3.1222 \(\int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=124 \[ \frac{\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 d}-\frac{\left (a^2-2 b^2\right ) \csc (c+d x)}{d}-\frac{a^2 \csc ^5(c+d x)}{5 d}-\frac{a b \csc ^4(c+d x)}{2 d}+\frac{2 a b \csc ^2(c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin (c+d x)}{d} \]

[Out]

-(((a^2 - 2*b^2)*Csc[c + d*x])/d) + (2*a*b*Csc[c + d*x]^2)/d + ((2*a^2 - b^2)*Csc[c + d*x]^3)/(3*d) - (a*b*Csc
[c + d*x]^4)/(2*d) - (a^2*Csc[c + d*x]^5)/(5*d) + (2*a*b*Log[Sin[c + d*x]])/d + (b^2*Sin[c + d*x])/d

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Rubi [A]  time = 0.140322, antiderivative size = 124, normalized size of antiderivative = 1., 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 = {2837, 12, 948} \[ \frac{\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 d}-\frac{\left (a^2-2 b^2\right ) \csc (c+d x)}{d}-\frac{a^2 \csc ^5(c+d x)}{5 d}-\frac{a b \csc ^4(c+d x)}{2 d}+\frac{2 a b \csc ^2(c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 2*b^2)*Csc[c + d*x])/d) + (2*a*b*Csc[c + d*x]^2)/d + ((2*a^2 - b^2)*Csc[c + d*x]^3)/(3*d) - (a*b*Csc
[c + d*x]^4)/(2*d) - (a^2*Csc[c + d*x]^5)/(5*d) + (2*a*b*Log[Sin[c + d*x]])/d + (b^2*Sin[c + d*x])/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 \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^6 (a+x)^2 \left (b^2-x^2\right )^2}{x^6} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (b^2-x^2\right )^2}{x^6} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \left (1+\frac{a^2 b^4}{x^6}+\frac{2 a b^4}{x^5}+\frac{-2 a^2 b^2+b^4}{x^4}-\frac{4 a b^2}{x^3}+\frac{a^2-2 b^2}{x^2}+\frac{2 a}{x}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\left (a^2-2 b^2\right ) \csc (c+d x)}{d}+\frac{2 a b \csc ^2(c+d x)}{d}+\frac{\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 d}-\frac{a b \csc ^4(c+d x)}{2 d}-\frac{a^2 \csc ^5(c+d x)}{5 d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.161695, size = 105, normalized size = 0.85 \[ \frac{10 \left (2 a^2-b^2\right ) \csc ^3(c+d x)-30 \left (a^2-2 b^2\right ) \csc (c+d x)-6 a^2 \csc ^5(c+d x)-15 a b \csc ^4(c+d x)+60 a b \csc ^2(c+d x)+30 b (2 a \log (\sin (c+d x))+b \sin (c+d x))}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*(a^2 - 2*b^2)*Csc[c + d*x] + 60*a*b*Csc[c + d*x]^2 + 10*(2*a^2 - b^2)*Csc[c + d*x]^3 - 15*a*b*Csc[c + d*x
]^4 - 6*a^2*Csc[c + d*x]^5 + 30*b*(2*a*Log[Sin[c + d*x]] + b*Sin[c + d*x]))/(30*d)

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Maple [B]  time = 0.089, size = 279, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}-{\frac{8\,{a}^{2}\sin \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{ab \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{ab \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{ab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}+{\frac{8\,{b}^{2}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{4\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x)

[Out]

-1/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^6+1/15/d*a^2/sin(d*x+c)^3*cos(d*x+c)^6-1/5/d*a^2/sin(d*x+c)*cos(d*x+c)^6-8/
15*a^2*sin(d*x+c)/d-1/5/d*a^2*sin(d*x+c)*cos(d*x+c)^4-4/15/d*a^2*sin(d*x+c)*cos(d*x+c)^2-1/2/d*a*b*cot(d*x+c)^
4+1/d*a*b*cot(d*x+c)^2+2*a*b*ln(sin(d*x+c))/d-1/3/d*b^2/sin(d*x+c)^3*cos(d*x+c)^6+1/d*b^2/sin(d*x+c)*cos(d*x+c
)^6+8/3*b^2*sin(d*x+c)/d+1/d*sin(d*x+c)*b^2*cos(d*x+c)^4+4/3/d*sin(d*x+c)*b^2*cos(d*x+c)^2

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Maxima [A]  time = 0.990565, size = 142, normalized size = 1.15 \begin{align*} \frac{60 \, a b \log \left (\sin \left (d x + c\right )\right ) + 30 \, b^{2} \sin \left (d x + c\right ) + \frac{60 \, a b \sin \left (d x + c\right )^{3} - 30 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} - 15 \, a b \sin \left (d x + c\right ) + 10 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(60*a*b*log(sin(d*x + c)) + 30*b^2*sin(d*x + c) + (60*a*b*sin(d*x + c)^3 - 30*(a^2 - 2*b^2)*sin(d*x + c)^
4 - 15*a*b*sin(d*x + c) + 10*(2*a^2 - b^2)*sin(d*x + c)^2 - 6*a^2)/sin(d*x + c)^5)/d

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Fricas [A]  time = 1.8117, size = 425, normalized size = 3.43 \begin{align*} -\frac{30 \, b^{2} \cos \left (d x + c\right )^{6} + 30 \,{\left (a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 40 \,{\left (a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 60 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 16 \, a^{2} - 80 \, b^{2} + 15 \,{\left (4 \, a b \cos \left (d x + c\right )^{2} - 3 \, a b\right )} \sin \left (d x + c\right )}{30 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(30*b^2*cos(d*x + c)^6 + 30*(a^2 - 5*b^2)*cos(d*x + c)^4 - 40*(a^2 - 5*b^2)*cos(d*x + c)^2 - 60*(a*b*cos
(d*x + c)^4 - 2*a*b*cos(d*x + c)^2 + a*b)*log(1/2*sin(d*x + c))*sin(d*x + c) + 16*a^2 - 80*b^2 + 15*(4*a*b*cos
(d*x + c)^2 - 3*a*b)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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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)**6*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.23539, size = 177, normalized size = 1.43 \begin{align*} \frac{60 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, b^{2} \sin \left (d x + c\right ) - \frac{137 \, a b \sin \left (d x + c\right )^{5} + 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, b^{2} \sin \left (d x + c\right )^{4} - 60 \, a b \sin \left (d x + c\right )^{3} - 20 \, a^{2} \sin \left (d x + c\right )^{2} + 10 \, b^{2} \sin \left (d x + c\right )^{2} + 15 \, a b \sin \left (d x + c\right ) + 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(60*a*b*log(abs(sin(d*x + c))) + 30*b^2*sin(d*x + c) - (137*a*b*sin(d*x + c)^5 + 30*a^2*sin(d*x + c)^4 -
60*b^2*sin(d*x + c)^4 - 60*a*b*sin(d*x + c)^3 - 20*a^2*sin(d*x + c)^2 + 10*b^2*sin(d*x + c)^2 + 15*a*b*sin(d*x
+ c) + 6*a^2)/sin(d*x + c)^5)/d