\(\int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1224]

Optimal result

Integrand size = 29, antiderivative size = 129 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {b^2 \csc (c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {\left (a^2-2 b^2\right ) \csc ^3(c+d x)}{3 d}+\frac {a b \csc ^4(c+d x)}{d}+\frac {\left (2 a^2-b^2\right ) \csc ^5(c+d x)}{5 d}-\frac {a b \csc ^6(c+d x)}{3 d}-\frac {a^2 \csc ^7(c+d x)}{7 d} \]

[Out]

-b^2*csc(d*x+c)/d-a*b*csc(d*x+c)^2/d-1/3*(a^2-2*b^2)*csc(d*x+c)^3/d+a*b*csc(d*x+c)^4/d+1/5*(2*a^2-b^2)*csc(d*x
+c)^5/d-1/3*a*b*csc(d*x+c)^6/d-1/7*a^2*csc(d*x+c)^7/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 962} \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \csc ^5(c+d x)}{5 d}-\frac {\left (a^2-2 b^2\right ) \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^7(c+d x)}{7 d}-\frac {a b \csc ^6(c+d x)}{3 d}+\frac {a b \csc ^4(c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {b^2 \csc (c+d x)}{d} \]

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 962

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]))

Rule 2916

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/b)*x)^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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {b^8 (a+x)^2 \left (b^2-x^2\right )^2}{x^8} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {b^3 \text {Subst}\left (\int \frac {(a+x)^2 \left (b^2-x^2\right )^2}{x^8} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^3 \text {Subst}\left (\int \left (\frac {a^2 b^4}{x^8}+\frac {2 a b^4}{x^7}+\frac {-2 a^2 b^2+b^4}{x^6}-\frac {4 a b^2}{x^5}+\frac {a^2-2 b^2}{x^4}+\frac {2 a}{x^3}+\frac {1}{x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {b^2 \csc (c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {\left (a^2-2 b^2\right ) \csc ^3(c+d x)}{3 d}+\frac {a b \csc ^4(c+d x)}{d}+\frac {\left (2 a^2-b^2\right ) \csc ^5(c+d x)}{5 d}-\frac {a b \csc ^6(c+d x)}{3 d}-\frac {a^2 \csc ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\csc (c+d x) \left (105 b^2+105 a b \csc (c+d x)+35 \left (a^2-2 b^2\right ) \csc ^2(c+d x)-105 a b \csc ^3(c+d x)+21 \left (-2 a^2+b^2\right ) \csc ^4(c+d x)+35 a b \csc ^5(c+d x)+15 a^2 \csc ^6(c+d x)\right )}{105 d} \]

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

-1/105*(Csc[c + d*x]*(105*b^2 + 105*a*b*Csc[c + d*x] + 35*(a^2 - 2*b^2)*Csc[c + d*x]^2 - 105*a*b*Csc[c + d*x]^
3 + 21*(-2*a^2 + b^2)*Csc[c + d*x]^4 + 35*a*b*Csc[c + d*x]^5 + 15*a^2*Csc[c + d*x]^6))/d

