3.1206 \(\int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=85 \[ \frac{a \sin (c+d x)}{d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{2 a \csc (c+d x)}{d}+\frac{b \sin ^2(c+d x)}{2 d}-\frac{b \csc ^2(c+d x)}{2 d}-\frac{2 b \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0850271, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2837, 12, 766} \[ \frac{a \sin (c+d x)}{d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{2 a \csc (c+d x)}{d}+\frac{b \sin ^2(c+d x)}{2 d}-\frac{b \csc ^2(c+d x)}{2 d}-\frac{2 b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Csc[c + d*x])/d - (b*Csc[c + d*x]^2)/(2*d) - (a*Csc[c + d*x]^3)/(3*d) - (2*b*Log[Sin[c + d*x]])/d + (a*Si
n[c + d*x])/d + (b*Sin[c + d*x]^2)/(2*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 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.145554, size = 76, normalized size = 0.89 \[ \frac{a \sin (c+d x)}{d}-\frac{a \csc ^3(c+d x)}{3 d}+\frac{2 a \csc (c+d x)}{d}-\frac{b \left (-\sin ^2(c+d x)+\csc ^2(c+d x)+4 \log (\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 159, normalized size = 1.9 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}+{\frac{8\,a\sin \left ( dx+c \right ) }{3\,d}}+{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}a}{d}}+{\frac{4\,a\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{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)),x)

[Out]

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

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Maxima [A]  time = 1.00204, size = 93, normalized size = 1.09 \begin{align*} \frac{3 \, b \sin \left (d x + c\right )^{2} - 12 \, b \log \left (\sin \left (d x + c\right )\right ) + 6 \, a \sin \left (d x + c\right ) + \frac{12 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73743, size = 300, normalized size = 3.53 \begin{align*} -\frac{12 \, a \cos \left (d x + c\right )^{4} - 48 \, a \cos \left (d x + c\right )^{2} + 24 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \,{\left (2 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 32 \, a}{12 \,{\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)),x, algorithm="fricas")

[Out]

-1/12*(12*a*cos(d*x + c)^4 - 48*a*cos(d*x + c)^2 + 24*(b*cos(d*x + c)^2 - b)*log(1/2*sin(d*x + c))*sin(d*x + c
) + 3*(2*b*cos(d*x + c)^4 - 3*b*cos(d*x + c)^2 - b)*sin(d*x + c) + 32*a)/((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)),x)

[Out]

Timed out

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Giac [A]  time = 1.19465, size = 109, normalized size = 1.28 \begin{align*} \frac{3 \, b \sin \left (d x + c\right )^{2} - 12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a \sin \left (d x + c\right ) + \frac{22 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, 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)),x, algorithm="giac")

[Out]

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