3.1225 \(\int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=138 \[ \frac{\left (2 a^2-b^2\right ) \csc ^6(c+d x)}{6 d}-\frac{\left (a^2-2 b^2\right ) \csc ^4(c+d x)}{4 d}-\frac{a^2 \csc ^8(c+d x)}{8 d}-\frac{2 a b \csc ^7(c+d x)}{7 d}+\frac{4 a b \csc ^5(c+d x)}{5 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{b^2 \csc ^2(c+d x)}{2 d} \]

[Out]

-(b^2*Csc[c + d*x]^2)/(2*d) - (2*a*b*Csc[c + d*x]^3)/(3*d) - ((a^2 - 2*b^2)*Csc[c + d*x]^4)/(4*d) + (4*a*b*Csc
[c + d*x]^5)/(5*d) + ((2*a^2 - b^2)*Csc[c + d*x]^6)/(6*d) - (2*a*b*Csc[c + d*x]^7)/(7*d) - (a^2*Csc[c + d*x]^8
)/(8*d)

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Rubi [A]  time = 0.161012, antiderivative size = 138, 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} \[ \frac{\left (2 a^2-b^2\right ) \csc ^6(c+d x)}{6 d}-\frac{\left (a^2-2 b^2\right ) \csc ^4(c+d x)}{4 d}-\frac{a^2 \csc ^8(c+d x)}{8 d}-\frac{2 a b \csc ^7(c+d x)}{7 d}+\frac{4 a b \csc ^5(c+d x)}{5 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{b^2 \csc ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

-(b^2*Csc[c + d*x]^2)/(2*d) - (2*a*b*Csc[c + d*x]^3)/(3*d) - ((a^2 - 2*b^2)*Csc[c + d*x]^4)/(4*d) + (4*a*b*Csc
[c + d*x]^5)/(5*d) + ((2*a^2 - b^2)*Csc[c + d*x]^6)/(6*d) - (2*a*b*Csc[c + d*x]^7)/(7*d) - (a^2*Csc[c + d*x]^8
)/(8*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 ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b^9 (a+x)^2 \left (b^2-x^2\right )^2}{x^9} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (b^2-x^2\right )^2}{x^9} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \left (\frac{a^2 b^4}{x^9}+\frac{2 a b^4}{x^8}+\frac{-2 a^2 b^2+b^4}{x^7}-\frac{4 a b^2}{x^6}+\frac{a^2-2 b^2}{x^5}+\frac{2 a}{x^4}+\frac{1}{x^3}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{b^2 \csc ^2(c+d x)}{2 d}-\frac{2 a b \csc ^3(c+d x)}{3 d}-\frac{\left (a^2-2 b^2\right ) \csc ^4(c+d x)}{4 d}+\frac{4 a b \csc ^5(c+d x)}{5 d}+\frac{\left (2 a^2-b^2\right ) \csc ^6(c+d x)}{6 d}-\frac{2 a b \csc ^7(c+d x)}{7 d}-\frac{a^2 \csc ^8(c+d x)}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.218315, size = 108, normalized size = 0.78 \[ -\frac{\csc ^2(c+d x) \left (-140 \left (2 a^2-b^2\right ) \csc ^4(c+d x)+210 \left (a^2-2 b^2\right ) \csc ^2(c+d x)+105 a^2 \csc ^6(c+d x)+240 a b \csc ^5(c+d x)-672 a b \csc ^3(c+d x)+560 a b \csc (c+d x)+420 b^2\right )}{840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

-(Csc[c + d*x]^2*(420*b^2 + 560*a*b*Csc[c + d*x] + 210*(a^2 - 2*b^2)*Csc[c + d*x]^2 - 672*a*b*Csc[c + d*x]^3 -
140*(2*a^2 - b^2)*Csc[c + d*x]^4 + 240*a*b*Csc[c + d*x]^5 + 105*a^2*Csc[c + d*x]^6))/(840*d)

