3.295 \(\int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=117 \[ \frac {a^4 \cos ^5(c+d x)}{5 d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos (c+d x)}{d}-\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{d}+\frac {5 a^4 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a^4 x}{2} \]

[Out]

5/2*a^4*x-a^4*arctanh(cos(d*x+c))/d+a^4*cos(d*x+c)/d-7/3*a^4*cos(d*x+c)^3/d+1/5*a^4*cos(d*x+c)^5/d+5/2*a^4*cos
(d*x+c)*sin(d*x+c)/d-a^4*cos(d*x+c)^3*sin(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2873, 2635, 8, 2592, 321, 206, 2565, 30, 2568, 14} \[ \frac {a^4 \cos ^5(c+d x)}{5 d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos (c+d x)}{d}-\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{d}+\frac {5 a^4 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a^4 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(5*a^4*x)/2 - (a^4*ArcTanh[Cos[c + d*x]])/d + (a^4*Cos[c + d*x])/d - (7*a^4*Cos[c + d*x]^3)/(3*d) + (a^4*Cos[c
+ d*x]^5)/(5*d) + (5*a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (a^4*Cos[c + d*x]^3*Sin[c + d*x])/d

Rule 8

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

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 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 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*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))^4 \, dx &=\int \left (4 a^4 \cos ^2(c+d x)+a^4 \cos (c+d x) \cot (c+d x)+6 a^4 \cos ^2(c+d x) \sin (c+d x)+4 a^4 \cos ^2(c+d x) \sin ^2(c+d x)+a^4 \cos ^2(c+d x) \sin ^3(c+d x)\right ) \, dx\\ &=a^4 \int \cos (c+d x) \cot (c+d x) \, dx+a^4 \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^2(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^2(c+d x) \sin (c+d x) \, dx\\ &=\frac {2 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}+a^4 \int \cos ^2(c+d x) \, dx+\left (2 a^4\right ) \int 1 \, dx-\frac {a^4 \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^4 \operatorname {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (6 a^4\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=2 a^4 x+\frac {a^4 \cos (c+d x)}{d}-\frac {2 a^4 \cos ^3(c+d x)}{d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}+\frac {1}{2} a^4 \int 1 \, dx-\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^4 \operatorname {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a^4 x}{2}-\frac {a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^4 \cos (c+d x)}{d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos ^5(c+d x)}{5 d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 95, normalized size = 0.81 \[ \frac {a^4 \left (-150 \cos (c+d x)-125 \cos (3 (c+d x))+3 \cos (5 (c+d x))+30 \left (8 \sin (2 (c+d x))-\sin (4 (c+d x))+8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+20 c+20 d x\right )\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(-150*Cos[c + d*x] - 125*Cos[3*(c + d*x)] + 3*Cos[5*(c + d*x)] + 30*(20*c + 20*d*x - 8*Log[Cos[(c + d*x)/
2]] + 8*Log[Sin[(c + d*x)/2]] + 8*Sin[2*(c + d*x)] - Sin[4*(c + d*x)])))/(240*d)

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fricas [A]  time = 0.54, size = 115, normalized size = 0.98 \[ \frac {6 \, a^{4} \cos \left (d x + c\right )^{5} - 70 \, a^{4} \cos \left (d x + c\right )^{3} + 75 \, a^{4} d x + 30 \, a^{4} \cos \left (d x + c\right ) - 15 \, a^{4} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, a^{4} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{3} - 5 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \]

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))^4,x, algorithm="fricas")

[Out]

1/30*(6*a^4*cos(d*x + c)^5 - 70*a^4*cos(d*x + c)^3 + 75*a^4*d*x + 30*a^4*cos(d*x + c) - 15*a^4*log(1/2*cos(d*x
+ c) + 1/2) + 15*a^4*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a^4*cos(d*x + c)^3 - 5*a^4*cos(d*x + c))*sin(d*x +
c))/d

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giac [A]  time = 0.24, size = 181, normalized size = 1.55 \[ \frac {75 \, {\left (d x + c\right )} a^{4} + 30 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 210 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 300 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 210 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 34 \, a^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{30 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/30*(75*(d*x + c)*a^4 + 30*a^4*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(45*a^4*tan(1/2*d*x + 1/2*c)^9 + 150*a^4*ta
n(1/2*d*x + 1/2*c)^8 + 210*a^4*tan(1/2*d*x + 1/2*c)^7 + 300*a^4*tan(1/2*d*x + 1/2*c)^6 + 40*a^4*tan(1/2*d*x +
1/2*c)^4 - 210*a^4*tan(1/2*d*x + 1/2*c)^3 + 20*a^4*tan(1/2*d*x + 1/2*c)^2 - 45*a^4*tan(1/2*d*x + 1/2*c) + 34*a
^4)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 0.37, size = 135, normalized size = 1.15 \[ -\frac {a^{4} \left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5 d}-\frac {32 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{15 d}-\frac {a^{4} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}+\frac {5 a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{4} x}{2}+\frac {5 a^{4} c}{2 d}+\frac {a^{4} \cos \left (d x +c \right )}{d}+\frac {a^{4} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

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))^4,x)

[Out]

-1/5/d*a^4*cos(d*x+c)^3*sin(d*x+c)^2-32/15*a^4*cos(d*x+c)^3/d-a^4*cos(d*x+c)^3*sin(d*x+c)/d+5/2*a^4*cos(d*x+c)
*sin(d*x+c)/d+5/2*a^4*x+5/2/d*a^4*c+a^4*cos(d*x+c)/d+1/d*a^4*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.33, size = 125, normalized size = 1.07 \[ -\frac {240 \, a^{4} \cos \left (d x + c\right )^{3} - 8 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{4} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 60 \, a^{4} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

-1/120*(240*a^4*cos(d*x + c)^3 - 8*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^4 - 15*(4*d*x + 4*c - sin(4*d*x + 4
*c))*a^4 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 - 60*a^4*(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 = 10.42, size = 295, normalized size = 2.52 \[ \frac {a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a^4\,\mathrm {atan}\left (\frac {25\,a^8}{10\,a^8-25\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {10\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{10\,a^8-25\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+14\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-14\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {34\,a^4}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*log(tan(c/2 + (d*x)/2)))/d + (5*a^4*atan((25*a^8)/(10*a^8 - 25*a^8*tan(c/2 + (d*x)/2)) + (10*a^8*tan(c/2
+ (d*x)/2))/(10*a^8 - 25*a^8*tan(c/2 + (d*x)/2))))/d - ((4*a^4*tan(c/2 + (d*x)/2)^2)/3 - 14*a^4*tan(c/2 + (d*x
)/2)^3 + (8*a^4*tan(c/2 + (d*x)/2)^4)/3 + 20*a^4*tan(c/2 + (d*x)/2)^6 + 14*a^4*tan(c/2 + (d*x)/2)^7 + 10*a^4*t
an(c/2 + (d*x)/2)^8 + 3*a^4*tan(c/2 + (d*x)/2)^9 + (34*a^4)/15 - 3*a^4*tan(c/2 + (d*x)/2))/(d*(5*tan(c/2 + (d*
x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 +
1))

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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)*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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