- 2*b^2)*Sin[c + d*x] + 3*a*b*Sin[c + d*x]^2 + b^2*Sin[c + d*x]^3)/(3*d)

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Maple [B]  time = 0.087, size = 255, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}+{\frac{8\,{a}^{2}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{4\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-4\,{\frac{ab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{8\,{b}^{2}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{4\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4*(a+b*sin(d*x+c))^2,x)

[Out]

-1/3/d*a^2/sin(d*x+c)^3*cos(d*x+c)^6+1/d*a^2/sin(d*x+c)*cos(d*x+c)^6+8/3*a^2*sin(d*x+c)/d+1/d*a^2*sin(d*x+c)*c
os(d*x+c)^4+4/3/d*a^2*sin(d*x+c)*cos(d*x+c)^2-1/d*a*b/sin(d*x+c)^2*cos(d*x+c)^6-1/d*a*b*cos(d*x+c)^4-2/d*a*b*c
os(d*x+c)^2-4*a*b*ln(sin(d*x+c))/d-1/d*b^2/sin(d*x+c)*cos(d*x+c)^6-8/3*b^2*sin(d*x+c)/d-1/d*sin(d*x+c)*b^2*cos
(d*x+c)^4-4/3/d*sin(d*x+c)*b^2*cos(d*x+c)^2

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Maxima [A]  time = 0.992184, size = 139, normalized size = 1.16 \begin{align*} \frac{b^{2} \sin \left (d x + c\right )^{3} + 3 \, a b \sin \left (d x + c\right )^{2} - 12 \, a b \log \left (\sin \left (d x + c\right )\right ) + 3 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac{3 \, a b \sin \left (d x + c\right ) - 3 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(b^2*sin(d*x + c)^3 + 3*a*b*sin(d*x + c)^2 - 12*a*b*log(sin(d*x + c)) + 3*(a^2 - 2*b^2)*sin(d*x + c) - (3*
a*b*sin(d*x + c) - 3*(2*a^2 - b^2)*sin(d*x + c)^2 + a^2)/sin(d*x + c)^3)/d

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Fricas [A]  time = 1.77982, size = 382, normalized size = 3.18 \begin{align*} \frac{2 \, b^{2} \cos \left (d x + c\right )^{6} - 6 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 24 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 24 \,{\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 16 \, a^{2} + 16 \, b^{2} - 3 \,{\left (2 \, a b \cos \left (d x + c\right )^{4} - 3 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(2*b^2*cos(d*x + c)^6 - 6*(a^2 - b^2)*cos(d*x + c)^4 + 24*(a^2 - b^2)*cos(d*x + c)^2 - 24*(a*b*cos(d*x + c
)^2 - a*b)*log(1/2*sin(d*x + c))*sin(d*x + c) - 16*a^2 + 16*b^2 - 3*(2*a*b*cos(d*x + c)^4 - 3*a*b*cos(d*x + c)
^2 - a*b)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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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)**5*csc(d*x+c)**4*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19856, size = 171, normalized size = 1.42 \begin{align*} \frac{b^{2} \sin \left (d x + c\right )^{3} + 3 \, a b \sin \left (d x + c\right )^{2} - 12 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 3 \, a^{2} \sin \left (d x + c\right ) - 6 \, b^{2} \sin \left (d x + c\right ) + \frac{22 \, a b \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} - 3 \, a b \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(b^2*sin(d*x + c)^3 + 3*a*b*sin(d*x + c)^2 - 12*a*b*log(abs(sin(d*x + c))) + 3*a^2*sin(d*x + c) - 6*b^2*si
n(d*x + c) + (22*a*b*sin(d*x + c)^3 + 6*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^2 - 3*a*b*sin(d*x + c) - a^2)/
sin(d*x + c)^3)/d