3.1221 \(\int \cot ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=126 \[ \frac{\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 d}+\frac{\left (a^2-2 b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^4(c+d x)}{4 d}+\frac{2 a b \sin (c+d x)}{d}-\frac{2 a b \csc ^3(c+d x)}{3 d}+\frac{4 a b \csc (c+d x)}{d}+\frac{b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0919727, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 948} \[ \frac{\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 d}+\frac{\left (a^2-2 b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 \csc ^4(c+d x)}{4 d}+\frac{2 a b \sin (c+d x)}{d}-\frac{2 a b \csc ^3(c+d x)}{3 d}+\frac{4 a b \csc (c+d x)}{d}+\frac{b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A]  time = 0.760486, size = 107, normalized size = 0.85 \[ \frac{6 \left (2 a^2-b^2\right ) \csc ^2(c+d x)+6 \left (2 \left (a^2-2 b^2\right ) \log (\sin (c+d x))+4 a b \sin (c+d x)+b^2 \sin ^2(c+d x)\right )-3 a^2 \csc ^4(c+d x)-8 a b \csc ^3(c+d x)+48 a b \csc (c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.088, size = 220, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+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 ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\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)^5*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.02934, size = 142, normalized size = 1.13 \begin{align*} \frac{6 \, b^{2} \sin \left (d x + c\right )^{2} + 24 \, a b \sin \left (d x + c\right ) + 12 \,{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac{48 \, a b \sin \left (d x + c\right )^{3} - 8 \, a b \sin \left (d x + c\right ) + 6 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{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)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(6*b^2*sin(d*x + c)^2 + 24*a*b*sin(d*x + c) + 12*(a^2 - 2*b^2)*log(sin(d*x + c)) + (48*a*b*sin(d*x + c)^3
- 8*a*b*sin(d*x + c) + 6*(2*a^2 - b^2)*sin(d*x + c)^2 - 3*a^2)/sin(d*x + c)^4)/d

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Fricas [A]  time = 1.76401, size = 437, normalized size = 3.47 \begin{align*} -\frac{6 \, b^{2} \cos \left (d x + c\right )^{6} - 15 \, b^{2} \cos \left (d x + c\right )^{4} + 6 \,{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 9 \, a^{2} + 3 \, b^{2} - 12 \,{\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} - 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (3 \, a b \cos \left (d x + c\right )^{4} - 12 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(6*b^2*cos(d*x + c)^6 - 15*b^2*cos(d*x + c)^4 + 6*(2*a^2 + b^2)*cos(d*x + c)^2 - 9*a^2 + 3*b^2 - 12*((a^
2 - 2*b^2)*cos(d*x + c)^4 - 2*(a^2 - 2*b^2)*cos(d*x + c)^2 + a^2 - 2*b^2)*log(1/2*sin(d*x + c)) - 8*(3*a*b*cos
(d*x + c)^4 - 12*a*b*cos(d*x + c)^2 + 8*a*b)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*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)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27178, size = 186, normalized size = 1.48 \begin{align*} \frac{6 \, b^{2} \sin \left (d x + c\right )^{2} + 24 \, a b \sin \left (d x + c\right ) + 12 \,{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{25 \, a^{2} \sin \left (d x + c\right )^{4} - 50 \, b^{2} \sin \left (d x + c\right )^{4} - 48 \, a b \sin \left (d x + c\right )^{3} - 12 \, a^{2} \sin \left (d x + c\right )^{2} + 6 \, b^{2} \sin \left (d x + c\right )^{2} + 8 \, a b \sin \left (d x + c\right ) + 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{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)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(6*b^2*sin(d*x + c)^2 + 24*a*b*sin(d*x + c) + 12*(a^2 - 2*b^2)*log(abs(sin(d*x + c))) - (25*a^2*sin(d*x +
c)^4 - 50*b^2*sin(d*x + c)^4 - 48*a*b*sin(d*x + c)^3 - 12*a^2*sin(d*x + c)^2 + 6*b^2*sin(d*x + c)^2 + 8*a*b*s
in(d*x + c) + 3*a^2)/sin(d*x + c)^4)/d