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(d*x+c)/d+2*a*b*csc(d*x+c)^2/d+1/3*(2*a^2-b^2)*csc(d*x+c)^3/d-1/2*a*b*csc(d*x+c)^4/d-1/5*a^2*c
sc(d*x+c)^5/d+2*a*b*ln(sin(d*x+c))/d+b^2*sin(d*x+c)/d

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Rubi [A]  time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 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 (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*}

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Mathematica [A]  time = 0.19, 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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fricas [A]  time = 0.64, size = 166, normalized size = 1.34 \[ -\frac {30 \, b^{2} \cos \left (d x + c\right )^{6} + 30 \, {\left (a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 40 \, {\left (a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (a b \cos \left (d x + c\right )^{4} - 2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 16 \, a^{2} - 80 \, b^{2} + 15 \, {\left (4 \, a b \cos \left (d x + c\right )^{2} - 3 \, a b\right )} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.27, size = 131, normalized size = 1.06 \[ \frac {60 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, b^{2} \sin \left (d x + c\right ) - \frac {137 \, a b \sin \left (d x + c\right )^{5} + 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, b^{2} \sin \left (d x + c\right )^{4} - 60 \, a b \sin \left (d x + c\right )^{3} - 20 \, a^{2} \sin \left (d x + c\right )^{2} + 10 \, b^{2} \sin \left (d x + c\right )^{2} + 15 \, a b \sin \left (d x + c\right ) + 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

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

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maple [B]  time = 0.54, size = 279, normalized size = 2.25 \[ -\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}+\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}-\frac {8 a^{2} \sin \left (d x +c \right )}{15 d}-\frac {a^{2} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}-\frac {4 a^{2} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{15 d}-\frac {a b \left (\cot ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a b \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {b^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {b^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {8 b^{2} \sin \left (d x +c \right )}{3 d}+\frac {\sin \left (d x +c \right ) b^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{d}+\frac {4 \sin \left (d x +c \right ) b^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{3 d} \]

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 = 0.32, size = 105, normalized size = 0.85 \[ \frac {60 \, a b \log \left (\sin \left (d x + c\right )\right ) + 30 \, b^{2} \sin \left (d x + c\right ) + \frac {60 \, a b \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} - 15 \, a b \sin \left (d x + c\right ) + 10 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

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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mupad [B]  time = 11.71, size = 297, normalized size = 2.40 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,a^2}{96}-\frac {b^2}{24}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^2}{16}-\frac {7\,b^2}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (10\,a^2-92\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {22\,a^2}{15}-\frac {4\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {25\,a^2}{3}-\frac {80\,b^2}{3}\right )+\frac {a^2}{5}-11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}+\frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

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)^6,x)

[Out]

(tan(c/2 + (d*x)/2)^3*((5*a^2)/96 - b^2/24))/d - (tan(c/2 + (d*x)/2)*((5*a^2)/16 - (7*b^2)/8))/d - (tan(c/2 +
(d*x)/2)^6*(10*a^2 - 92*b^2) - tan(c/2 + (d*x)/2)^2*((22*a^2)/15 - (4*b^2)/3) + tan(c/2 + (d*x)/2)^4*((25*a^2)
/3 - (80*b^2)/3) + a^2/5 - 11*a*b*tan(c/2 + (d*x)/2)^3 - 12*a*b*tan(c/2 + (d*x)/2)^5 + a*b*tan(c/2 + (d*x)/2))
/(d*(32*tan(c/2 + (d*x)/2)^5 + 32*tan(c/2 + (d*x)/2)^7)) - (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) + (3*a*b*tan(c/2
+ (d*x)/2)^2)/(8*d) - (a*b*tan(c/2 + (d*x)/2)^4)/(32*d) + (2*a*b*log(tan(c/2 + (d*x)/2)))/d - (2*a*b*log(tan(
c/2 + (d*x)/2)^2 + 1))/d

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

[Out]

Timed out

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