3.284 \(\int \cot ^2(c+d x) \csc ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=124 \[ -\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {5 a^2 \cot (c+d x) \csc ^3(c+d x)}{24 d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{16 d} \]

[Out]

3/16*a^2*arctanh(cos(d*x+c))/d-2/3*a^2*cot(d*x+c)^3/d-2/5*a^2*cot(d*x+c)^5/d+3/16*a^2*cot(d*x+c)*csc(d*x+c)/d-
5/24*a^2*cot(d*x+c)*csc(d*x+c)^3/d-1/6*a^2*cot(d*x+c)*csc(d*x+c)^5/d

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Rubi [A]  time = 0.26, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2873, 2611, 3768, 3770, 2607, 14} \[ -\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {5 a^2 \cot (c+d x) \csc ^3(c+d x)}{24 d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(3*a^2*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a^2*Cot[c + d*x]^3)/(3*d) - (2*a^2*Cot[c + d*x]^5)/(5*d) + (3*a^2*Co
t[c + d*x]*Csc[c + d*x])/(16*d) - (5*a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(24*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^
5)/(6*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 ^5(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cot ^2(c+d x) \csc ^3(c+d x)+2 a^2 \cot ^2(c+d x) \csc ^4(c+d x)+a^2 \cot ^2(c+d x) \csc ^5(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx+a^2 \int \cot ^2(c+d x) \csc ^5(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^2(c+d x) \csc ^4(c+d x) \, dx\\ &=-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {1}{6} a^2 \int \csc ^5(c+d x) \, dx-\frac {1}{4} a^2 \int \csc ^3(c+d x) \, dx+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac {a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {5 a^2 \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {1}{8} a^2 \int \csc (c+d x) \, dx-\frac {1}{8} a^2 \int \csc ^3(c+d x) \, dx+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {2 a^2 \cot ^5(c+d x)}{5 d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{16 d}-\frac {5 a^2 \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {1}{16} a^2 \int \csc (c+d x) \, dx\\ &=\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}-\frac {2 a^2 \cot ^5(c+d x)}{5 d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{16 d}-\frac {5 a^2 \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 229, normalized size = 1.85 \[ \frac {a^2 \csc ^6(c+d x) \left (-960 \sin (2 (c+d x))-384 \sin (4 (c+d x))+64 \sin (6 (c+d x))-1500 \cos (c+d x)+130 \cos (3 (c+d x))+90 \cos (5 (c+d x))-450 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-675 \cos (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+270 \cos (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-45 \cos (6 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+450 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+675 \cos (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-270 \cos (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+45 \cos (6 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{7680 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Csc[c + d*x]^6*(-1500*Cos[c + d*x] + 130*Cos[3*(c + d*x)] + 90*Cos[5*(c + d*x)] + 450*Log[Cos[(c + d*x)/2
]] - 675*Cos[2*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 270*Cos[4*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 45*Cos[6*(c + d
*x)]*Log[Cos[(c + d*x)/2]] - 450*Log[Sin[(c + d*x)/2]] + 675*Cos[2*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 270*Cos[
4*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 45*Cos[6*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 960*Sin[2*(c + d*x)] - 384*Si
n[4*(c + d*x)] + 64*Sin[6*(c + d*x)]))/(7680*d)

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fricas [B]  time = 0.51, size = 227, normalized size = 1.83 \[ -\frac {90 \, a^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 90 \, a^{2} \cos \left (d x + c\right ) - 45 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 45 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 64 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{5} - 5 \, a^{2} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{480 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/480*(90*a^2*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^3 - 90*a^2*cos(d*x + c) - 45*(a^2*cos(d*x + c)^6 - 3*a^2*c
os(d*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2) + 45*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d
*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x + c) + 1/2) + 64*(2*a^2*cos(d*x + c)^5 - 5*a^2*cos(d*
x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)

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giac [B]  time = 0.23, size = 228, normalized size = 1.84 \[ \frac {5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 360 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {882 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/1920*(5*a^2*tan(1/2*d*x + 1/2*c)^6 + 24*a^2*tan(1/2*d*x + 1/2*c)^5 + 45*a^2*tan(1/2*d*x + 1/2*c)^4 + 40*a^2*
tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 360*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 240*a^2*tan(
1/2*d*x + 1/2*c) + (882*a^2*tan(1/2*d*x + 1/2*c)^6 + 240*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2
*c)^4 - 40*a^2*tan(1/2*d*x + 1/2*c)^3 - 45*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*a^2*tan(1/2*d*x + 1/2*c) - 5*a^2)/t
an(1/2*d*x + 1/2*c)^6)/d

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maple [A]  time = 0.34, size = 160, normalized size = 1.29 \[ -\frac {3 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{4}}-\frac {3 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{16 d \sin \left (d x +c \right )^{2}}-\frac {3 a^{2} \cos \left (d x +c \right )}{16 d}-\frac {3 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16 d}-\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {4 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x)

[Out]

-3/8/d*a^2/sin(d*x+c)^4*cos(d*x+c)^3-3/16/d*a^2/sin(d*x+c)^2*cos(d*x+c)^3-3/16*a^2*cos(d*x+c)/d-3/16/d*a^2*ln(
csc(d*x+c)-cot(d*x+c))-2/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^3-4/15/d*a^2/sin(d*x+c)^3*cos(d*x+c)^3-1/6/d*a^2/sin(
d*x+c)^6*cos(d*x+c)^3

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maxima [A]  time = 0.35, size = 187, normalized size = 1.51 \[ -\frac {5 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 30 \, a^{2} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {64 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{2}}{\tan \left (d x + c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/480*(5*a^2*(2*(3*cos(d*x + c)^5 - 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3
*cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) + 30*a^2*(2*(cos(d*x + c)^3 + cos(d*
x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 64*(5*tan(d
*x + c)^2 + 3)*a^2/tan(d*x + c)^5)/d

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mupad [B]  time = 9.66, size = 339, normalized size = 2.73 \[ -\frac {a^2\,\left (5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-24\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-45\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+240\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-240\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+45\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\right )}{1920\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^2)/sin(c + d*x)^7,x)

[Out]

-(a^2*(5*cos(c/2 + (d*x)/2)^12 - 5*sin(c/2 + (d*x)/2)^12 - 24*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^11 + 24*co
s(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2) - 45*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 - 40*cos(c/2 + (d*x)/2)
^3*sin(c/2 + (d*x)/2)^9 + 15*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 + 240*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d
*x)/2)^7 - 240*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5 - 15*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 + 40*c
os(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3 + 45*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2 + 360*log(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6))/(1920*d*cos(c/2 + (d*x)/2)^6*sin(c/2 +
(d*x)/2)^6)

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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)**7*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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