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

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Rubi [A]  time = 0.16, antiderivative size = 130, 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} \[ \frac {\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 d}-\frac {\left (a^2-2 b^2\right ) \csc ^2(c+d x)}{2 d}-\frac {a^2 \csc ^6(c+d x)}{6 d}-\frac {2 a b \csc ^5(c+d x)}{5 d}+\frac {4 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}+\frac {b^2 \log (\sin (c+d x))}{d} \]

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 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 ^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*}

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Mathematica [A]  time = 0.17, 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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fricas [A]  time = 1.02, size = 183, normalized size = 1.41 \[ \frac {30 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 15 \, {\left (2 \, a^{2} - 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 10 \, a^{2} - 45 \, b^{2} + 60 \, {\left (b^{2} \cos \left (d x + c\right )^{6} - 3 \, b^{2} \cos \left (d x + c\right )^{4} + 3 \, b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 8 \, {\left (15 \, a b \cos \left (d x + c\right )^{4} - 20 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

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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giac [A]  time = 0.30, size = 134, normalized size = 1.03 \[ \frac {60 \, b^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {147 \, b^{2} \sin \left (d x + c\right )^{6} + 120 \, a b \sin \left (d x + c\right )^{5} + 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, b^{2} \sin \left (d x + c\right )^{4} - 80 \, a b \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, b^{2} \sin \left (d x + c\right )^{2} + 24 \, a b \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \]

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

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maple [A]  time = 0.56, size = 196, normalized size = 1.51 \[ -\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}}-\frac {2 a b \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}+\frac {2 a b \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {2 a b \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}-\frac {16 a b \sin \left (d x +c \right )}{15 d}-\frac {2 a b \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}-\frac {8 a b \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{15 d}-\frac {b^{2} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {b^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d} \]

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*a*b*sin(d*x+c)*cos(d*x+c)^4-8/15/d*a*b*sin(d*x+
c)*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 = 0.34, size = 108, normalized size = 0.83 \[ \frac {60 \, b^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {120 \, a b \sin \left (d x + c\right )^{5} - 80 \, a b \sin \left (d x + c\right )^{3} + 30 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 24 \, a b \sin \left (d x + c\right ) - 15 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \]

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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mupad [B]  time = 11.87, size = 274, normalized size = 2.11 \[ \frac {b^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {5\,a^2}{2}-12\,b^2\right )+\frac {a^2}{6}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-b^2\right )-\frac {20\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+40\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )}{64\,d}-\frac {b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2}{64}-\frac {b^2}{64}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^2}{128}-\frac {3\,b^2}{16}\right )}{d}+\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80\,d}-\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \]

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)^7,x)

[Out]

(b^2*log(tan(c/2 + (d*x)/2)))/d - (a^2*tan(c/2 + (d*x)/2)^6)/(384*d) - (cot(c/2 + (d*x)/2)^6*(tan(c/2 + (d*x)/
2)^4*((5*a^2)/2 - 12*b^2) + a^2/6 - tan(c/2 + (d*x)/2)^2*(a^2 - b^2) - (20*a*b*tan(c/2 + (d*x)/2)^3)/3 + 40*a*
b*tan(c/2 + (d*x)/2)^5 + (4*a*b*tan(c/2 + (d*x)/2))/5))/(64*d) - (b^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d + (tan(
c/2 + (d*x)/2)^4*(a^2/64 - b^2/64))/d - (tan(c/2 + (d*x)/2)^2*((5*a^2)/128 - (3*b^2)/16))/d + (5*a*b*tan(c/2 +
(d*x)/2)^3)/(48*d) - (a*b*tan(c/2 + (d*x)/2)^5)/(80*d) - (5*a*b*tan(c/2 + (d*x)/2))/(8*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)**7*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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