3.686 \(\int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ -\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d}+\frac {2 \sin (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\csc (c+d x)}{a d}-\frac {2 \log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ -\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d}+\frac {2 \sin (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\csc (c+d x)}{a d}-\frac {2 \log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) - (2*Log[Sin[c + d*x]])/(a*d) + (2*Sin[c + d*x])/(a*d) + Sin[c + d
*x]^2/(2*a*d) - Sin[c + d*x]^3/(3*a*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^3 (a-x)^3 (a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^3 (a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 a^2+\frac {a^5}{x^3}-\frac {a^4}{x^2}-\frac {2 a^3}{x}+a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {2 \log (\sin (c+d x))}{a d}+\frac {2 \sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 66, normalized size = 0.68 \[ \frac {-2 \sin ^3(c+d x)+3 \sin ^2(c+d x)+12 \sin (c+d x)-3 \csc ^2(c+d x)+6 \csc (c+d x)-12 \log (\sin (c+d x))}{6 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

(6*Csc[c + d*x] - 3*Csc[c + d*x]^2 - 12*Log[Sin[c + d*x]] + 12*Sin[c + d*x] + 3*Sin[c + d*x]^2 - 2*Sin[c + d*x
]^3)/(6*a*d)

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fricas [A]  time = 0.51, size = 91, normalized size = 0.94 \[ -\frac {6 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*cos(d*x + c)^4 - 9*cos(d*x + c)^2 + 24*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 4*(cos(d*x + c)^4
+ 4*cos(d*x + c)^2 - 8)*sin(d*x + c) - 3)/(a*d*cos(d*x + c)^2 - a*d)

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giac [A]  time = 0.21, size = 94, normalized size = 0.97 \[ -\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \sin \left (d x + c\right )}{a^{3}} - \frac {3 \, {\left (6 \, \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.47, size = 94, normalized size = 0.97 \[ -\frac {\sin ^{3}\left (d x +c \right )}{3 d a}+\frac {\sin ^{2}\left (d x +c \right )}{2 a d}+\frac {2 \sin \left (d x +c \right )}{a d}+\frac {1}{d a \sin \left (d x +c \right )}-\frac {2 \ln \left (\sin \left (d x +c \right )\right )}{a d}-\frac {1}{2 a d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.36, size = 74, normalized size = 0.76 \[ -\frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {3 \, {\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.23, size = 231, normalized size = 2.38 \[ \frac {18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7/(sin(c + d*x)^3*(a + a*sin(c + d*x))),x)

[Out]

(2*tan(c/2 + (d*x)/2) - (3*tan(c/2 + (d*x)/2)^2)/2 + 22*tan(c/2 + (d*x)/2)^3 + (13*tan(c/2 + (d*x)/2)^4)/2 + (
82*tan(c/2 + (d*x)/2)^5)/3 + (15*tan(c/2 + (d*x)/2)^6)/2 + 18*tan(c/2 + (d*x)/2)^7 - 1/2)/(d*(4*a*tan(c/2 + (d
*x)/2)^2 + 12*a*tan(c/2 + (d*x)/2)^4 + 12*a*tan(c/2 + (d*x)/2)^6 + 4*a*tan(c/2 + (d*x)/2)^8)) - (2*log(tan(c/2
+ (d*x)/2)))/(a*d) - tan(c/2 + (d*x)/2)^2/(8*a*d) + tan(c/2 + (d*x)/2)/(2*a*d) + (2*log(tan(c/2 + (d*x)/2)^2
+ 1))/(a*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)**7*csc(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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