3.6.37 \(\int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [537]

Optimal. Leaf size=62 \[ -\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 62, 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 = {2915, 12, 76} \begin {gather*} \frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - Log[Sin[c + d*x]]/(a*d) - Sin[c + d*x]/(a*d) + Sin[c + d*x]^2/(2*a*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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 2915

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

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 45, normalized size = 0.73 \begin {gather*} \frac {6-2 \csc (c+d x)-2 \log (\sin (c+d x))-2 \sin (c+d x)+\sin ^2(c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6 - 2*Csc[c + d*x] - 2*Log[Sin[c + d*x]] - 2*Sin[c + d*x] + Sin[c + d*x]^2)/(2*a*d)

________________________________________________________________________________________

Maple [A]
time = 0.20, size = 46, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )}{d a}\)	\(46\)
default	\(\frac {-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )}{d a}\)	\(46\)
risch	\(\frac {i x}{a}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {2 i c}{a d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}\)	\(140\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {7 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {7 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\)	\(223\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 52, normalized size = 0.84 \begin {gather*} \frac {\frac {\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{a} - \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {2}{a \sin \left (d x + c\right )}}{2 \, d} \end {gather*}

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]

1/2*((sin(d*x + c)^2 - 2*sin(d*x + c))/a - 2*log(sin(d*x + c))/a - 2/(a*sin(d*x + c)))/d

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 65, normalized size = 1.05 \begin {gather*} \frac {4 \, \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8}{4 \, a d \sin \left (d x + c\right )} \end {gather*}

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]

1/4*(4*cos(d*x + c)^2 - (2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 4*log(1/2*sin(d*x + c))*sin(d*x + c) - 8)/(a*d*s
in(d*x + c))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 65, normalized size = 1.05 \begin {gather*} -\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{a^{2}} - \frac {2 \, {\left (\sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )}}{2 \, d} \end {gather*}

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]

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

________________________________________________________________________________________

Mupad [B]
time = 8.80, size = 146, normalized size = 2.35 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________