3.348 \(\int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=96 \[ \frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {a \sin (c+d x)}{b^2 d}-\frac {\csc (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 96, 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 = {2837, 12, 894} \[ \frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3 d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {a \sin (c+d x)}{b^2 d}-\frac {\csc (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - (b*Log[Sin[c + d*x]])/(a^2*d) + ((a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(a^2*b^3*d) -
(a*Sin[c + d*x])/(b^2*d) + Sin[c + d*x]^2/(2*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 894

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 2837

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

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Mathematica [A]  time = 0.21, size = 86, normalized size = 0.90 \[ \frac {\frac {2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^2 b^3}-\frac {2 b \log (\sin (c+d x))}{a^2}-\frac {2 a \sin (c+d x)}{b^2}-\frac {2 \csc (c+d x)}{a}+\frac {\sin ^2(c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 133, normalized size = 1.39 \[ \frac {4 \, a^{3} b \cos \left (d x + c\right )^{2} - 4 \, b^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 4 \, a^{3} b - 4 \, a b^{3} + 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{4 \, a^{2} b^{3} d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.91, size = 105, normalized size = 1.09 \[ -\frac {\frac {2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 124, normalized size = 1.29 \[ \frac {\sin ^{2}\left (d x +c \right )}{2 b d}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) a^{2}}{d \,b^{3}}-\frac {2 \ln \left (a +b \sin \left (d x +c \right )\right )}{b d}+\frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{2}}-\frac {1}{d a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 91, normalized size = 0.95 \[ -\frac {\frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} + \frac {2}{a \sin \left (d x + c\right )} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.05, size = 233, normalized size = 2.43 \[ \frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,{\left (a^2-b^2\right )}^2}{a^2\,b^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2-2\,b^2\right )}{b^3\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+b^2\right )}{b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+b^2\right )}{b^2}-\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b}+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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