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

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Rubi [A]  time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [A]  time = 0.67, size = 117, normalized size = 1.38 \[ -\frac {12 \, a \cos \left (d x + c\right )^{4} - 48 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 32 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.21, size = 81, normalized size = 0.95 \[ \frac {3 \, b \sin \left (d x + c\right )^{2} - 12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a \sin \left (d x + c\right ) + \frac {22 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, d} \]

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

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maple [A]  time = 0.38, size = 159, normalized size = 1.87 \[ -\frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {8 a \sin \left (d x +c \right )}{3 d}+\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{d}+\frac {4 a \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}-\frac {b \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {b \left (\cos ^{4}\left (d x +c \right )\right )}{2 d}-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {2 b \ln \left (\sin \left (d x +c \right )\right )}{d} \]

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*b*cos(d*x+c)^4/d-b*cos(d*x+c)^2/d-2*
b*ln(sin(d*x+c))/d

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maxima [A]  time = 0.33, size = 69, normalized size = 0.81 \[ \frac {3 \, b \sin \left (d x + c\right )^{2} - 12 \, b \log \left (\sin \left (d x + c\right )\right ) + 6 \, a \sin \left (d x + c\right ) + \frac {12 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a}{\sin \left (d x + c\right )^{3}}}{6 \, d} \]

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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mupad [B]  time = 11.91, size = 218, normalized size = 2.56 \[ \frac {23\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {89\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {7\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {2\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((19*a*tan(c/2 + (d*x)/2)^2)/3 - b*tan(c/2 + (d*x)/2) - a/3 + (89*a*tan(c/2 + (d*x)/2)^4)/3 + 23*a*tan(c/2 + (
d*x)/2)^6 - 2*b*tan(c/2 + (d*x)/2)^3 + 15*b*tan(c/2 + (d*x)/2)^5)/(d*(8*tan(c/2 + (d*x)/2)^3 + 16*tan(c/2 + (d
*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^7)) + (7*a*tan(c/2 + (d*x)/2))/(8*d) + (2*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d -
(a*tan(c/2 + (d*x)/2)^3)/(24*d) - (b*tan(c/2 + (d*x)/2)^2)/(8*d) - (2*b*log(tan(c/2 + (d*x)/2)))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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