3.1072 \(\int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=102 \[ -\frac {a^3 \cot (c+d x)}{d}+a^3 (-x)+\frac {3 a^2 b \cos (c+d x)}{d}-\frac {3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2722, 2635, 8, 2592, 321, 206, 3473, 2565, 30} \[ \frac {3 a^2 b \cos (c+d x)}{d}-\frac {3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x)}{d}+a^3 (-x)+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

-(a^3*x) + (3*a*b^2*x)/2 - (3*a^2*b*ArcTanh[Cos[c + d*x]])/d + (3*a^2*b*Cos[c + d*x])/d - (b^3*Cos[c + d*x]^3)
/(3*d) - (a^3*Cot[c + d*x])/d + (3*a*b^2*Cos[c + d*x]*Sin[c + d*x])/(2*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 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 1.32, size = 143, normalized size = 1.40 \[ \frac {-6 a^3 \cot \left (\frac {1}{2} (c+d x)\right )+\left (36 a^2 b-3 b^3\right ) \cos (c+d x)+6 a \left (a^2 \tan \left (\frac {1}{2} (c+d x)\right )-2 a^2 c-2 a^2 d x+6 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 b^2 c+3 b^2 d x\right )+9 a b^2 \sin (2 (c+d x))-b^3 \cos (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

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

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fricas [A]  time = 0.71, size = 143, normalized size = 1.40 \[ -\frac {9 \, a b^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, a^{2} b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) + {\left (2 \, b^{3} \cos \left (d x + c\right )^{3} - 18 \, a^{2} b \cos \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x\right )} \sin \left (d x + c\right )}{6 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/6*(9*a*b^2*cos(d*x + c)^3 + 9*a^2*b*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*a^2*b*log(-1/2*cos(d*x + c
) + 1/2)*sin(d*x + c) + 3*(2*a^3 - 3*a*b^2)*cos(d*x + c) + (2*b^3*cos(d*x + c)^3 - 18*a^2*b*cos(d*x + c) + 3*(
2*a^3 - 3*a*b^2)*d*x)*sin(d*x + c))/(d*sin(d*x + c))

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giac [B]  time = 0.24, size = 199, normalized size = 1.95 \[ \frac {18 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {3 \, {\left (6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(18*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^3*tan(1/2*d*x + 1/2*c) - 3*(2*a^3 - 3*a*b^2)*(d*x + c) - 3*
(6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) - 2*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1
/2*d*x + 1/2*c)^4 + 6*b^3*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2
*c) - 18*a^2*b + 2*b^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.40, size = 125, normalized size = 1.23 \[ -a^{3} x -\frac {a^{3} \cot \left (d x +c \right )}{d}-\frac {a^{3} c}{d}+\frac {3 a^{2} b \cos \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a \,b^{2} x}{2}+\frac {3 a \,b^{2} c}{2 d}-\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 95, normalized size = 0.93 \[ -\frac {4 \, b^{3} \cos \left (d x + c\right )^{3} + 12 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} - 18 \, a^{2} b {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(4*b^3*cos(d*x + c)^3 + 12*(d*x + c + 1/tan(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^2 - 1
8*a^2*b*(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 = 9.47, size = 289, normalized size = 2.83 \[ \frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,\left (\frac {a\,b^2\,3{}\mathrm {i}}{2}-a^3\,1{}\mathrm {i}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2\,b-\frac {4\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3+6\,a\,b^2\right )-3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a\,b^2-3\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (12\,a^2\,b-4\,b^3\right )-a^3+24\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2))/(2*d) - (log(tan(c/2 + (d*x)/2) - 1i)*((a*b^2*3i)/2 - a^3*1i))/d + (tan(c/2 + (d*x)/2
)*(12*a^2*b - (4*b^3)/3) - tan(c/2 + (d*x)/2)^6*(6*a*b^2 + a^3) - 3*a^3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)
/2)^2*(6*a*b^2 - 3*a^3) + tan(c/2 + (d*x)/2)^5*(12*a^2*b - 4*b^3) - a^3 + 24*a^2*b*tan(c/2 + (d*x)/2)^3)/(d*(2
*tan(c/2 + (d*x)/2) + 6*tan(c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)/2)^5 + 2*tan(c/2 + (d*x)/2)^7)) + (3*a^2*b*lo
g(tan(c/2 + (d*x)/2)))/d - (a*log(tan(c/2 + (d*x)/2) + 1i)*(2*a^2 - 3*b^2)*1i)/(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)**2*csc(d*x+c)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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