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

Optimal. Leaf size=98 \[ \frac {3 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {5 a^3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635} \[ \frac {3 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^3*x)/2 - (5*a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (3*a^3*Cos[c + d*x])/d - (3*a^3*Cot[c + d*x])/d - (a^3*Co
t[c + d*x]*Csc[c + d*x])/(2*d) + (a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

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 \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\int \left (-2 a^5+2 a^5 \csc (c+d x)+3 a^5 \csc ^2(c+d x)+a^5 \csc ^3(c+d x)-3 a^5 \sin (c+d x)-a^5 \sin ^2(c+d x)\right ) \, dx}{a^2}\\ &=-2 a^3 x+a^3 \int \csc ^3(c+d x) \, dx-a^3 \int \sin ^2(c+d x) \, dx+\left (2 a^3\right ) \int \csc (c+d x) \, dx+\left (3 a^3\right ) \int \csc ^2(c+d x) \, dx-\left (3 a^3\right ) \int \sin (c+d x) \, dx\\ &=-2 a^3 x-\frac {2 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^3 \int 1 \, dx+\frac {1}{2} a^3 \int \csc (c+d x) \, dx-\frac {\left (3 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac {5 a^3 x}{2}-\frac {5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 112, normalized size = 1.14 \[ \frac {a^3 \left (2 \sin (2 (c+d x))+24 \cos (c+d x)+12 \tan \left (\frac {1}{2} (c+d x)\right )-12 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-20 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-20 c-20 d x\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-20*c - 20*d*x + 24*Cos[c + d*x] - 12*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 20*Log[Cos[(c + d*x)/2]] +
20*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + 2*Sin[2*(c + d*x)] + 12*Tan[(c + d*x)/2]))/(8*d)

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fricas [A]  time = 0.54, size = 159, normalized size = 1.62 \[ -\frac {10 \, a^{3} d x \cos \left (d x + c\right )^{2} - 12 \, a^{3} \cos \left (d x + c\right )^{3} - 10 \, a^{3} d x + 10 \, a^{3} \cos \left (d x + c\right ) + 5 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 5 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + 5 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(10*a^3*d*x*cos(d*x + c)^2 - 12*a^3*cos(d*x + c)^3 - 10*a^3*d*x + 10*a^3*cos(d*x + c) + 5*(a^3*cos(d*x +
c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2) - 5*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*c
os(d*x + c)^3 + 5*a^3*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [B]  time = 0.24, size = 184, normalized size = 1.88 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, {\left (d x + c\right )} a^{3} + 20 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 20*(d*x + c)*a^3 + 20*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^3*tan(1/2*d*
x + 1/2*c) - (10*a^3*tan(1/2*d*x + 1/2*c)^6 + 20*a^3*tan(1/2*d*x + 1/2*c)^5 - 27*a^3*tan(1/2*d*x + 1/2*c)^4 +
16*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/(tan(1/2*d*
x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2)/d

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maple [A]  time = 0.42, size = 113, normalized size = 1.15 \[ \frac {a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {5 a^{3} x}{2}-\frac {5 a^{3} c}{2 d}+\frac {5 a^{3} \cos \left (d x +c \right )}{2 d}+\frac {5 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \cot \left (d x +c \right )}{d}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 124, normalized size = 1.27 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{3} + a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.68, size = 259, normalized size = 2.64 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {5\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {47\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-23\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {5\,a^3\,\mathrm {atan}\left (\frac {25\,a^6}{25\,a^6+25\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {25\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{25\,a^6+25\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (5*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) - (8*a^3*tan(c/2 + (d*x)/2)^3 - 23*a^
3*tan(c/2 + (d*x)/2)^2 - (47*a^3*tan(c/2 + (d*x)/2)^4)/2 + 10*a^3*tan(c/2 + (d*x)/2)^5 + a^3/2 + 6*a^3*tan(c/2
+ (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 8*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6)) + (5*a^3*atan((25*
a^6)/(25*a^6 + 25*a^6*tan(c/2 + (d*x)/2)) - (25*a^6*tan(c/2 + (d*x)/2))/(25*a^6 + 25*a^6*tan(c/2 + (d*x)/2))))
/d + (3*a^3*tan(c/2 + (d*x)/2))/(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)**2*csc(d*x+c)**3*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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