3.578 \(\int \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ -\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {5 a \cos (c+d x)}{2 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {15 a x}{8} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2838, 2592, 288, 302, 206, 2591, 321, 203} \[ -\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {5 a \cos (c+d x)}{2 d}-\frac {15 a \cot (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {15 a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a*x)/8 + (5*a*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a*Cos[c + d*x])/(2*d) - (5*a*Cos[c + d*x]^3)/(6*d) - (15*
a*Cot[c + d*x])/(8*d) + (5*a*Cos[c + d*x]^2*Cot[c + d*x])/(8*d) + (a*Cos[c + d*x]^4*Cot[c + d*x])/(4*d) - (a*C
os[c + d*x]^3*Cot[c + d*x]^2)/(2*d)

Rule 203

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

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 2591

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

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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \cot ^2(c+d x) \, dx+a \int \cos ^3(c+d x) \cot ^3(c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{4 d}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac {(15 a) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}+\frac {(5 a) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac {5 a \cos (c+d x)}{2 d}-\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {(15 a) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac {15 a x}{8}+\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {5 a \cos (c+d x)}{2 d}-\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 2.75, size = 117, normalized size = 0.87 \[ -\frac {a \left (216 \cos (c+d x)+8 \cos (3 (c+d x))+3 \left (16 \sin (2 (c+d x))+\sin (4 (c+d x))+32 \cot (c+d x)+4 \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 \sec ^2\left (\frac {1}{2} (c+d x)\right )+80 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-80 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+60 c+60 d x\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(a*(216*Cos[c + d*x] + 8*Cos[3*(c + d*x)] + 3*(60*c + 60*d*x + 32*Cot[c + d*x] + 4*Csc[(c + d*x)/2]^2 -
80*Log[Cos[(c + d*x)/2]] + 80*Log[Sin[(c + d*x)/2]] - 4*Sec[(c + d*x)/2]^2 + 16*Sin[2*(c + d*x)] + Sin[4*(c +
d*x)])))/d

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fricas [A]  time = 0.71, size = 162, normalized size = 1.21 \[ -\frac {8 \, a \cos \left (d x + c\right )^{5} + 45 \, a d x \cos \left (d x + c\right )^{2} + 40 \, a \cos \left (d x + c\right )^{3} - 45 \, a d x - 60 \, a \cos \left (d x + c\right ) - 30 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 30 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (2 \, a \cos \left (d x + c\right )^{5} + 5 \, a \cos \left (d x + c\right )^{3} - 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/24*(8*a*cos(d*x + c)^5 + 45*a*d*x*cos(d*x + c)^2 + 40*a*cos(d*x + c)^3 - 45*a*d*x - 60*a*cos(d*x + c) - 30*
(a*cos(d*x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2) + 30*(a*cos(d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2) +
3*(2*a*cos(d*x + c)^5 + 5*a*cos(d*x + c)^3 - 15*a*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.20, size = 214, normalized size = 1.60 \[ \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, {\left (d x + c\right )} a - 60 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, {\left (30 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left (27 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 168 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 152 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 56 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*a*tan(1/2*d*x + 1/2*c)^2 - 45*(d*x + c)*a - 60*a*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a*tan(1/2*d*x + 1
/2*c) + 3*(30*a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^2 + 2*(27*a*tan(1/
2*d*x + 1/2*c)^7 - 72*a*tan(1/2*d*x + 1/2*c)^6 + 3*a*tan(1/2*d*x + 1/2*c)^5 - 168*a*tan(1/2*d*x + 1/2*c)^4 - 3
*a*tan(1/2*d*x + 1/2*c)^3 - 152*a*tan(1/2*d*x + 1/2*c)^2 - 27*a*tan(1/2*d*x + 1/2*c) - 56*a)/(tan(1/2*d*x + 1/
2*c)^2 + 1)^4)/d

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maple [A]  time = 0.37, size = 177, normalized size = 1.32 \[ -\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {a \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}-\frac {15 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {15 a x}{8}-\frac {15 c a}{8 d}-\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{2 d}-\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {5 a \cos \left (d x +c \right )}{2 d}-\frac {5 a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x)

[Out]

-1/d*a/sin(d*x+c)*cos(d*x+c)^7-a*cos(d*x+c)^5*sin(d*x+c)/d-5/4*a*cos(d*x+c)^3*sin(d*x+c)/d-15/8*a*cos(d*x+c)*s
in(d*x+c)/d-15/8*a*x-15/8/d*c*a-1/2/d*a/sin(d*x+c)^2*cos(d*x+c)^7-1/2*a*cos(d*x+c)^5/d-5/6*a*cos(d*x+c)^3/d-5/
2*a*cos(d*x+c)/d-5/2/d*a*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.57, size = 131, normalized size = 0.98 \[ -\frac {2 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a + 3 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/24*(2*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1))*a + 3*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^5
+ 2*tan(d*x + c)^3 + tan(d*x + c)))*a)/d

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mupad [B]  time = 8.86, size = 321, normalized size = 2.40 \[ \frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {49\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+58\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {161\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {62\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {15\,a\,\mathrm {atan}\left (\frac {225\,a^2}{16\,\left (\frac {75\,a^2}{4}-\frac {225\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {75\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {75\,a^2}{4}-\frac {225\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x)))/sin(c + d*x)^3,x)

[Out]

(a*tan(c/2 + (d*x)/2))/(2*d) + (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (5*a*log(tan(c/2 + (d*x)/2)))/(2*d) - (a/2 + 2
*a*tan(c/2 + (d*x)/2) + (62*a*tan(c/2 + (d*x)/2)^2)/3 + 17*a*tan(c/2 + (d*x)/2)^3 + (161*a*tan(c/2 + (d*x)/2)^
4)/3 + 13*a*tan(c/2 + (d*x)/2)^5 + 58*a*tan(c/2 + (d*x)/2)^6 + 7*a*tan(c/2 + (d*x)/2)^7 + (49*a*tan(c/2 + (d*x
)/2)^8)/2 - 7*a*tan(c/2 + (d*x)/2)^9)/(d*(4*tan(c/2 + (d*x)/2)^2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x
)/2)^6 + 16*tan(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) - (15*a*atan((225*a^2)/(16*((75*a^2)/4 - (225*a^2
*tan(c/2 + (d*x)/2))/16)) + (75*a^2*tan(c/2 + (d*x)/2))/(4*((75*a^2)/4 - (225*a^2*tan(c/2 + (d*x)/2))/16))))/(
4*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)**6*csc(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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