3.514 \(\int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=119 \[ \frac {a^2 \sin ^6(c+d x)}{6 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^4(c+d x)}{4 d}-\frac {4 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

[Out]

a^2*ln(sin(d*x+c))/d+2*a^2*sin(d*x+c)/d-1/2*a^2*sin(d*x+c)^2/d-4/3*a^2*sin(d*x+c)^3/d-1/4*a^2*sin(d*x+c)^4/d+2
/5*a^2*sin(d*x+c)^5/d+1/6*a^2*sin(d*x+c)^6/d

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Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ \frac {a^2 \sin ^6(c+d x)}{6 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^4(c+d x)}{4 d}-\frac {4 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Mathematica [A]  time = 0.08, size = 78, normalized size = 0.66 \[ \frac {a^2 \left (10 \sin ^6(c+d x)+24 \sin ^5(c+d x)-15 \sin ^4(c+d x)-80 \sin ^3(c+d x)-30 \sin ^2(c+d x)+120 \sin (c+d x)+60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(60*Log[Sin[c + d*x]] + 120*Sin[c + d*x] - 30*Sin[c + d*x]^2 - 80*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 24
*Sin[c + d*x]^5 + 10*Sin[c + d*x]^6))/(60*d)

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fricas [A]  time = 0.76, size = 99, normalized size = 0.83 \[ -\frac {10 \, a^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(10*a^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c)^2 - 60*a^2*log(1/2*sin(d*x + c)) -
8*(3*a^2*cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 + 8*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.24, size = 95, normalized size = 0.80 \[ \frac {10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*s
in(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) + 120*a^2*sin(d*x + c))/d

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maple [A]  time = 0.47, size = 122, normalized size = 1.03 \[ -\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6 d}+\frac {16 a^{2} \sin \left (d x +c \right )}{15 d}+\frac {2 \sin \left (d x +c \right ) a^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}+\frac {8 \sin \left (d x +c \right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{15 d}+\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{2}}{4 d}+\frac {a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*a^2*cos(d*x+c)^6+16/15*a^2*sin(d*x+c)/d+2/5/d*sin(d*x+c)*a^2*cos(d*x+c)^4+8/15/d*sin(d*x+c)*a^2*cos(d*x
+c)^2+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

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maxima [A]  time = 0.46, size = 94, normalized size = 0.79 \[ \frac {10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*s
in(d*x + c)^2 + 60*a^2*log(sin(d*x + c)) + 120*a^2*sin(d*x + c))/d

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mupad [B]  time = 9.07, size = 132, normalized size = 1.11 \[ \frac {5\,a^2\,\sin \left (c+d\,x\right )}{4\,d}-\frac {a^2\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {19\,a^2\,\cos \left (2\,c+2\,d\,x\right )}{64\,d}-\frac {a^2\,\cos \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {5\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{24\,d}+\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{40\,d} \]

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),x)

[Out]

(5*a^2*sin(c + d*x))/(4*d) - (a^2*log(1/cos(c/2 + (d*x)/2)^2))/d + (a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)
/2)))/d + (19*a^2*cos(2*c + 2*d*x))/(64*d) - (a^2*cos(6*c + 6*d*x))/(192*d) + (5*a^2*sin(3*c + 3*d*x))/(24*d)
+ (a^2*sin(5*c + 5*d*x))/(40*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)*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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