3.1244 \(\int \cos ^5(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=157 \[ \frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a b \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac {5 a b \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {5 a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a b x}{8}-\frac {b^2 \cos ^7(c+d x)}{7 d} \]

[Out]

5/8*a*b*x-a^2*arctanh(cos(d*x+c))/d+a^2*cos(d*x+c)/d+1/3*a^2*cos(d*x+c)^3/d+1/5*a^2*cos(d*x+c)^5/d-1/7*b^2*cos
(d*x+c)^7/d+5/8*a*b*cos(d*x+c)*sin(d*x+c)/d+5/12*a*b*cos(d*x+c)^3*sin(d*x+c)/d+1/3*a*b*cos(d*x+c)^5*sin(d*x+c)
/d

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Rubi [A]  time = 0.21, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2911, 2635, 8, 14, 207} \[ \frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a b \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac {5 a b \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac {5 a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a b x}{8}-\frac {b^2 \cos ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(5*a*b*x)/8 - (a^2*ArcTanh[Cos[c + d*x]])/d + (a^2*Cos[c + d*x])/d + (a^2*Cos[c + d*x]^3)/(3*d) + (a^2*Cos[c +
d*x]^5)/(5*d) - (b^2*Cos[c + d*x]^7)/(7*d) + (5*a*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (5*a*b*Cos[c + d*x]^3*
Sin[c + d*x])/(12*d) + (a*b*Cos[c + d*x]^5*Sin[c + d*x])/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 207

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

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 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^6(c+d x) \, dx+\int \cos ^5(c+d x) \cot (c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a b \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} (5 a b) \int \cos ^4(c+d x) \, dx-\frac {\operatorname {Subst}\left (\int x^6 \left (b^2-\frac {a^2}{-1+x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac {1}{4} (5 a b) \int \cos ^2(c+d x) \, dx-\frac {\operatorname {Subst}\left (\int \left (-a^2-a^2 x^2-a^2 x^4+b^2 x^6-\frac {a^2}{-1+x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {b^2 \cos ^7(c+d x)}{7 d}+\frac {5 a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac {1}{8} (5 a b) \int 1 \, dx+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a b x}{8}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {b^2 \cos ^7(c+d x)}{7 d}+\frac {5 a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 166, normalized size = 1.06 \[ \frac {105 \left (88 a^2-5 b^2\right ) \cos (c+d x)+35 \left (28 a^2-9 b^2\right ) \cos (3 (c+d x))+84 a^2 \cos (5 (c+d x))+6720 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6720 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3150 a b \sin (2 (c+d x))+630 a b \sin (4 (c+d x))+70 a b \sin (6 (c+d x))+4200 a b c+4200 a b d x-105 b^2 \cos (5 (c+d x))-15 b^2 \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(4200*a*b*c + 4200*a*b*d*x + 105*(88*a^2 - 5*b^2)*Cos[c + d*x] + 35*(28*a^2 - 9*b^2)*Cos[3*(c + d*x)] + 84*a^2
*Cos[5*(c + d*x)] - 105*b^2*Cos[5*(c + d*x)] - 15*b^2*Cos[7*(c + d*x)] - 6720*a^2*Log[Cos[(c + d*x)/2]] + 6720
*a^2*Log[Sin[(c + d*x)/2]] + 3150*a*b*Sin[2*(c + d*x)] + 630*a*b*Sin[4*(c + d*x)] + 70*a*b*Sin[6*(c + d*x)])/(
6720*d)

