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

Optimal. Leaf size=83 \[ \frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 83, 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} \[ \frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 83, normalized size = 1.00 \[ \frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 91, normalized size = 1.10 \[ \frac {32 \, a \cos \left (d x + c\right )^{4} + 128 \, a \cos \left (d x + c\right )^{2} + 96 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (8 \, a \cos \left (d x + c\right )^{4} + 16 \, a \cos \left (d x + c\right )^{2} - 11 \, a\right )} \sin \left (d x + c\right ) - 256 \, a}{96 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*a*cos(d*x + c)^4 + 128*a*cos(d*x + c)^2 + 96*a*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*a*cos(d*x +
c)^4 + 16*a*cos(d*x + c)^2 - 11*a)*sin(d*x + c) - 256*a)/(d*sin(d*x + c))

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giac [A]  time = 0.19, size = 79, normalized size = 0.95 \[ \frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, {\left (a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*a*sin(d*x + c)^2 + 12*a*log(abs(sin(d*x + c))) - 24*a*sin(d
*x + c) - 12*(a*sin(d*x + c) + a)/sin(d*x + c))/d

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maple [A]  time = 0.31, size = 116, normalized size = 1.40 \[ \frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {8 a \sin \left (d x +c \right )}{3 d}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{d}-\frac {4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 69, normalized size = 0.83 \[ \frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, a}{\sin \left (d x + c\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*a*sin(d*x + c)^2 + 12*a*log(sin(d*x + c)) - 24*a*sin(d*x +
c) - 12*a/sin(d*x + c))/d

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mupad [B]  time = 8.88, size = 250, normalized size = 3.01 \[ \frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}-\frac {4\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {4\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {9\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {20\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x)/2)^2))/d - (4*a*cos(c/2 + (d*x)/2)
^2)/d + (8*a*cos(c/2 + (d*x)/2)^4)/d - (8*a*cos(c/2 + (d*x)/2)^6)/d + (4*a*cos(c/2 + (d*x)/2)^8)/d - (9*a*cos(
c/2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a*sin(c/2 + (d*x)/2))/(2*d*cos(c/2 + (d*x)/2)) + (20*a*cos(c/2 + (
d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) - (16*a*cos(c/2 + (d*x)/2)^5)/(3*d*sin(c/2 + (d*x)/2)) + (8*a*cos(c/2 + (d
*x)/2)^7)/(3*d*sin(c/2 + (d*x)/2))

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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)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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