3.1097 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=83 \[ -\frac {3 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 a x}{2}+\frac {b \cos ^3(c+d x)}{3 d}+\frac {b \cos (c+d x)}{d}-\frac {b \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2838, 2591, 288, 321, 203, 2592, 302, 206} \[ -\frac {3 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 a x}{2}+\frac {b \cos ^3(c+d x)}{3 d}+\frac {b \cos (c+d x)}{d}-\frac {b \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.33, size = 105, normalized size = 1.27 \[ -\frac {3 a (c+d x)}{2 d}-\frac {a \sin (2 (c+d x))}{4 d}-\frac {a \cot (c+d x)}{d}+\frac {5 b \cos (c+d x)}{4 d}+\frac {b \cos (3 (c+d x))}{12 d}+\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*(c + d*x))/(2*d) + (5*b*Cos[c + d*x])/(4*d) + (b*Cos[3*(c + d*x)])/(12*d) - (a*Cot[c + d*x])/d - (b*Log[
Cos[(c + d*x)/2]])/d + (b*Log[Sin[(c + d*x)/2]])/d - (a*Sin[2*(c + d*x)])/(4*d)

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 107, normalized size = 1.29 \[ \frac {3 \, a \cos \left (d x + c\right )^{3} - 3 \, b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, a \cos \left (d x + c\right ) + {\left (2 \, b \cos \left (d x + c\right )^{3} - 9 \, a d x + 6 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.21, size = 142, normalized size = 1.71 \[ -\frac {9 \, {\left (d x + c\right )} a - 6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, {\left (2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(9*(d*x + c)*a - 6*b*log(abs(tan(1/2*d*x + 1/2*c))) - 3*a*tan(1/2*d*x + 1/2*c) + 3*(2*b*tan(1/2*d*x + 1/2
*c) + a)/tan(1/2*d*x + 1/2*c) - 2*(3*a*tan(1/2*d*x + 1/2*c)^5 + 12*b*tan(1/2*d*x + 1/2*c)^4 + 12*b*tan(1/2*d*x
+ 1/2*c)^2 - 3*a*tan(1/2*d*x + 1/2*c) + 8*b)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.39, size = 119, normalized size = 1.43 \[ -\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {a \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {3 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {3 a x}{2}-\frac {3 c a}{2 d}+\frac {b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {b \cos \left (d x +c \right )}{d}+\frac {b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 91, normalized size = 1.10 \[ -\frac {3 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a - {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 9.39, size = 241, normalized size = 2.90 \[ \frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {16\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-a}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {3\,a\,\mathrm {atan}\left (\frac {9\,a^2}{9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,b\,a}-\frac {6\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,b\,a}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((16*b*tan(c/2 + (d*x)/2))/3 - a - 5*a*tan(c/2 + (d*x)/2)^2 - 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^
6 + 8*b*tan(c/2 + (d*x)/2)^3 + 8*b*tan(c/2 + (d*x)/2)^5)/(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)) + (a*tan(c/2 + (d*x)/2))/(2*d) + (b*log(tan(c/2 + (d*x)/2)))/
d + (3*a*atan((9*a^2)/(6*a*b + 9*a^2*tan(c/2 + (d*x)/2)) - (6*a*b*tan(c/2 + (d*x)/2))/(6*a*b + 9*a^2*tan(c/2 +
(d*x)/2))))/d

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**2*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cos(c + d*x)**4*csc(c + d*x)**2, x)

________________________________________________________________________________________