3.1315 \(\int \frac{\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=105 \[ -\frac{\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}-\frac{\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\sin (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.179845, antiderivative size = 105, normalized size of antiderivative = 1., 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} \[ -\frac{\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}-\frac{\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d}+\frac{\sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]))

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^3 \left (b^2-x^2\right )^2}{x^3 (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^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{b^4}{a x^3}-\frac{b^4}{a^2 x^2}+\frac{-2 a^2 b^2+b^4}{a^3 x}-\frac{\left (a^2-b^2\right )^2}{a^3 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d}-\frac{\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}+\frac{\sin (c+d x)}{b d}\\ \end{align*}

Mathematica [A]  time = 0.3884, size = 97, normalized size = 0.92 \[ \frac{\frac{\frac{2 b^2 \left (b^2-2 a^2\right ) \log (\sin (c+d x))-2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3}+2 b \sin (c+d x)}{b^2}+\frac{2 b \csc (c+d x)}{a^2}-\frac{\csc ^2(c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.086, size = 140, normalized size = 1.3 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{bd}}-{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{2}}}+2\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{da}}-{\frac{{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}-{\frac{1}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}+{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}+{\frac{b}{d{a}^{2}\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.00201, size = 134, normalized size = 1.28 \begin{align*} \frac{\frac{2 \, \sin \left (d x + c\right )}{b} - \frac{2 \,{\left (2 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \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^{3} b^{2}} + \frac{2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74236, size = 382, normalized size = 3.64 \begin{align*} \frac{a^{2} b^{2} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \,{\left (2 \, a^{2} b^{2} - b^{4} -{\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 2 \,{\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{3} b^{2} d \cos \left (d x + c\right )^{2} - a^{3} b^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.21629, size = 176, normalized size = 1.68 \begin{align*} \frac{\frac{2 \, \sin \left (d x + c\right )}{b} - \frac{2 \,{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \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^{3} b^{2}} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*sin(d*x + c)/b - 2*(2*a^2 - b^2)*log(abs(sin(d*x + c)))/a^3 - 2*(a^4 - 2*a^2*b^2 + b^4)*log(abs(b*sin(d
*x + c) + a))/(a^3*b^2) + (6*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^2 + 2*a*b*sin(d*x + c) - a^2)/(a^3*sin(d*
x + c)^2))/d