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Maple [A]  time = 0.092, size = 173, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{8}}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24\, \left ( \sin \left ( dx+c \right ) \right ) ^{6}}} \right ) +2\,ab \left ( -1/7\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}}-1/35\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{105\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-1/35\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{\sin \left ( dx+c \right ) }}-1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{6}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/8/sin(d*x+c)^8*cos(d*x+c)^6-1/24/sin(d*x+c)^6*cos(d*x+c)^6)+2*a*b*(-1/7/sin(d*x+c)^7*cos(d*x+c)^6
-1/35/sin(d*x+c)^5*cos(d*x+c)^6+1/105/sin(d*x+c)^3*cos(d*x+c)^6-1/35/sin(d*x+c)*cos(d*x+c)^6-1/35*(8/3+cos(d*x
+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/6*b^2/sin(d*x+c)^6*cos(d*x+c)^6)

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Maxima [A]  time = 0.984058, size = 143, normalized size = 1.04 \begin{align*} -\frac{420 \, b^{2} \sin \left (d x + c\right )^{6} + 560 \, a b \sin \left (d x + c\right )^{5} - 672 \, a b \sin \left (d x + c\right )^{3} + 210 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 240 \, a b \sin \left (d x + c\right ) - 140 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + 105 \, a^{2}}{840 \, d \sin \left (d x + c\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/840*(420*b^2*sin(d*x + c)^6 + 560*a*b*sin(d*x + c)^5 - 672*a*b*sin(d*x + c)^3 + 210*(a^2 - 2*b^2)*sin(d*x +
c)^4 + 240*a*b*sin(d*x + c) - 140*(2*a^2 - b^2)*sin(d*x + c)^2 + 105*a^2)/(d*sin(d*x + c)^8)

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Fricas [A]  time = 1.71055, size = 379, normalized size = 2.75 \begin{align*} \frac{420 \, b^{2} \cos \left (d x + c\right )^{6} - 210 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 140 \,{\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 35 \, a^{2} - 140 \, b^{2} - 16 \,{\left (35 \, a b \cos \left (d x + c\right )^{4} - 28 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{840 \,{\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, 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)^9*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/840*(420*b^2*cos(d*x + c)^6 - 210*(a^2 + 4*b^2)*cos(d*x + c)^4 + 140*(a^2 + 4*b^2)*cos(d*x + c)^2 - 35*a^2 -
140*b^2 - 16*(35*a*b*cos(d*x + c)^4 - 28*a*b*cos(d*x + c)^2 + 8*a*b)*sin(d*x + c))/(d*cos(d*x + c)^8 - 4*d*co
s(d*x + c)^6 + 6*d*cos(d*x + c)^4 - 4*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)**9*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24006, size = 159, normalized size = 1.15 \begin{align*} -\frac{420 \, b^{2} \sin \left (d x + c\right )^{6} + 560 \, a b \sin \left (d x + c\right )^{5} + 210 \, a^{2} \sin \left (d x + c\right )^{4} - 420 \, b^{2} \sin \left (d x + c\right )^{4} - 672 \, a b \sin \left (d x + c\right )^{3} - 280 \, a^{2} \sin \left (d x + c\right )^{2} + 140 \, b^{2} \sin \left (d x + c\right )^{2} + 240 \, a b \sin \left (d x + c\right ) + 105 \, a^{2}}{840 \, d \sin \left (d x + c\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/840*(420*b^2*sin(d*x + c)^6 + 560*a*b*sin(d*x + c)^5 + 210*a^2*sin(d*x + c)^4 - 420*b^2*sin(d*x + c)^4 - 67
2*a*b*sin(d*x + c)^3 - 280*a^2*sin(d*x + c)^2 + 140*b^2*sin(d*x + c)^2 + 240*a*b*sin(d*x + c) + 105*a^2)/(d*si
n(d*x + c)^8)