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{7}\left (d x +c \right )\right ) a^{2}}{7}+\frac {a b \left (\csc ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\csc ^{5}\left (d x +c \right )\right )}{5}-a b \left (\csc ^{4}\left (d x +c \right )\right )+\frac {\left (a^{2}-2 b^{2}\right ) \left (\csc ^{3}\left (d x +c \right )\right )}{3}+a b \left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right ) b^{2}}{d}\)	\(103\)
default	\(-\frac {\frac {\left (\csc ^{7}\left (d x +c \right )\right ) a^{2}}{7}+\frac {a b \left (\csc ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\csc ^{5}\left (d x +c \right )\right )}{5}-a b \left (\csc ^{4}\left (d x +c \right )\right )+\frac {\left (a^{2}-2 b^{2}\right ) \left (\csc ^{3}\left (d x +c \right )\right )}{3}+a b \left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right ) b^{2}}{d}\)	\(103\)
parallelrisch	\(-\frac {\left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\cos \left (2 d x +2 c \right )+\frac {5 \cos \left (4 d x +4 c \right )}{4}+\frac {57}{28}\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {825 b \left (\cos \left (2 d x +2 c \right )+\frac {42 \cos \left (4 d x +4 c \right )}{55}+\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {14}{11}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{128}-20 b^{2} \left (\cos \left (2 d x +2 c \right )-\frac {3 \cos \left (4 d x +4 c \right )}{4}-\frac {29}{20}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3840 d}\)	\(159\)
risch	\(-\frac {2 i \left (105 b^{2} {\mathrm e}^{13 i \left (d x +c \right )}-140 a^{2} {\mathrm e}^{11 i \left (d x +c \right )}-350 b^{2} {\mathrm e}^{11 i \left (d x +c \right )}+700 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}-112 a^{2} {\mathrm e}^{9 i \left (d x +c \right )}+791 b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-210 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-456 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-1092 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+210 i a b \,{\mathrm e}^{12 i \left (d x +c \right )}-112 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+791 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-210 i a b \,{\mathrm e}^{10 i \left (d x +c \right )}-140 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-350 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+210 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+105 b^{2} {\mathrm e}^{i \left (d x +c \right )}-700 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}\)	\(273\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/d*(1/7*csc(d*x+c)^7*a^2+1/3*a*b*csc(d*x+c)^6+1/5*(-2*a^2+b^2)*csc(d*x+c)^5-a*b*csc(d*x+c)^4+1/3*(a^2-2*b^2)
*csc(d*x+c)^3+a*b*csc(d*x+c)^2+csc(d*x+c)*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.13 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {105 \, b^{2} \cos \left (d x + c\right )^{6} - 35 \, {\left (a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 28 \, {\left (a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \, a^{2} - 56 \, b^{2} - 35 \, {\left (3 \, a b \cos \left (d x + c\right )^{4} - 3 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(105*b^2*cos(d*x + c)^6 - 35*(a^2 + 7*b^2)*cos(d*x + c)^4 + 28*(a^2 + 7*b^2)*cos(d*x + c)^2 - 8*a^2 - 5
6*b^2 - 35*(3*a*b*cos(d*x + c)^4 - 3*a*b*cos(d*x + c)^2 + 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)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**8*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {105 \, b^{2} \sin \left (d x + c\right )^{6} + 105 \, a b \sin \left (d x + c\right )^{5} - 105 \, a b \sin \left (d x + c\right )^{3} + 35 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 35 \, a b \sin \left (d x + c\right ) - 21 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + 15 \, a^{2}}{105 \, d \sin \left (d x + c\right )^{7}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(105*b^2*sin(d*x + c)^6 + 105*a*b*sin(d*x + c)^5 - 105*a*b*sin(d*x + c)^3 + 35*(a^2 - 2*b^2)*sin(d*x +
c)^4 + 35*a*b*sin(d*x + c) - 21*(2*a^2 - b^2)*sin(d*x + c)^2 + 15*a^2)/(d*sin(d*x + c)^7)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {105 \, b^{2} \sin \left (d x + c\right )^{6} + 105 \, a b \sin \left (d x + c\right )^{5} + 35 \, a^{2} \sin \left (d x + c\right )^{4} - 70 \, b^{2} \sin \left (d x + c\right )^{4} - 105 \, a b \sin \left (d x + c\right )^{3} - 42 \, a^{2} \sin \left (d x + c\right )^{2} + 21 \, b^{2} \sin \left (d x + c\right )^{2} + 35 \, a b \sin \left (d x + c\right ) + 15 \, a^{2}}{105 \, d \sin \left (d x + c\right )^{7}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/105*(105*b^2*sin(d*x + c)^6 + 105*a*b*sin(d*x + c)^5 + 35*a^2*sin(d*x + c)^4 - 70*b^2*sin(d*x + c)^4 - 105*
a*b*sin(d*x + c)^3 - 42*a^2*sin(d*x + c)^2 + 21*b^2*sin(d*x + c)^2 + 35*a*b*sin(d*x + c) + 15*a^2)/(d*sin(d*x
+ c)^7)

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\frac {a^2}{7}+{\sin \left (c+d\,x\right )}^4\,\left (\frac {a^2}{3}-\frac {2\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {2\,a^2}{5}-\frac {b^2}{5}\right )+b^2\,{\sin \left (c+d\,x\right )}^6+\frac {a\,b\,\sin \left (c+d\,x\right )}{3}-a\,b\,{\sin \left (c+d\,x\right )}^3+a\,b\,{\sin \left (c+d\,x\right )}^5}{d\,{\sin \left (c+d\,x\right )}^7} \]

[In]

int((cos(c + d*x)^5*(a + b*sin(c + d*x))^2)/sin(c + d*x)^8,x)

[Out]

-(a^2/7 + sin(c + d*x)^4*(a^2/3 - (2*b^2)/3) - sin(c + d*x)^2*((2*a^2)/5 - b^2/5) + b^2*sin(c + d*x)^6 + (a*b*
sin(c + d*x))/3 - a*b*sin(c + d*x)^3 + a*b*sin(c + d*x)^5)/(d*sin(c + d*x)^7)