3.1224 \(\int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=129 \[ \frac {\left (2 a^2-b^2\right ) \csc ^5(c+d x)}{5 d}-\frac {\left (a^2-2 b^2\right ) \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^7(c+d x)}{7 d}-\frac {a b \csc ^6(c+d x)}{3 d}+\frac {a b \csc ^4(c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {b^2 \csc (c+d x)}{d} \]

[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

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Rubi [A]  time = 0.16, antiderivative size = 129, normalized size of antiderivative = 1.00, 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} \[ \frac {\left (2 a^2-b^2\right ) \csc ^5(c+d x)}{5 d}-\frac {\left (a^2-2 b^2\right ) \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^7(c+d x)}{7 d}-\frac {a b \csc ^6(c+d x)}{3 d}+\frac {a b \csc ^4(c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {b^2 \csc (c+d x)}{d} \]

Antiderivative was successfully verified.

[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 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]))

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]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {\operatorname {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 \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.20, size = 104, normalized size = 0.81 \[ -\frac {\csc (c+d x) \left (21 \left (b^2-2 a^2\right ) \csc ^4(c+d x)+35 \left (a^2-2 b^2\right ) \csc ^2(c+d x)+15 a^2 \csc ^6(c+d x)+35 a b \csc ^5(c+d x)-105 a b \csc ^3(c+d x)+105 a b \csc (c+d x)+105 b^2\right )}{105 d} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.86, size = 146, normalized size = 1.13 \[ -\frac {105 \, b^{2} \cos \left (d x + c\right )^{6} - 35 \, {\left (a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 28 \, {\left (a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \, a^{2} - 56 \, b^{2} - 35 \, {\left (3 \, a b \cos \left (d x + c\right )^{4} - 3 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.28, size = 118, normalized size = 0.91 \[ -\frac {105 \, b^{2} \sin \left (d x + c\right )^{6} + 105 \, a b \sin \left (d x + c\right )^{5} + 35 \, a^{2} \sin \left (d x + c\right )^{4} - 70 \, b^{2} \sin \left (d x + c\right )^{4} - 105 \, a b \sin \left (d x + c\right )^{3} - 42 \, a^{2} \sin \left (d x + c\right )^{2} + 21 \, b^{2} \sin \left (d x + c\right )^{2} + 35 \, a b \sin \left (d x + c\right ) + 15 \, a^{2}}{105 \, d \sin \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.56, size = 218, normalized size = 1.69 \[ \frac {a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{7 \sin \left (d x +c \right )^{7}}-\frac {\cos ^{6}\left (d x +c \right )}{35 \sin \left (d x +c \right )^{5}}+\frac {\cos ^{6}\left (d x +c \right )}{105 \sin \left (d x +c \right )^{3}}-\frac {\cos ^{6}\left (d x +c \right )}{35 \sin \left (d x +c \right )}-\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {a b \left (\cos ^{6}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{6}}+b^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos ^{6}\left (d x +c \right )}{15 \sin \left (d x +c \right )^{3}}-\frac {\cos ^{6}\left (d x +c \right )}{5 \sin \left (d x +c \right )}-\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 106, normalized size = 0.82 \[ -\frac {105 \, b^{2} \sin \left (d x + c\right )^{6} + 105 \, a b \sin \left (d x + c\right )^{5} - 105 \, a b \sin \left (d x + c\right )^{3} + 35 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 35 \, a b \sin \left (d x + c\right ) - 21 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + 15 \, a^{2}}{105 \, d \sin \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 11.79, size = 105, normalized size = 0.81 \[ -\frac {\frac {a^2}{7}+{\sin \left (c+d\,x\right )}^4\,\left (\frac {a^2}{3}-\frac {2\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {2\,a^2}{5}-\frac {b^2}{5}\right )+b^2\,{\sin \left (c+d\,x\right )}^6+\frac {a\,b\,\sin \left (c+d\,x\right )}{3}-a\,b\,{\sin \left (c+d\,x\right )}^3+a\,b\,{\sin \left (c+d\,x\right )}^5}{d\,{\sin \left (c+d\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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