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

Optimal. Leaf size=116 \[ \frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^4(c+d x)}{4 d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {4 a^2 \csc (c+d x)}{d}-\frac {a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2707, 88} \[ \frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^4(c+d x)}{4 d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}+\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {4 a^2 \csc (c+d x)}{d}-\frac {a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.45, size = 76, normalized size = 0.66 \[ \frac {a^2 \left (6 \sin ^2(c+d x)+24 \sin (c+d x)-3 \csc ^4(c+d x)-8 \csc ^3(c+d x)+6 \csc ^2(c+d x)+48 \csc (c+d x)-12 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(48*Csc[c + d*x] + 6*Csc[c + d*x]^2 - 8*Csc[c + d*x]^3 - 3*Csc[c + d*x]^4 - 12*Log[Sin[c + d*x]] + 24*Sin
[c + d*x] + 6*Sin[c + d*x]^2))/(12*d)

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fricas [A]  time = 0.83, size = 152, normalized size = 1.31 \[ -\frac {6 \, a^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} + 18 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} + 12 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} - 12 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*a^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 + 18*a^2*cos(d*x + c)^2 - 6*a^2 + 12*(a^2*cos(d*x + c)^4 -
2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*sin(d*x + c)) - 8*(3*a^2*cos(d*x + c)^4 - 12*a^2*cos(d*x + c)^2 + 8*a^2)*
sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.28, size = 108, normalized size = 0.93 \[ \frac {6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 24 \, a^{2} \sin \left (d x + c\right ) + \frac {25 \, a^{2} \sin \left (d x + c\right )^{4} + 48 \, a^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{2} \sin \left (d x + c\right ) - 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) + 24*a^2*sin(d*x + c) + (25*a^2*sin(d*x + c)^4 + 48
*a^2*sin(d*x + c)^3 + 6*a^2*sin(d*x + c)^2 - 8*a^2*sin(d*x + c) - 3*a^2)/sin(d*x + c)^4)/d

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maple [A]  time = 0.42, size = 211, normalized size = 1.82 \[ -\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{2}}{2 d}-\frac {a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {2 a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {16 a^{2} \sin \left (d x +c \right )}{3 d}+\frac {2 \sin \left (d x +c \right ) a^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{d}+\frac {8 \sin \left (d x +c \right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 94, normalized size = 0.81 \[ \frac {6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 24 \, a^{2} \sin \left (d x + c\right ) + \frac {48 \, a^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{2} \sin \left (d x + c\right ) - 3 \, a^{2}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) + 24*a^2*sin(d*x + c) + (48*a^2*sin(d*x + c)^3 + 6*a^2*s
in(d*x + c)^2 - 8*a^2*sin(d*x + c) - 3*a^2)/sin(d*x + c)^4)/d

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mupad [B]  time = 8.80, size = 276, normalized size = 2.38 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d}+\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {92\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+33\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {356\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {76\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )} \]

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)^5,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(16*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(12*d) - (a^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (a
^2*log(tan(c/2 + (d*x)/2)))/d + (7*a^2*tan(c/2 + (d*x)/2))/(4*d) + (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d + ((a
^2*tan(c/2 + (d*x)/2)^2)/2 + (76*a^2*tan(c/2 + (d*x)/2)^3)/3 + (7*a^2*tan(c/2 + (d*x)/2)^4)/4 + (356*a^2*tan(c
/2 + (d*x)/2)^5)/3 + 33*a^2*tan(c/2 + (d*x)/2)^6 + 92*a^2*tan(c/2 + (d*x)/2)^7 - a^2/4 - (4*a^2*tan(c/2 + (d*x
)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 32*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^8))

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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)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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