3.1232 \(\int \frac{\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=147 \[ -\frac{\left (a^2-b^2\right )^2}{a^4 b d (a+b \sin (c+d x))}+\frac{\left (2 a^2-3 b^2\right ) \csc (c+d x)}{a^4 d}+\frac{4 b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^5 d}-\frac{4 b \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}+\frac{b \csc ^2(c+d x)}{a^3 d}-\frac{\csc ^3(c+d x)}{3 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.205735, antiderivative size = 147, 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, 894} \[ -\frac{\left (a^2-b^2\right )^2}{a^4 b d (a+b \sin (c+d x))}+\frac{\left (2 a^2-3 b^2\right ) \csc (c+d x)}{a^4 d}+\frac{4 b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^5 d}-\frac{4 b \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}+\frac{b \csc ^2(c+d x)}{a^3 d}-\frac{\csc ^3(c+d x)}{3 a^2 d} \]

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

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 894

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] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 1.94143, size = 127, normalized size = 0.86 \[ \frac{-\frac{3 a \left (a^2-b^2\right )^2}{b (a+b \sin (c+d x))}+3 a \left (2 a^2-3 b^2\right ) \csc (c+d x)+3 a^2 b \csc ^2(c+d x)-a^3 \csc ^3(c+d x)+12 b (a-b) (a+b) \log (\sin (c+d x))-12 b (a-b) (a+b) \log (a+b \sin (c+d x))}{3 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(2*a^2 - 3*b^2)*Csc[c + d*x] + 3*a^2*b*Csc[c + d*x]^2 - a^3*Csc[c + d*x]^3 + 12*(a - b)*b*(a + b)*Log[Sin
[c + d*x]] - 12*(a - b)*b*(a + b)*Log[a + b*Sin[c + d*x]] - (3*a*(a^2 - b^2)^2)/(b*(a + b*Sin[c + d*x])))/(3*a
^5*d)

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Maple [A]  time = 0.152, size = 209, normalized size = 1.4 \begin{align*} -{\frac{1}{bd \left ( a+b\sin \left ( dx+c \right ) \right ) }}+2\,{\frac{b}{d{a}^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{{b}^{3}}{d{a}^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-4\,{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}+4\,{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{5}}}-{\frac{1}{3\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+2\,{\frac{1}{d{a}^{2}\sin \left ( dx+c \right ) }}-3\,{\frac{{b}^{2}}{d{a}^{4}\sin \left ( dx+c \right ) }}+{\frac{b}{d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}-4\,{\frac{{b}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{5}}} \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/b/d/(a+b*sin(d*x+c))+2/d*b/a^2/(a+b*sin(d*x+c))-1/d/a^4*b^3/(a+b*sin(d*x+c))-4/d/a^3*b*ln(a+b*sin(d*x+c))+4
/d*b^3/a^5*ln(a+b*sin(d*x+c))-1/3/d/a^2/sin(d*x+c)^3+2/d/a^2/sin(d*x+c)-3/d/a^4/sin(d*x+c)*b^2+1/d/a^3*b/sin(d
*x+c)^2+4*b*ln(sin(d*x+c))/a^3/d-4/d*b^3/a^5*ln(sin(d*x+c))

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Maxima [A]  time = 1.02349, size = 213, normalized size = 1.45 \begin{align*} \frac{\frac{2 \, a^{2} b^{2} \sin \left (d x + c\right ) - a^{3} b - 3 \,{\left (a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{4}\right )} \sin \left (d x + c\right )^{3} + 6 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2}}{a^{4} b^{2} \sin \left (d x + c\right )^{4} + a^{5} b \sin \left (d x + c\right )^{3}} - \frac{12 \,{\left (a^{2} b - b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} + \frac{12 \,{\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}}}{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*((2*a^2*b^2*sin(d*x + c) - a^3*b - 3*(a^4 - 4*a^2*b^2 + 4*b^4)*sin(d*x + c)^3 + 6*(a^3*b - a*b^3)*sin(d*x
+ c)^2)/(a^4*b^2*sin(d*x + c)^4 + a^5*b*sin(d*x + c)^3) - 12*(a^2*b - b^3)*log(b*sin(d*x + c) + a)/a^5 + 12*(a
^2*b - b^3)*log(sin(d*x + c))/a^5)/d

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Fricas [B]  time = 2.00698, size = 859, normalized size = 5.84 \begin{align*} \frac{5 \, a^{4} b - 6 \, a^{2} b^{3} - 6 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left (a^{2} b^{3} - b^{5} +{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{3} b^{2} - a b^{4} -{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \,{\left (a^{2} b^{3} - b^{5} +{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{3} b^{2} - a b^{4} -{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) -{\left (3 \, a^{5} - 14 \, a^{3} b^{2} + 12 \, a b^{4} - 3 \,{\left (a^{5} - 4 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{5} b^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b^{2} d \cos \left (d x + c\right )^{2} + a^{5} b^{2} d -{\left (a^{6} b d \cos \left (d x + c\right )^{2} - a^{6} b d\right )} \sin \left (d x + c\right )\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]

1/3*(5*a^4*b - 6*a^2*b^3 - 6*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 12*(a^2*b^3 - b^5 + (a^2*b^3 - b^5)*cos(d*x +
c)^4 - 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 + (a^3*b^2 - a*b^4 - (a^3*b^2 - a*b^4)*cos(d*x + c)^2)*sin(d*x + c))*l
og(b*sin(d*x + c) + a) + 12*(a^2*b^3 - b^5 + (a^2*b^3 - b^5)*cos(d*x + c)^4 - 2*(a^2*b^3 - b^5)*cos(d*x + c)^2
+ (a^3*b^2 - a*b^4 - (a^3*b^2 - a*b^4)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) - (3*a^5 - 14*a^3*
b^2 + 12*a*b^4 - 3*(a^5 - 4*a^3*b^2 + 4*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(a^5*b^2*d*cos(d*x + c)^4 - 2*a^5
*b^2*d*cos(d*x + c)^2 + a^5*b^2*d - (a^6*b*d*cos(d*x + c)^2 - a^6*b*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 = 1.19374, size = 285, normalized size = 1.94 \begin{align*} \frac{\frac{12 \,{\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac{12 \,{\left (a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} + \frac{3 \,{\left (4 \, a^{2} b^{3} \sin \left (d x + c\right ) - 4 \, b^{5} \sin \left (d x + c\right ) - a^{5} + 6 \, a^{3} b^{2} - 5 \, a b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{5} b} - \frac{22 \, a^{2} b \sin \left (d x + c\right )^{3} - 22 \, b^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} b \sin \left (d x + c\right ) + a^{3}}{a^{5} \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*(12*(a^2*b - b^3)*log(abs(sin(d*x + c)))/a^5 - 12*(a^2*b^2 - b^4)*log(abs(b*sin(d*x + c) + a))/(a^5*b) + 3
*(4*a^2*b^3*sin(d*x + c) - 4*b^5*sin(d*x + c) - a^5 + 6*a^3*b^2 - 5*a*b^4)/((b*sin(d*x + c) + a)*a^5*b) - (22*
a^2*b*sin(d*x + c)^3 - 22*b^3*sin(d*x + c)^3 - 6*a^3*sin(d*x + c)^2 + 9*a*b^2*sin(d*x + c)^2 - 3*a^2*b*sin(d*x
+ c) + a^3)/(a^5*sin(d*x + c)^3))/d