3.4.32 \(\int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [332]

Optimal. Leaf size=59 \[ \frac {\log (\sin (c+d x))}{a d}+\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a b^2 d}-\frac {\sin (c+d x)}{b d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 59, 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 = {2916, 12, 908} \begin {gather*} \frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a b^2 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Sin[c + d*x]]/(a*d) + ((a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(a*b^2*d) - Sin[c + d*x]/(b*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 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {b \left (b^2-x^2\right )}{x (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\text {Subst}\left (\int \frac {b^2-x^2}{x (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac {\text {Subst}\left (\int \left (-1+\frac {b^2}{a x}+\frac {a^2-b^2}{a (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac {\log (\sin (c+d x))}{a d}+\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a b^2 d}-\frac {\sin (c+d x)}{b d}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 53, normalized size = 0.90 \begin {gather*} \frac {b^2 \log (\sin (c+d x))+\left (a^2-b^2\right ) \log (a+b \sin (c+d x))-a b \sin (c+d x)}{a b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.18, size = 55, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} a}}{d}\)	\(55\)
default	\(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} a}}{d}\)	\(55\)
risch	\(-\frac {i a x}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {2 i a c}{b^{2} d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 54, normalized size = 0.92 \begin {gather*} \frac {\frac {\log \left (\sin \left (d x + c\right )\right )}{a} - \frac {\sin \left (d x + c\right )}{b} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a b^{2}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 55, normalized size = 0.93 \begin {gather*} \frac {b^{2} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - a b \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2*cot(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [A]
time = 7.95, size = 56, normalized size = 0.95 \begin {gather*} \frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {\sin \left (d x + c\right )}{b} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a b^{2}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.69, size = 98, normalized size = 1.66 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\sin \left (c+d\,x\right )}{b\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (\frac {a}{b^2}-\frac {1}{a}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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