3.1098 \(\int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=94 \[ -\frac{3 a \cos (c+d x)}{2 d}-\frac{a \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 b \cot (c+d x)}{2 d}+\frac{b \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{3 b x}{2} \]

[Out]

(-3*b*x)/2 + (3*a*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a*Cos[c + d*x])/(2*d) - (3*b*Cot[c + d*x])/(2*d) + (b*Cos[
c + d*x]^2*Cot[c + d*x])/(2*d) - (a*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.121987, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2838, 2592, 288, 321, 206, 2591, 203} \[ -\frac{3 a \cos (c+d x)}{2 d}-\frac{a \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 b \cot (c+d x)}{2 d}+\frac{b \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{3 b x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*x)/2 + (3*a*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a*Cos[c + d*x])/(2*d) - (3*b*Cot[c + d*x])/(2*d) + (b*Cos[
c + d*x]^2*Cot[c + d*x])/(2*d) - (a*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d)

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 1.53349, size = 132, normalized size = 1.4 \[ -\frac{a \cos (c+d x)}{d}-\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{3 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{3 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{3 b (c+d x)}{2 d}-\frac{b \sin (2 (c+d x))}{4 d}-\frac{b \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(c + d*x))/(2*d) - (a*Cos[c + d*x])/d - (b*Cot[c + d*x])/d - (a*Csc[(c + d*x)/2]^2)/(8*d) + (3*a*Log[Cos
[(c + d*x)/2]])/(2*d) - (3*a*Log[Sin[(c + d*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d) - (b*Sin[2*(c + d*x)]
)/(4*d)

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Maple [A]  time = 0.056, size = 143, normalized size = 1.5 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{3\,\cos \left ( dx+c \right ) a}{2\,d}}-{\frac{3\,a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-{\frac{3\,b\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,bx}{2}}-{\frac{3\,cb}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c)),x)

[Out]

-1/2/d*a/sin(d*x+c)^2*cos(d*x+c)^5-1/2*a*cos(d*x+c)^3/d-3/2*a*cos(d*x+c)/d-3/2/d*a*ln(csc(d*x+c)-cot(d*x+c))-1
/d*b/sin(d*x+c)*cos(d*x+c)^5-b*cos(d*x+c)^3*sin(d*x+c)/d-3/2*b*cos(d*x+c)*sin(d*x+c)/d-3/2*b*x-3/2*b*c/d

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Maxima [A]  time = 1.71044, size = 136, normalized size = 1.45 \begin{align*} -\frac{2 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} b - a{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(2*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*b - a*(2*cos(d*x + c)/(cos(d*x
+ c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.83968, size = 365, normalized size = 3.88 \begin{align*} -\frac{6 \, b d x \cos \left (d x + c\right )^{2} + 4 \, a \cos \left (d x + c\right )^{3} - 6 \, b d x - 6 \, a \cos \left (d x + c\right ) - 3 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (b \cos \left (d x + c\right )^{3} - 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(6*b*d*x*cos(d*x + c)^2 + 4*a*cos(d*x + c)^3 - 6*b*d*x - 6*a*cos(d*x + c) - 3*(a*cos(d*x + c)^2 - a)*log(
1/2*cos(d*x + c) + 1/2) + 3*(a*cos(d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2) + 2*(b*cos(d*x + c)^3 - 3*b*co
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - 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*csc(d*x+c)**3*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.44834, size = 220, normalized size = 2.34 \begin{align*} \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \,{\left (d x + c\right )} b - 12 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{6 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(a*tan(1/2*d*x + 1/2*c)^2 - 12*(d*x + c)*b - 12*a*log(abs(tan(1/2*d*x + 1/2*c))) + 4*b*tan(1/2*d*x + 1/2*c
) + (6*a*tan(1/2*d*x + 1/2*c)^6 + 4*b*tan(1/2*d*x + 1/2*c)^5 - 5*a*tan(1/2*d*x + 1/2*c)^4 - 16*b*tan(1/2*d*x +
1/2*c)^3 - 12*a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c) - a)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x
+ 1/2*c))^2)/d