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

Optimal. Leaf size=99 \[ -\frac {a^3 \cos ^3(c+d x)}{d}+\frac {a^3 \cos (c+d x)}{d}-\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {13 a^3 x}{8} \]

[Out]

13/8*a^3*x-a^3*arctanh(cos(d*x+c))/d+a^3*cos(d*x+c)/d-a^3*cos(d*x+c)^3/d+13/8*a^3*cos(d*x+c)*sin(d*x+c)/d-1/4*
a^3*cos(d*x+c)^3*sin(d*x+c)/d

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Rubi [A]  time = 0.17, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2873, 2635, 8, 2592, 321, 206, 2565, 30, 2568} \[ -\frac {a^3 \cos ^3(c+d x)}{d}+\frac {a^3 \cos (c+d x)}{d}-\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {13 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13*a^3*x)/8 - (a^3*ArcTanh[Cos[c + d*x]])/d + (a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/d + (13*a^3*Cos[c +
d*x]*Sin[c + d*x])/(8*d) - (a^3*Cos[c + d*x]^3*Sin[c + d*x])/(4*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 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))^3 \, dx &=\int \left (3 a^3 \cos ^2(c+d x)+a^3 \cos (c+d x) \cot (c+d x)+3 a^3 \cos ^2(c+d x) \sin (c+d x)+a^3 \cos ^2(c+d x) \sin ^2(c+d x)\right ) \, dx\\ &=a^3 \int \cos (c+d x) \cot (c+d x) \, dx+a^3 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin (c+d x) \, dx\\ &=\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} a^3 \int \cos ^2(c+d x) \, dx+\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac {a^3 \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 a^3 x}{2}+\frac {a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{d}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} a^3 \int 1 \, dx-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {13 a^3 x}{8}-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{d}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 82, normalized size = 0.83 \[ \frac {a^3 \left (24 \sin (2 (c+d x))-\sin (4 (c+d x))+8 \cos (c+d x)-8 \cos (3 (c+d x))+32 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-32 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+52 c+52 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(52*c + 52*d*x + 8*Cos[c + d*x] - 8*Cos[3*(c + d*x)] - 32*Log[Cos[(c + d*x)/2]] + 32*Log[Sin[(c + d*x)/2]
] + 24*Sin[2*(c + d*x)] - Sin[4*(c + d*x)]))/(32*d)

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fricas [A]  time = 0.50, size = 101, normalized size = 1.02 \[ -\frac {8 \, a^{3} \cos \left (d x + c\right )^{3} - 13 \, a^{3} d x - 8 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} - 13 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

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))^3,x, algorithm="fricas")

[Out]

-1/8*(8*a^3*cos(d*x + c)^3 - 13*a^3*d*x - 8*a^3*cos(d*x + c) + 4*a^3*log(1/2*cos(d*x + c) + 1/2) - 4*a^3*log(-
1/2*cos(d*x + c) + 1/2) + (2*a^3*cos(d*x + c)^3 - 13*a^3*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.21, size = 144, normalized size = 1.45 \[ \frac {13 \, {\left (d x + c\right )} a^{3} + 8 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (11 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 16 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/8*(13*(d*x + c)*a^3 + 8*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(11*a^3*tan(1/2*d*x + 1/2*c)^7 + 16*a^3*tan(1
/2*d*x + 1/2*c)^6 + 19*a^3*tan(1/2*d*x + 1/2*c)^5 - 19*a^3*tan(1/2*d*x + 1/2*c)^3 - 16*a^3*tan(1/2*d*x + 1/2*c
)^2 - 11*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.36, size = 111, normalized size = 1.12 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}+\frac {13 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {13 a^{3} x}{8}+\frac {13 a^{3} c}{8 d}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {a^{3} \cos \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

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))^3,x)

[Out]

-1/4*a^3*cos(d*x+c)^3*sin(d*x+c)/d+13/8*a^3*cos(d*x+c)*sin(d*x+c)/d+13/8*a^3*x+13/8/d*a^3*c-a^3*cos(d*x+c)^3/d
+a^3*cos(d*x+c)/d+1/d*a^3*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.32, size = 99, normalized size = 1.00 \[ -\frac {32 \, a^{3} \cos \left (d x + c\right )^{3} - {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 16 \, a^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{32 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/32*(32*a^3*cos(d*x + c)^3 - (4*d*x + 4*c - sin(4*d*x + 4*c))*a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3
- 16*a^3*(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 = 10.36, size = 244, normalized size = 2.46 \[ \frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {13\,a^3\,\mathrm {atan}\left (\frac {169\,a^6}{16\,\left (\frac {13\,a^6}{2}-\frac {169\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {13\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {13\,a^6}{2}-\frac {169\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {-\frac {11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {11\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^3)/sin(c + d*x),x)

[Out]

(a^3*log(tan(c/2 + (d*x)/2)))/d + (13*a^3*atan((169*a^6)/(16*((13*a^6)/2 - (169*a^6*tan(c/2 + (d*x)/2))/16)) +
(13*a^6*tan(c/2 + (d*x)/2))/(2*((13*a^6)/2 - (169*a^6*tan(c/2 + (d*x)/2))/16))))/(4*d) + (4*a^3*tan(c/2 + (d*
x)/2)^2 + (19*a^3*tan(c/2 + (d*x)/2)^3)/4 - (19*a^3*tan(c/2 + (d*x)/2)^5)/4 - 4*a^3*tan(c/2 + (d*x)/2)^6 - (11
*a^3*tan(c/2 + (d*x)/2)^7)/4 + (11*a^3*tan(c/2 + (d*x)/2))/4)/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2
)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

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))**3,x)

[Out]

a**3*(Integral(cos(c + d*x)**2*csc(c + d*x), x) + Integral(3*sin(c + d*x)*cos(c + d*x)**2*csc(c + d*x), x) + I
ntegral(3*sin(c + d*x)**2*cos(c + d*x)**2*csc(c + d*x), x) + Integral(sin(c + d*x)**3*cos(c + d*x)**2*csc(c +
d*x), x))

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