3.579 \(\int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=130 \[ -\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {5 a \cos (c+d x)}{2 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {5 a \cot (c+d x)}{2 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {5 a x}{2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2838, 2591, 288, 302, 203, 2592, 206} \[ -\frac {5 a \cos ^3(c+d x)}{6 d}-\frac {5 a \cos (c+d x)}{2 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {5 a \cot (c+d x)}{2 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac {5 a \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {5 a x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(5*a*x)/2 + (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) + (5*a*C
ot[c + d*x])/(2*d) - (a*Cos[c + d*x]^3*Cot[c + d*x]^2)/(2*d) - (5*a*Cot[c + d*x]^3)/(6*d) + (a*Cos[c + d*x]^2*
Cot[c + d*x]^3)/(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 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 ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^3(c+d x) \cot ^3(c+d x) \, dx+a \int \cos ^2(c+d x) \cot ^4(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 )^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 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) \operatorname {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 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) \operatorname {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\cot (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 {5 a \cot (c+d x)}{2 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac {5 a x}{2}+\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 {5 a \cot (c+d x)}{2 d}-\frac {a \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}-\frac {5 a \cot ^3(c+d x)}{6 d}+\frac {a \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 6.10, size = 174, normalized size = 1.34 \[ \frac {5 a (c+d x)}{2 d}+\frac {a \sin (2 (c+d x))}{4 d}-\frac {9 a \cos (c+d x)}{4 d}-\frac {a \cos (3 (c+d x))}{12 d}+\frac {7 a \cot (c+d x)}{3 d}-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {5 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {5 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {a \cot (c+d x) \csc ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(5*a*(c + d*x))/(2*d) - (9*a*Cos[c + d*x])/(4*d) - (a*Cos[3*(c + d*x)])/(12*d) + (7*a*Cot[c + d*x])/(3*d) - (a
*Csc[(c + d*x)/2]^2)/(8*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) + (5*a*Log[Cos[(c + d*x)/2]])/(2*d) - (5*a*
Log[Sin[(c + d*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d) + (a*Sin[2*(c + d*x)])/(4*d)

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fricas [A]  time = 0.84, size = 182, normalized size = 1.40 \[ -\frac {6 \, a \cos \left (d x + c\right )^{5} - 40 \, a \cos \left (d x + c\right )^{3} - 15 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, a \cos \left (d x + c\right ) + 2 \, {\left (2 \, a \cos \left (d x + c\right )^{5} - 15 \, a d x \cos \left (d x + c\right )^{2} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a d x - 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*a*cos(d*x + c)^5 - 40*a*cos(d*x + c)^3 - 15*(a*cos(d*x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2)*sin(d*
x + c) + 15*(a*cos(d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 30*a*cos(d*x + c) + 2*(2*a*cos(
d*x + c)^5 - 15*a*d*x*cos(d*x + c)^2 + 10*a*cos(d*x + c)^3 + 15*a*d*x - 15*a*cos(d*x + c))*sin(d*x + c))/((d*c
os(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.21, size = 220, normalized size = 1.69 \[ \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 180 \, {\left (d x + c\right )} a - 180 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 81 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {110 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 111 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 273 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 306 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 253 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3}}}{72 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/72*(3*a*tan(1/2*d*x + 1/2*c)^3 + 9*a*tan(1/2*d*x + 1/2*c)^2 + 180*(d*x + c)*a - 180*a*log(abs(tan(1/2*d*x +
1/2*c))) - 81*a*tan(1/2*d*x + 1/2*c) + (110*a*tan(1/2*d*x + 1/2*c)^9 + 9*a*tan(1/2*d*x + 1/2*c)^8 - 111*a*tan(
1/2*d*x + 1/2*c)^7 + 240*a*tan(1/2*d*x + 1/2*c)^6 - 273*a*tan(1/2*d*x + 1/2*c)^5 + 306*a*tan(1/2*d*x + 1/2*c)^
4 - 253*a*tan(1/2*d*x + 1/2*c)^3 + 72*a*tan(1/2*d*x + 1/2*c)^2 - 9*a*tan(1/2*d*x + 1/2*c) - 3*a)/(tan(1/2*d*x
+ 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^3)/d

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maple [A]  time = 0.32, size = 199, normalized size = 1.53 \[ -\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{2 d}-\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {5 a \cos \left (d x +c \right )}{2 d}-\frac {5 a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {4 a \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {4 a \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a x}{2}+\frac {5 c a}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.60, size = 122, normalized size = 0.94 \[ -\frac {{\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a - 2 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*((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 - 2*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^5 +
tan(d*x + c)^3))*a)/d

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mupad [B]  time = 8.97, size = 310, normalized size = 2.38 \[ \frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+49\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {80\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+67\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-34\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {121\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {9\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {5\,a\,\mathrm {atan}\left (\frac {25\,a^2}{25\,a^2+25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{25\,a^2+25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x)))/sin(c + d*x)^4,x)

[Out]

(a*tan(c/2 + (d*x)/2)^2)/(8*d) - (a/3 + a*tan(c/2 + (d*x)/2) - 8*a*tan(c/2 + (d*x)/2)^2 + (121*a*tan(c/2 + (d*
x)/2)^3)/3 - 34*a*tan(c/2 + (d*x)/2)^4 + 67*a*tan(c/2 + (d*x)/2)^5 - (80*a*tan(c/2 + (d*x)/2)^6)/3 + 49*a*tan(
c/2 + (d*x)/2)^7 - a*tan(c/2 + (d*x)/2)^8)/(d*(8*tan(c/2 + (d*x)/2)^3 + 24*tan(c/2 + (d*x)/2)^5 + 24*tan(c/2 +
(d*x)/2)^7 + 8*tan(c/2 + (d*x)/2)^9)) - (9*a*tan(c/2 + (d*x)/2))/(8*d) + (a*tan(c/2 + (d*x)/2)^3)/(24*d) - (5
*a*log(tan(c/2 + (d*x)/2)))/(2*d) - (5*a*atan((25*a^2)/(25*a^2 + 25*a^2*tan(c/2 + (d*x)/2)) - (25*a^2*tan(c/2
+ (d*x)/2))/(25*a^2 + 25*a^2*tan(c/2 + (d*x)/2))))/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)**4*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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