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

Optimal. Leaf size=73 \[ -\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}-a^2 x \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}-a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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

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Mathematica [A]  time = 0.58, size = 140, normalized size = 1.92 \[ -\frac {a^2 \left (-8 \tan \left (\frac {1}{2} (c+d x)\right )+8 \cot \left (\frac {1}{2} (c+d x)\right )+6 \csc ^2\left (\frac {1}{2} (c+d x)\right )-6 \sec ^2\left (\frac {1}{2} (c+d x)\right )+24 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-24 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )-8 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+24 c+24 d x\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.53, size = 166, normalized size = 2.27 \[ -\frac {4 \, a^{2} \cos \left (d x + c\right )^{3} - 6 \, a^{2} \cos \left (d x + c\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a^{2} d x \cos \left (d x + c\right )^{2} - a^{2} d x - a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 141, normalized size = 1.93 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, {\left (d x + c\right )} a^{2} - 24 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {44 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*(d*x + c)*a^2 - 24*a^2*log(abs(tan(1/2*d*
x + 1/2*c))) + 9*a^2*tan(1/2*d*x + 1/2*c) + (44*a^2*tan(1/2*d*x + 1/2*c)^3 - 9*a^2*tan(1/2*d*x + 1/2*c)^2 - 6*
a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.32, size = 117, normalized size = 1.60 \[ -a^{2} x -\frac {a^{2} \cot \left (d x +c \right )}{d}-\frac {a^{2} c}{d}-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \cos \left (d x +c \right )}{d}-\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 83, normalized size = 1.14 \[ -\frac {6 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, a^{2}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.94, size = 193, normalized size = 2.64 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,a^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \]

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)^4,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(4*d) - (a^2*cot(c/2 + (d*x)/2)^3)/(24*d) - (a^2*cot(c/2 + (d*x)/2)^2)/(4*d) + (a^2
*tan(c/2 + (d*x)/2)^3)/(24*d) - (2*a^2*atan((cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2) - si
n(c/2 + (d*x)/2))))/d - (a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (3*a^2*cot(c/2 + (d*x)/2))/(8*d)
+ (3*a^2*tan(c/2 + (d*x)/2))/(8*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)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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