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

Optimal. Leaf size=71 \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+a^2 x \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2873, 2635, 8, 2592, 321, 206, 2565, 30} \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x - (a^2*ArcTanh[Cos[c + d*x]])/d + (a^2*Cos[c + d*x])/d - (a^2*Cos[c + d*x]^3)/(3*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 30

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 71, normalized size = 1.00 \[ \frac {a^2 \left (9 \cos (c+d x)-\cos (3 (c+d x))+6 \left (\sin (2 (c+d x))+2 \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(9*Cos[c + d*x] - Cos[3*(c + d*x)] + 6*(2*(c + d*x - Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]]) + Sin
[2*(c + d*x)])))/(12*d)

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fricas [A]  time = 0.53, size = 86, normalized size = 1.21 \[ -\frac {2 \, a^{2} \cos \left (d x + c\right )^{3} - 6 \, a^{2} d x - 6 \, a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 101, normalized size = 1.42 \[ \frac {3 \, {\left (d x + c\right )} a^{2} + 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a^2 + 3*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*
d*x + 1/2*c)^2 - 3*a^2*tan(1/2*d*x + 1/2*c) - 2*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.34, size = 86, normalized size = 1.21 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+a^{2} x +\frac {a^{2} c}{d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a^{2} \cos \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 75, normalized size = 1.06 \[ -\frac {2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 3 \, a^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a^2*cos(d*x + c)^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 3*a^2*(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 = 9.04, size = 188, normalized size = 2.65 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,a^2}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a^2\,\mathrm {atan}\left (\frac {4\,a^4}{4\,a^4-4\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {4\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^4-4\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

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),x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (4*a^2*tan(c/2 + (d*x)/2)^2 - 2*a^2*tan(c/2 + (d*x)/2)^5 + (4*a^2)/3 + 2*a^2
*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*a^2
*atan((4*a^4)/(4*a^4 - 4*a^4*tan(c/2 + (d*x)/2)) + (4*a^4*tan(c/2 + (d*x)/2))/(4*a^4 - 4*a^4*tan(c/2 + (d*x)/2
))))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(cos(c + d*x)**2*csc(c + d*x), x) + Integral(2*sin(c + d*x)*cos(c + d*x)**2*csc(c + d*x), x) + I
ntegral(sin(c + d*x)**2*cos(c + d*x)**2*csc(c + d*x), x))

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