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

Optimal. Leaf size=116 \[ \frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin ^2(c+d x)}{2 d}-\frac {4 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \log (\sin (c+d x))}{d} \]

[Out]

-2*a^2*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d-a^2*ln(sin(d*x+c))/d-4*a^2*sin(d*x+c)/d-1/2*a^2*sin(d*x+c)^2/d+2/3*
a^2*sin(d*x+c)^3/d+1/4*a^2*sin(d*x+c)^4/d

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Rubi [A]  time = 0.12, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin ^2(c+d x)}{2 d}-\frac {4 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - (a^2*Log[Sin[c + d*x]])/d - (4*a^2*Sin[c + d*x])/d - (a
^2*Sin[c + d*x]^2)/(2*d) + (2*a^2*Sin[c + d*x]^3)/(3*d) + (a^2*Sin[c + d*x]^4)/(4*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 76, normalized size = 0.66 \[ -\frac {a^2 \left (-3 \sin ^4(c+d x)-8 \sin ^3(c+d x)+6 \sin ^2(c+d x)+48 \sin (c+d x)+6 \csc ^2(c+d x)+24 \csc (c+d x)+12 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(a^2*(24*Csc[c + d*x] + 6*Csc[c + d*x]^2 + 12*Log[Sin[c + d*x]] + 48*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 8
*Sin[c + d*x]^3 - 3*Sin[c + d*x]^4))/d

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fricas [A]  time = 0.74, size = 131, normalized size = 1.13 \[ \frac {24 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, a^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{2} \cos \left (d x + c\right )^{2} + 57 \, a^{2} - 96 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 64 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/96*(24*a^2*cos(d*x + c)^6 - 24*a^2*cos(d*x + c)^4 - 9*a^2*cos(d*x + c)^2 + 57*a^2 - 96*(a^2*cos(d*x + c)^2 -
a^2)*log(1/2*sin(d*x + c)) - 64*(a^2*cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 - 8*a^2)*sin(d*x + c))/(d*cos(d*x
+ c)^2 - d)

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giac [A]  time = 0.26, size = 109, normalized size = 0.94 \[ \frac {3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 48 \, a^{2} \sin \left (d x + c\right ) + \frac {6 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{2} \sin \left (d x + c\right ) - a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) - 48*
a^2*sin(d*x + c) + 6*(3*a^2*sin(d*x + c)^2 - 4*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^2)/d

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maple [A]  time = 0.47, size = 155, normalized size = 1.34 \[ -\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{2}}{4 d}-\frac {a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {16 a^{2} \sin \left (d x +c \right )}{3 d}-\frac {2 \sin \left (d x +c \right ) a^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{d}-\frac {8 \sin \left (d x +c \right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d*cos(d*x+c)^4*a^2-1/2/d*a^2*cos(d*x+c)^2-a^2*ln(sin(d*x+c))/d-2/d*a^2/sin(d*x+c)*cos(d*x+c)^6-16/3*a^2*s
in(d*x+c)/d-2/d*sin(d*x+c)*a^2*cos(d*x+c)^4-8/3/d*sin(d*x+c)*a^2*cos(d*x+c)^2-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c
)^6

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maxima [A]  time = 0.56, size = 93, normalized size = 0.80 \[ \frac {3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 48 \, a^{2} \sin \left (d x + c\right ) - \frac {6 \, {\left (4 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) - 48*a^2*s
in(d*x + c) - 6*(4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d

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mupad [B]  time = 8.94, size = 297, normalized size = 2.56 \[ \frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {272\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {296\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+48\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x)^3,x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a^2*log(tan(c/2 + (d*x)/2)))/d - (2*a^2*tan(c/2 + (d*x)/2)^2 + 48*a^2
*tan(c/2 + (d*x)/2)^3 + 11*a^2*tan(c/2 + (d*x)/2)^4 + (296*a^2*tan(c/2 + (d*x)/2)^5)/3 + 2*a^2*tan(c/2 + (d*x)
/2)^6 + (272*a^2*tan(c/2 + (d*x)/2)^7)/3 + (17*a^2*tan(c/2 + (d*x)/2)^8)/2 + 36*a^2*tan(c/2 + (d*x)/2)^9 + a^2
/2 + 4*a^2*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6
+ 16*tan(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) - (a^2*tan(c/2 + (d*x)/2))/d - (a^2*tan(c/2 + (d*x)/2)^2
)/(8*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)**5*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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