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

[Out]

(2*a^2-3*b^2)*csc(d*x+c)/a^4/d+b*csc(d*x+c)^2/a^3/d-1/3*csc(d*x+c)^3/a^2/d+4*b*(a^2-b^2)*ln(sin(d*x+c))/a^5/d-
4*b*(a^2-b^2)*ln(a+b*sin(d*x+c))/a^5/d-(a^2-b^2)^2/a^4/b/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.21, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 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]))

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 \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*}

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Mathematica [A]  time = 1.90, size = 127, normalized size = 0.86 \[ \frac {-a^3 \csc ^3(c+d x)-\frac {3 a \left (a^2-b^2\right )^2}{b (a+b \sin (c+d x))}+3 a \left (2 a^2-3 b^2\right ) \csc (c+d x)+3 a^2 b \csc ^2(c+d x)+12 b (a-b) (a+b) \log (\sin (c+d x))-12 b (a-b) (a+b) \log (a+b \sin (c+d x))}{3 a^5 d} \]

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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fricas [B]  time = 0.85, size = 401, normalized size = 2.73 \[ \frac {5 \, a^{4} b - 6 \, a^{2} b^{3} - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4} - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left (a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4} - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (3 \, a^{5} - 14 \, a^{3} b^{2} + 12 \, a b^{4} - 3 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{5} b^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b^{2} d \cos \left (d x + c\right )^{2} + a^{5} b^{2} d - {\left (a^{6} b d \cos \left (d x + c\right )^{2} - a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.25, size = 211, normalized size = 1.44 \[ \frac {\frac {12 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac {12 \, {\left (a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} + \frac {3 \, {\left (4 \, a^{2} b^{3} \sin \left (d x + c\right ) - 4 \, b^{5} \sin \left (d x + c\right ) - a^{5} + 6 \, a^{3} b^{2} - 5 \, a b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{5} b} - \frac {22 \, a^{2} b \sin \left (d x + c\right )^{3} - 22 \, b^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} b \sin \left (d x + c\right ) + a^{3}}{a^{5} \sin \left (d x + c\right )^{3}}}{3 \, d} \]

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

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maple [A]  time = 0.84, size = 209, normalized size = 1.42 \[ -\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 b}{d \,a^{2} \left (a +b \sin \left (d x +c \right )\right )}-\frac {b^{3}}{d \,a^{4} \left (a +b \sin \left (d x +c \right )\right )}-\frac {4 b \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{3}}+\frac {4 b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{5}}-\frac {1}{3 d \,a^{2} \sin \left (d x +c \right )^{3}}+\frac {2}{d \,a^{2} \sin \left (d x +c \right )}-\frac {3 b^{2}}{d \,a^{4} \sin \left (d x +c \right )}+\frac {b}{d \,a^{3} \sin \left (d x +c \right )^{2}}+\frac {4 b \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {4 b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{5}} \]

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 = 0.32, size = 158, normalized size = 1.07 \[ \frac {\frac {2 \, a^{2} b^{2} \sin \left (d x + c\right ) - a^{3} b - 3 \, {\left (a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{4}\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2}}{a^{4} b^{2} \sin \left (d x + c\right )^{4} + a^{5} b \sin \left (d x + c\right )^{3}} - \frac {12 \, {\left (a^{2} b - b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} + \frac {12 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}}}{3 \, d} \]

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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mupad [B]  time = 11.72, size = 319, normalized size = 2.17 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (16\,a^2\,b-24\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (8\,a\,b^2-\frac {20\,a^3}{3}\right )-\frac {a^3}{3}+\frac {4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (23\,a^4-44\,a^2\,b^2+16\,b^4\right )}{a}}{d\,\left (8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,b\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {\frac {a^2}{4}+\frac {b^2}{2}}{a^4}+\frac {5}{8\,a^2}-\frac {2\,b^2}{a^4}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,a^2\,b-4\,b^3\right )}{a^5\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^3\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (4\,a^2\,b-4\,b^3\right )}{a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*(16*a^2*b - 24*b^3) - tan(c/2 + (d*x)/2)^2*(8*a*b^2 - (20*a^3)/3) - a^3/3 + (4*a^2*b*tan
(c/2 + (d*x)/2))/3 + (tan(c/2 + (d*x)/2)^4*(23*a^4 + 16*b^4 - 44*a^2*b^2))/a)/(d*(8*a^5*tan(c/2 + (d*x)/2)^3 +
8*a^5*tan(c/2 + (d*x)/2)^5 + 16*a^4*b*tan(c/2 + (d*x)/2)^4)) - tan(c/2 + (d*x)/2)^3/(24*a^2*d) + (tan(c/2 + (
d*x)/2)*((a^2/4 + b^2/2)/a^4 + 5/(8*a^2) - (2*b^2)/a^4))/d + (log(tan(c/2 + (d*x)/2))*(4*a^2*b - 4*b^3))/(a^5*
d) + (b*tan(c/2 + (d*x)/2)^2)/(4*a^3*d) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(4*a^2*b -
4*b^3))/(a^5*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))**2,x)

[Out]

Timed out

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