3.18 \(\int \cot ^7(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=132 \[ -\frac{a^2 \sin ^2(c+d x)}{2 d}-\frac{2 a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^6(c+d x)}{6 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}+\frac{a^2 \csc ^4(c+d x)}{2 d}+\frac{2 a^2 \csc ^3(c+d x)}{d}-\frac{6 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0754306, antiderivative size = 132, normalized size of antiderivative = 1., 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} \[ -\frac{a^2 \sin ^2(c+d x)}{2 d}-\frac{2 a^2 \sin (c+d x)}{d}-\frac{a^2 \csc ^6(c+d x)}{6 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}+\frac{a^2 \csc ^4(c+d x)}{2 d}+\frac{2 a^2 \csc ^3(c+d x)}{d}-\frac{6 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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]))

Rubi steps

\begin{align*} \int \cot ^7(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^3 (a+x)^5}{x^7} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2 a+\frac{a^8}{x^7}+\frac{2 a^7}{x^6}-\frac{2 a^6}{x^5}-\frac{6 a^5}{x^4}+\frac{6 a^3}{x^2}+\frac{2 a^2}{x}-x\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{6 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \csc ^3(c+d x)}{d}+\frac{a^2 \csc ^4(c+d x)}{2 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{a^2 \csc ^6(c+d x)}{6 d}+\frac{2 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*}

Mathematica [A]  time = 0.215678, size = 86, normalized size = 0.65 \[ -\frac{a^2 \left (15 \sin ^2(c+d x)+60 \sin (c+d x)+5 \csc ^6(c+d x)+12 \csc ^5(c+d x)-15 \csc ^4(c+d x)-60 \csc ^3(c+d x)+180 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*(180*Csc[c + d*x] - 60*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 5*Csc[c + d*x]^6 - 60*Lo
g[Sin[c + d*x]] + 60*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(30*d)

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Maple [B]  time = 0.049, size = 313, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d}}+{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+2\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{d\sin \left ( dx+c \right ) }}-{\frac{32\,{a}^{2}\sin \left ( dx+c \right ) }{5\,d}}-2\,{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d}}-{\frac{12\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{16\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{5\,d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{6}}{6\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^7*(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d*a^2/sin(d*x+c)^4*cos(d*x+c)^8+1/2/d*a^2/sin(d*x+c)^2*cos(d*x+c)^8+1/2/d*a^2*cos(d*x+c)^6+3/4/d*a^2*cos(
d*x+c)^4+3/2/d*a^2*cos(d*x+c)^2+2*a^2*ln(sin(d*x+c))/d-2/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^8+2/5/d*a^2/sin(d*x+c
)^3*cos(d*x+c)^8-2/d*a^2/sin(d*x+c)*cos(d*x+c)^8-32/5*a^2*sin(d*x+c)/d-2/d*a^2*sin(d*x+c)*cos(d*x+c)^6-12/5/d*
a^2*sin(d*x+c)*cos(d*x+c)^4-16/5/d*a^2*cos(d*x+c)^2*sin(d*x+c)-1/6/d*a^2*cot(d*x+c)^6+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 = 1.08562, size = 144, normalized size = 1.09 \begin{align*} -\frac{15 \, a^{2} \sin \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{2} \sin \left (d x + c\right ) + \frac{180 \, a^{2} \sin \left (d x + c\right )^{5} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*a^2*sin(d*x + c)^2 - 60*a^2*log(sin(d*x + c)) + 60*a^2*sin(d*x + c) + (180*a^2*sin(d*x + c)^5 - 60*a
^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 12*a^2*sin(d*x + c) + 5*a^2)/sin(d*x + c)^6)/d

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Fricas [A]  time = 1.76805, size = 508, normalized size = 3.85 \begin{align*} \frac{30 \, a^{2} \cos \left (d x + c\right )^{8} - 105 \, a^{2} \cos \left (d x + c\right )^{6} + 135 \, a^{2} \cos \left (d x + c\right )^{4} - 45 \, a^{2} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 120 \,{\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 ) - 24 \,{\left (5 \, a^{2} \cos \left (d x + c\right )^{6} - 30 \, a^{2} \cos \left (d x + c\right )^{4} + 40 \, a^{2} \cos \left (d x + c\right )^{2} - 16 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(30*a^2*cos(d*x + c)^8 - 105*a^2*cos(d*x + c)^6 + 135*a^2*cos(d*x + c)^4 - 45*a^2*cos(d*x + c)^2 - 5*a^2
+ 120*(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)) - 24*(5*a
^2*cos(d*x + c)^6 - 30*a^2*cos(d*x + c)^4 + 40*a^2*cos(d*x + c)^2 - 16*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**7*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.49559, size = 163, normalized size = 1.23 \begin{align*} -\frac{15 \, a^{2} \sin \left (d x + c\right )^{2} - 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{2} \sin \left (d x + c\right ) + \frac{147 \, a^{2} \sin \left (d x + c\right )^{6} + 180 \, a^{2} \sin \left (d x + c\right )^{5} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*a^2*sin(d*x + c)^2 - 60*a^2*log(abs(sin(d*x + c))) + 60*a^2*sin(d*x + c) + (147*a^2*sin(d*x + c)^6 +
180*a^2*sin(d*x + c)^5 - 60*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 12*a^2*sin(d*x + c) + 5*a^2)/sin(d*x
+ c)^6)/d