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

Optimal. Leaf size=119 \[ -\frac {a^2 \csc ^6(c+d x)}{6 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}+\frac {a^2 \csc ^4(c+d x)}{4 d}+\frac {4 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 119, 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 = {2836, 12, 88} \[ -\frac {a^2 \csc ^6(c+d x)}{6 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}+\frac {a^2 \csc ^4(c+d x)}{4 d}+\frac {4 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Csc[c + d*x])/d + (a^2*Csc[c + d*x]^2)/(2*d) + (4*a^2*Csc[c + d*x]^3)/(3*d) + (a^2*Csc[c + d*x]^4)/(4*
d) - (2*a^2*Csc[c + d*x]^5)/(5*d) - (a^2*Csc[c + d*x]^6)/(6*d) + (a^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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^7 (a-x)^2 (a+x)^4}{x^7} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {a^2 \operatorname {Subst}\left (\int \frac {(a-x)^2 (a+x)^4}{x^7} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^2 \operatorname {Subst}\left (\int \left (\frac {a^6}{x^7}+\frac {2 a^5}{x^6}-\frac {a^4}{x^5}-\frac {4 a^3}{x^4}-\frac {a^2}{x^3}+\frac {2 a}{x^2}+\frac {1}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {4 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^4(c+d x)}{4 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^6(c+d x)}{6 d}+\frac {a^2 \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 102, normalized size = 0.86 \[ a^2 \left (-\frac {\csc ^6(c+d x)}{6 d}-\frac {2 \csc ^5(c+d x)}{5 d}+\frac {\csc ^4(c+d x)}{4 d}+\frac {4 \csc ^3(c+d x)}{3 d}+\frac {\csc ^2(c+d x)}{2 d}-\frac {2 \csc (c+d x)}{d}+\frac {\log (\sin (c+d x))}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 167, normalized size = 1.40 \[ -\frac {30 \, a^{2} \cos \left (d x + c\right )^{4} - 75 \, a^{2} \cos \left (d x + c\right )^{2} + 35 \, a^{2} - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{4} - 20 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(30*a^2*cos(d*x + c)^4 - 75*a^2*cos(d*x + c)^2 + 35*a^2 - 60*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d*x + c)^4
+ 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c)) - 8*(15*a^2*cos(d*x + c)^4 - 20*a^2*cos(d*x + c)^2 + 8*a^2
)*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 = 0.29, size = 111, normalized size = 0.93 \[ \frac {60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {147 \, a^{2} \sin \left (d x + c\right )^{6} + 120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*a^2*log(abs(sin(d*x + c))) - (147*a^2*sin(d*x + c)^6 + 120*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4
- 80*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 24*a^2*sin(d*x + c) + 10*a^2)/sin(d*x + c)^6)/d

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maple [A]  time = 0.44, size = 202, normalized size = 1.70 \[ -\frac {a^{2} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}+\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}-\frac {16 a^{2} \sin \left (d x +c \right )}{15 d}-\frac {2 \sin \left (d x +c \right ) a^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}-\frac {8 \sin \left (d x +c \right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{15 d}-\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d*a^2*cot(d*x+c)^4+1/2/d*a^2*cot(d*x+c)^2+a^2*ln(sin(d*x+c))/d-2/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^6+2/15/d
*a^2/sin(d*x+c)^3*cos(d*x+c)^6-2/5/d*a^2/sin(d*x+c)*cos(d*x+c)^6-16/15*a^2*sin(d*x+c)/d-2/5/d*sin(d*x+c)*a^2*c
os(d*x+c)^4-8/15/d*sin(d*x+c)*a^2*cos(d*x+c)^2-1/6/d*a^2/sin(d*x+c)^6*cos(d*x+c)^6

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maxima [A]  time = 0.66, size = 97, normalized size = 0.82 \[ \frac {60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {120 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} \sin \left (d x + c\right ) + 10 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/60*(60*a^2*log(sin(d*x + c)) - (120*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4 - 80*a^2*sin(d*x + c)^3 - 15*
a^2*sin(d*x + c)^2 + 24*a^2*sin(d*x + c) + 10*a^2)/sin(d*x + c)^6)/d

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mupad [B]  time = 8.98, size = 217, normalized size = 1.82 \[ \frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{128\,d}+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,d}+\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (40\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {a^2}{6}\right )}{64\,d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x)^7,x)

[Out]

(19*a^2*tan(c/2 + (d*x)/2)^2)/(128*d) + (5*a^2*tan(c/2 + (d*x)/2)^3)/(48*d) - (a^2*tan(c/2 + (d*x)/2)^5)/(80*d
) - (a^2*tan(c/2 + (d*x)/2)^6)/(384*d) + (a^2*log(tan(c/2 + (d*x)/2)))/d - (cot(c/2 + (d*x)/2)^6*(40*a^2*tan(c
/2 + (d*x)/2)^5 - (19*a^2*tan(c/2 + (d*x)/2)^4)/2 - (20*a^2*tan(c/2 + (d*x)/2)^3)/3 + a^2/6 + (4*a^2*tan(c/2 +
(d*x)/2))/5))/(64*d) - (5*a^2*tan(c/2 + (d*x)/2))/(8*d) - (a^2*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)**7*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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