3.384 \(\int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=98 \[ \frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ \frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-4 a^6-a^6 \csc (c+d x)+2 a^6 \csc ^2(c+d x)+a^6 \csc ^3(c+d x)-a^6 \sin (c+d x)+2 a^6 \sin ^2(c+d x)+a^6 \sin ^3(c+d x)\right ) \, dx}{a^4}\\ &=-4 a^2 x-a^2 \int \csc (c+d x) \, dx+a^2 \int \csc ^3(c+d x) \, dx-a^2 \int \sin (c+d x) \, dx+a^2 \int \sin ^3(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^2(c+d x) \, dx+\left (2 a^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-4 a^2 x+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{d}+\frac {1}{2} a^2 \int \csc (c+d x) \, dx+a^2 \int 1 \, dx-\frac {a^2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-3 a^2 x+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 2.12, size = 158, normalized size = 1.61 \[ \frac {a^2 (\sin (c+d x)+1)^2 \left (6 \cos (c+d x)+2 \cos (3 (c+d x))+3 \left (-4 \sin (2 (c+d x))+8 \tan \left (\frac {1}{2} (c+d x)\right )-8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-24 c-24 d x\right )\right )}{24 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(6*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + 3*(-24*c - 24*d*x - 8*Cot[(c + d*x)/2] - Csc[
(c + d*x)/2]^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 4*Sin[2*(c + d*x)] +
8*Tan[(c + d*x)/2])))/(24*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [A]  time = 0.58, size = 172, normalized size = 1.76 \[ \frac {4 \, a^{2} \cos \left (d x + c\right )^{5} - 36 \, a^{2} d x \cos \left (d x + c\right )^{2} - 4 \, a^{2} \cos \left (d x + c\right )^{3} + 36 \, a^{2} d x + 6 \, a^{2} \cos \left (d x + c\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, {\left (a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(4*a^2*cos(d*x + c)^5 - 36*a^2*d*x*cos(d*x + c)^2 - 4*a^2*cos(d*x + c)^3 + 36*a^2*d*x + 6*a^2*cos(d*x + c
) + 3*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2) - 3*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x +
c) + 1/2) - 12*(a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.24, size = 178, normalized size = 1.82 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, {\left (d x + c\right )} a^{2} - 12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {16 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a^2 - 12*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 24*a^2*tan(1/2
*d*x + 1/2*c) + 3*(6*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 1
6*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a^2*tan(1/2*d*x + 1/2*c)^4 - 3*a^2*tan(1/2*d*x + 1/2*c) + a^2)/(tan(1/2*d*
x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.49, size = 161, normalized size = 1.64 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {a^{2} \cos \left (d x +c \right )}{2 d}-\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {3 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}-3 a^{2} x -\frac {3 a^{2} c}{d}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

-1/6*a^2*cos(d*x+c)^3/d-1/2*a^2*cos(d*x+c)/d-1/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/d*a^2/sin(d*x+c)*cos(d*x+c)
^5-2*a^2*cos(d*x+c)^3*sin(d*x+c)/d-3*a^2*cos(d*x+c)*sin(d*x+c)/d-3*a^2*x-3/d*a^2*c-1/2/d*a^2/sin(d*x+c)^2*cos(
d*x+c)^5

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maxima [A]  time = 0.46, size = 151, normalized size = 1.54 \[ \frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 12 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} + 3 \, a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(2*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a^2 - 12*(3*d*
x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^2 + 3*a^2*(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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mupad [B]  time = 8.91, size = 303, normalized size = 3.09 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {6\,a^2\,\mathrm {atan}\left (\frac {36\,a^4}{6\,a^4-36\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {6\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6\,a^4-36\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6}+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^2)/sin(c + d*x)^3,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (a^2*log(tan(c/2 + (d*x)/2)))/(2*d) - (6*a^2*atan((36*a^4)/(6*a^4 - 36*a^4*
tan(c/2 + (d*x)/2)) + (6*a^4*tan(c/2 + (d*x)/2))/(6*a^4 - 36*a^4*tan(c/2 + (d*x)/2))))/d - (20*a^2*tan(c/2 + (
d*x)/2)^3 - (7*a^2*tan(c/2 + (d*x)/2)^2)/6 + (3*a^2*tan(c/2 + (d*x)/2)^4)/2 + 12*a^2*tan(c/2 + (d*x)/2)^5 - (1
5*a^2*tan(c/2 + (d*x)/2)^6)/2 - 4*a^2*tan(c/2 + (d*x)/2)^7 + a^2/2 + 4*a^2*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 +
(d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)^6 + 4*tan(c/2 + (d*x)/2)^8)) + (a^2*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)**4*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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