3.1219 \(\int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=127 \[ \frac{\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 d}-\frac{\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}+\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{4 a b \sin (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}+\frac{b^2 \sin ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.157214, antiderivative size = 127, 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, 948} \[ \frac{\left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 d}-\frac{\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^2(c+d x)}{2 d}+\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{4 a b \sin (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}+\frac{b^2 \sin ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - ((2*a^2 - b^2)*Log[Sin[c + d*x]])/d - (4*a*b*Sin[c + d*
x])/d + ((a^2 - 2*b^2)*Sin[c + d*x]^2)/(2*d) + (2*a*b*Sin[c + d*x]^3)/(3*d) + (b^2*Sin[c + d*x]^4)/(4*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 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 0.282604, size = 103, normalized size = 0.81 \[ \frac{6 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+12 \left (b^2-2 a^2\right ) \log (\sin (c+d x))-6 a^2 \csc ^2(c+d x)+8 a b \sin ^3(c+d x)-48 a b \sin (c+d x)-24 a b \csc (c+d x)+3 b^2 \sin ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a*b*Csc[c + d*x] - 6*a^2*Csc[c + d*x]^2 + 12*(-2*a^2 + b^2)*Log[Sin[c + d*x]] - 48*a*b*Sin[c + d*x] + 6*(
a^2 - 2*b^2)*Sin[c + d*x]^2 + 8*a*b*Sin[c + d*x]^3 + 3*b^2*Sin[c + d*x]^4)/(12*d)

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Maple [A]  time = 0.087, size = 197, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{16\,ab\sin \left ( dx+c \right ) }{3\,d}}-2\,{\frac{ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{8\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \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))^2,x)

[Out]

-1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c)^6-1/2/d*a^2*cos(d*x+c)^4-1/d*a^2*cos(d*x+c)^2-2*a^2*ln(sin(d*x+c))/d-2/d*a*
b/sin(d*x+c)*cos(d*x+c)^6-16/3*a*b*sin(d*x+c)/d-2/d*sin(d*x+c)*a*b*cos(d*x+c)^4-8/3/d*sin(d*x+c)*a*b*cos(d*x+c
)^2+1/4/d*b^2*cos(d*x+c)^4+1/2/d*b^2*cos(d*x+c)^2+b^2*ln(sin(d*x+c))/d

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Maxima [A]  time = 0.995812, size = 140, normalized size = 1.1 \begin{align*} \frac{3 \, b^{2} \sin \left (d x + c\right )^{4} + 8 \, a b \sin \left (d x + c\right )^{3} - 48 \, a b \sin \left (d x + c\right ) + 6 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - 12 \,{\left (2 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) - \frac{6 \,{\left (4 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, 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))^2,x, algorithm="maxima")

[Out]

1/12*(3*b^2*sin(d*x + c)^4 + 8*a*b*sin(d*x + c)^3 - 48*a*b*sin(d*x + c) + 6*(a^2 - 2*b^2)*sin(d*x + c)^2 - 12*
(2*a^2 - b^2)*log(sin(d*x + c)) - 6*(4*a*b*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d

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Fricas [A]  time = 1.79847, size = 381, normalized size = 3. \begin{align*} \frac{24 \, b^{2} \cos \left (d x + c\right )^{6} - 24 \,{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 9 \,{\left (8 \, a^{2} - 9 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, a^{2} + 33 \, b^{2} - 96 \,{\left ({\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} + b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 64 \,{\left (a b \cos \left (d x + c\right )^{4} + 4 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b\right )} \sin \left (d x + c\right )}{96 \,{\left (d \cos \left (d x + c\right )^{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))^2,x, algorithm="fricas")

[Out]

1/96*(24*b^2*cos(d*x + c)^6 - 24*(2*a^2 - b^2)*cos(d*x + c)^4 + 9*(8*a^2 - 9*b^2)*cos(d*x + c)^2 + 24*a^2 + 33
*b^2 - 96*((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 + b^2)*log(1/2*sin(d*x + c)) - 64*(a*b*cos(d*x + c)^4 + 4*a*b*
cos(d*x + c)^2 - 8*a*b)*sin(d*x + c))/(d*cos(d*x + c)^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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.30277, size = 189, normalized size = 1.49 \begin{align*} \frac{3 \, b^{2} \sin \left (d x + c\right )^{4} + 8 \, a b \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, b^{2} \sin \left (d x + c\right )^{2} - 48 \, a b \sin \left (d x + c\right ) - 12 \,{\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac{6 \,{\left (6 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} - 4 \, a b \sin \left (d x + c\right ) - a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, 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))^2,x, algorithm="giac")

[Out]

1/12*(3*b^2*sin(d*x + c)^4 + 8*a*b*sin(d*x + c)^3 + 6*a^2*sin(d*x + c)^2 - 12*b^2*sin(d*x + c)^2 - 48*a*b*sin(
d*x + c) - 12*(2*a^2 - b^2)*log(abs(sin(d*x + c))) + 6*(6*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^2 - 4*a*b*si
n(d*x + c) - a^2)/sin(d*x + c)^2)/d