3.1223 \(\int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=130 \[ \frac{\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 d}-\frac{\left (a^2-2 b^2\right ) \csc ^2(c+d x)}{2 d}-\frac{a^2 \csc ^6(c+d x)}{6 d}-\frac{2 a b \csc ^5(c+d x)}{5 d}+\frac{4 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{b^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.158191, antiderivative size = 130, 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} \[ \frac{\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 d}-\frac{\left (a^2-2 b^2\right ) \csc ^2(c+d x)}{2 d}-\frac{a^2 \csc ^6(c+d x)}{6 d}-\frac{2 a b \csc ^5(c+d x)}{5 d}+\frac{4 a b \csc ^3(c+d x)}{3 d}-\frac{2 a b \csc (c+d x)}{d}+\frac{b^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*Csc[c + d*x])/d - ((a^2 - 2*b^2)*Csc[c + d*x]^2)/(2*d) + (4*a*b*Csc[c + d*x]^3)/(3*d) + ((2*a^2 - b^2)
*Csc[c + d*x]^4)/(4*d) - (2*a*b*Csc[c + d*x]^5)/(5*d) - (a^2*Csc[c + d*x]^6)/(6*d) + (b^2*Log[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 ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^7 (a+x)^2 \left (b^2-x^2\right )^2}{x^7} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (b^2-x^2\right )^2}{x^7} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \left (\frac{a^2 b^4}{x^7}+\frac{2 a b^4}{x^6}+\frac{-2 a^2 b^2+b^4}{x^5}-\frac{4 a b^2}{x^4}+\frac{a^2-2 b^2}{x^3}+\frac{2 a}{x^2}+\frac{1}{x}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{2 a b \csc (c+d x)}{d}-\frac{\left (a^2-2 b^2\right ) \csc ^2(c+d x)}{2 d}+\frac{4 a b \csc ^3(c+d x)}{3 d}+\frac{\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 d}-\frac{2 a b \csc ^5(c+d x)}{5 d}-\frac{a^2 \csc ^6(c+d x)}{6 d}+\frac{b^2 \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.165615, size = 107, normalized size = 0.82 \[ \frac{15 \left (2 a^2-b^2\right ) \csc ^4(c+d x)-30 \left (a^2-2 b^2\right ) \csc ^2(c+d x)-10 a^2 \csc ^6(c+d x)-24 a b \csc ^5(c+d x)+80 a b \csc ^3(c+d x)-120 a b \csc (c+d x)+60 b^2 \log (\sin (c+d x))}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*a*b*Csc[c + d*x] - 30*(a^2 - 2*b^2)*Csc[c + d*x]^2 + 80*a*b*Csc[c + d*x]^3 + 15*(2*a^2 - b^2)*Csc[c + d*
x]^4 - 24*a*b*Csc[c + d*x]^5 - 10*a^2*Csc[c + d*x]^6 + 60*b^2*Log[Sin[c + d*x]])/(60*d)

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Maple [A]  time = 0.095, size = 196, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}}-{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5\,d\sin \left ( dx+c \right ) }}-{\frac{16\,ab\sin \left ( dx+c \right ) }{15\,d}}-{\frac{2\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{8\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{b}^{2} \left ( \cot \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)^7*(a+b*sin(d*x+c))^2,x)

[Out]

-1/6/d*a^2/sin(d*x+c)^6*cos(d*x+c)^6-2/5/d*a*b/sin(d*x+c)^5*cos(d*x+c)^6+2/15/d*a*b/sin(d*x+c)^3*cos(d*x+c)^6-
2/5/d*a*b/sin(d*x+c)*cos(d*x+c)^6-16/15*a*b*sin(d*x+c)/d-2/5/d*sin(d*x+c)*a*b*cos(d*x+c)^4-8/15/d*sin(d*x+c)*a
*b*cos(d*x+c)^2-1/4/d*b^2*cot(d*x+c)^4+1/2/d*b^2*cot(d*x+c)^2+b^2*ln(sin(d*x+c))/d

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Maxima [A]  time = 0.988596, size = 146, normalized size = 1.12 \begin{align*} \frac{60 \, b^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac{120 \, a b \sin \left (d x + c\right )^{5} - 80 \, a b \sin \left (d x + c\right )^{3} + 30 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 24 \, a b \sin \left (d x + c\right ) - 15 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/60*(60*b^2*log(sin(d*x + c)) - (120*a*b*sin(d*x + c)^5 - 80*a*b*sin(d*x + c)^3 + 30*(a^2 - 2*b^2)*sin(d*x +
c)^4 + 24*a*b*sin(d*x + c) - 15*(2*a^2 - b^2)*sin(d*x + c)^2 + 10*a^2)/sin(d*x + c)^6)/d

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Fricas [A]  time = 1.80379, size = 448, normalized size = 3.45 \begin{align*} \frac{30 \,{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 15 \,{\left (2 \, a^{2} - 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 10 \, a^{2} - 45 \, b^{2} + 60 \,{\left (b^{2} \cos \left (d x + c\right )^{6} - 3 \, b^{2} \cos \left (d x + c\right )^{4} + 3 \, b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 8 \,{\left (15 \, a b \cos \left (d x + c\right )^{4} - 20 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{60 \,{\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, 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)^7*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(30*(a^2 - 2*b^2)*cos(d*x + c)^4 - 15*(2*a^2 - 7*b^2)*cos(d*x + c)^2 + 10*a^2 - 45*b^2 + 60*(b^2*cos(d*x
+ c)^6 - 3*b^2*cos(d*x + c)^4 + 3*b^2*cos(d*x + c)^2 - b^2)*log(1/2*sin(d*x + c)) + 8*(15*a*b*cos(d*x + c)^4 -
20*a*b*cos(d*x + c)^2 + 8*a*b)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*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)**7*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2441, size = 181, normalized size = 1.39 \begin{align*} \frac{60 \, b^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{147 \, b^{2} \sin \left (d x + c\right )^{6} + 120 \, 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} - 80 \, a b \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, b^{2} \sin \left (d x + c\right )^{2} + 24 \, a b \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*b^2*log(abs(sin(d*x + c))) - (147*b^2*sin(d*x + c)^6 + 120*a*b*sin(d*x + c)^5 + 30*a^2*sin(d*x + c)^4
- 60*b^2*sin(d*x + c)^4 - 80*a*b*sin(d*x + c)^3 - 30*a^2*sin(d*x + c)^2 + 15*b^2*sin(d*x + c)^2 + 24*a*b*sin(
d*x + c) + 10*a^2)/sin(d*x + c)^6)/d