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fricas [A]  time = 1.08, size = 137, normalized size = 0.87 \[ -\frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, a^{2} \cos \left (d x + c\right )^{5} - 280 \, a^{2} \cos \left (d x + c\right )^{3} - 525 \, a b d x - 840 \, a^{2} \cos \left (d x + c\right ) + 420 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 420 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} + 10 \, a b \cos \left (d x + c\right )^{3} + 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/840*(120*b^2*cos(d*x + c)^7 - 168*a^2*cos(d*x + c)^5 - 280*a^2*cos(d*x + c)^3 - 525*a*b*d*x - 840*a^2*cos(d
*x + c) + 420*a^2*log(1/2*cos(d*x + c) + 1/2) - 420*a^2*log(-1/2*cos(d*x + c) + 1/2) - 35*(8*a*b*cos(d*x + c)^
5 + 10*a*b*cos(d*x + c)^3 + 15*a*b*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.27, size = 291, normalized size = 1.85 \[ \frac {525 \, {\left (d x + c\right )} a b + 840 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (1155 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 2520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 840 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 980 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 10080 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 2975 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 20440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 24640 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2975 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16968 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2520 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 980 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6496 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1155 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1288 \, a^{2} + 120 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/840*(525*(d*x + c)*a*b + 840*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(1155*a*b*tan(1/2*d*x + 1/2*c)^13 - 2520
*a^2*tan(1/2*d*x + 1/2*c)^12 + 840*b^2*tan(1/2*d*x + 1/2*c)^12 + 980*a*b*tan(1/2*d*x + 1/2*c)^11 - 10080*a^2*t
an(1/2*d*x + 1/2*c)^10 + 2975*a*b*tan(1/2*d*x + 1/2*c)^9 - 20440*a^2*tan(1/2*d*x + 1/2*c)^8 + 4200*b^2*tan(1/2
*d*x + 1/2*c)^8 - 24640*a^2*tan(1/2*d*x + 1/2*c)^6 - 2975*a*b*tan(1/2*d*x + 1/2*c)^5 - 16968*a^2*tan(1/2*d*x +
1/2*c)^4 + 2520*b^2*tan(1/2*d*x + 1/2*c)^4 - 980*a*b*tan(1/2*d*x + 1/2*c)^3 - 6496*a^2*tan(1/2*d*x + 1/2*c)^2
- 1155*a*b*tan(1/2*d*x + 1/2*c) - 1288*a^2 + 120*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d

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maple [A]  time = 0.55, size = 160, normalized size = 1.02 \[ \frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a b \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{12 d}+\frac {5 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {5 a b x}{8}+\frac {5 a b c}{8 d}-\frac {b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)*(a+b*sin(d*x+c))^2,x)

[Out]

1/5*a^2*cos(d*x+c)^5/d+1/3*a^2*cos(d*x+c)^3/d+a^2*cos(d*x+c)/d+1/d*a^2*ln(csc(d*x+c)-cot(d*x+c))+1/3*a*b*cos(d
*x+c)^5*sin(d*x+c)/d+5/12*a*b*cos(d*x+c)^3*sin(d*x+c)/d+5/8*a*b*cos(d*x+c)*sin(d*x+c)/d+5/8*a*b*x+5/8/d*a*b*c-
1/7*b^2*cos(d*x+c)^7/d

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maxima [A]  time = 0.55, size = 122, normalized size = 0.78 \[ -\frac {480 \, b^{2} \cos \left (d x + c\right )^{7} - 112 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3360*(480*b^2*cos(d*x + c)^7 - 112*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x
+ c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 + 35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 4
8*sin(2*d*x + 2*c))*a*b)/d

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mupad [B]  time = 13.71, size = 415, normalized size = 2.64 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (6\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {146\,a^2}{3}-10\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {202\,a^2}{5}-6\,b^2\right )+\frac {232\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {176\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {46\,a^2}{15}-\frac {2\,b^2}{7}+\frac {7\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {85\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {85\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}-\frac {7\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {11\,a\,b\,\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 )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a\,b\,\mathrm {atan}\left (\frac {25\,a^2\,b^2}{16\,\left (\frac {5\,a^3\,b}{2}-\frac {25\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {5\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {5\,a^3\,b}{2}-\frac {25\,a^2\,b^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 + b*sin(c + d*x))^2)/sin(c + d*x),x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^12*(6*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^8*((146*a^2)/3 -
10*b^2) + tan(c/2 + (d*x)/2)^4*((202*a^2)/5 - 6*b^2) + (232*a^2*tan(c/2 + (d*x)/2)^2)/15 + (176*a^2*tan(c/2 +
(d*x)/2)^6)/3 + 24*a^2*tan(c/2 + (d*x)/2)^10 + (46*a^2)/15 - (2*b^2)/7 + (7*a*b*tan(c/2 + (d*x)/2)^3)/3 + (85
*a*b*tan(c/2 + (d*x)/2)^5)/12 - (85*a*b*tan(c/2 + (d*x)/2)^9)/12 - (7*a*b*tan(c/2 + (d*x)/2)^11)/3 - (11*a*b*t
an(c/2 + (d*x)/2)^13)/4 + (11*a*b*tan(c/2 + (d*x)/2))/4)/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4
+ 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan
(c/2 + (d*x)/2)^14 + 1)) + (5*a*b*atan((25*a^2*b^2)/(16*((5*a^3*b)/2 - (25*a^2*b^2*tan(c/2 + (d*x)/2))/16)) +
(5*a^3*b*tan(c/2 + (d*x)/2))/(2*((5*a^3*b)/2 - (25*a^2*b^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)*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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