3.388 \(\int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=132 \[ -\frac{2 a^2 \cot ^5(c+d x)}{5 d}-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{8 d}+\frac{5 a^2 \cot (c+d x) \csc (c+d x)}{16 d} \]

[Out]

(-7*a^2*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a^2*Cot[c + d*x]^5)/(5*d) + (5*a^2*Cot[c + d*x]*Csc[c + d*x])/(16*d
) - (a^2*Cot[c + d*x]^3*Csc[c + d*x])/(4*d) + (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(8*d) - (a^2*Cot[c + d*x]^3*Cs
c[c + d*x]^3)/(6*d)

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Rubi [A]  time = 0.261761, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2873, 2611, 3770, 2607, 30, 3768} \[ -\frac{2 a^2 \cot ^5(c+d x)}{5 d}-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{8 d}+\frac{5 a^2 \cot (c+d x) \csc (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(-7*a^2*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a^2*Cot[c + d*x]^5)/(5*d) + (5*a^2*Cot[c + d*x]*Csc[c + d*x])/(16*d
) - (a^2*Cot[c + d*x]^3*Csc[c + d*x])/(4*d) + (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(8*d) - (a^2*Cot[c + d*x]^3*Cs
c[c + d*x]^3)/(6*d)

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

\begin{align*} \int \cot ^4(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cot ^4(c+d x) \csc (c+d x)+2 a^2 \cot ^4(c+d x) \csc ^2(c+d x)+a^2 \cot ^4(c+d x) \csc ^3(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^4(c+d x) \csc (c+d x) \, dx+a^2 \int \cot ^4(c+d x) \csc ^3(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^4(c+d x) \csc ^2(c+d x) \, dx\\ &=-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}-\frac{1}{2} a^2 \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx-\frac{1}{4} \left (3 a^2\right ) \int \cot ^2(c+d x) \csc (c+d x) \, dx+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=-\frac{2 a^2 \cot ^5(c+d x)}{5 d}+\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac{1}{8} a^2 \int \csc ^3(c+d x) \, dx+\frac{1}{8} \left (3 a^2\right ) \int \csc (c+d x) \, dx\\ &=-\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{2 a^2 \cot ^5(c+d x)}{5 d}+\frac{5 a^2 \cot (c+d x) \csc (c+d x)}{16 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac{1}{16} a^2 \int \csc (c+d x) \, dx\\ &=-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{2 a^2 \cot ^5(c+d x)}{5 d}+\frac{5 a^2 \cot (c+d x) \csc (c+d x)}{16 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac{a^2 \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 0.103325, size = 267, normalized size = 2.02 \[ a^2 \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{5 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{5 d}-\frac{\csc ^6\left (\frac{1}{2} (c+d x)\right )}{384 d}+\frac{9 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{\sec ^6\left (\frac{1}{2} (c+d x)\right )}{384 d}-\frac{9 \sec ^2\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{7 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{16 d}-\frac{7 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{16 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{80 d}+\frac{7 \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{80 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )}{80 d}-\frac{7 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{80 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

a^2*(-Cot[(c + d*x)/2]/(5*d) + (9*Csc[(c + d*x)/2]^2)/(64*d) + (7*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(80*d)
- (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(80*d) - Csc[(c + d*x)/2]^6/(384*d) - (7*Log[Cos[(c + d*x)/2]])/(16*d)
+ (7*Log[Sin[(c + d*x)/2]])/(16*d) - (9*Sec[(c + d*x)/2]^2)/(64*d) + Sec[(c + d*x)/2]^6/(384*d) + Tan[(c + d*
x)/2]/(5*d) - (7*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(80*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(80*d))

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Maple [A]  time = 0.081, size = 152, normalized size = 1.2 \begin{align*} -{\frac{7\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{24\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{7\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{48\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{48\,d}}+{\frac{7\,{a}^{2}\cos \left ( dx+c \right ) }{16\,d}}+{\frac{7\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{16\,d}}-{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x)

[Out]

-7/24/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5+7/48/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5+7/48*a^2*cos(d*x+c)^3/d+7/16*a^2*co
s(d*x+c)/d+7/16/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/5/d*a^2/sin(d*x+c)^5*cos(d*x+c)^5-1/6/d*a^2/sin(d*x+c)^6*cos
(d*x+c)^5

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Maxima [A]  time = 1.1125, size = 244, normalized size = 1.85 \begin{align*} \frac{5 \, a^{2}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 30 \, a^{2}{\left (\frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac{192 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/480*(5*a^2*(2*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*
cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 30*a^2*(2*(5*cos(d*x + c)^3 - 3*cos
(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) - 192*
a^2/tan(d*x + c)^5)/d

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Fricas [A]  time = 1.52329, size = 537, normalized size = 4.07 \begin{align*} \frac{192 \, a^{2} \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) - 270 \, a^{2} \cos \left (d x + c\right )^{5} + 560 \, a^{2} \cos \left (d x + c\right )^{3} - 210 \, a^{2} \cos \left (d x + c\right ) - 105 \,{\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} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 105 \,{\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} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{480 \,{\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(cos(d*x+c)^4*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/480*(192*a^2*cos(d*x + c)^5*sin(d*x + c) - 270*a^2*cos(d*x + c)^5 + 560*a^2*cos(d*x + c)^3 - 210*a^2*cos(d*x
+ c) - 105*(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*cos(d*x + c) + 1/
2) + 105*(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*cos(d*x + c) + 1/2)
)/(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(cos(d*x+c)**4*csc(d*x+c)**7*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.35044, size = 309, normalized size = 2.34 \begin{align*} \frac{5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 255 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 840 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 240 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{2058 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 240 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 255 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/1920*(5*a^2*tan(1/2*d*x + 1/2*c)^6 + 24*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2*c)^4 - 120*a^2
*tan(1/2*d*x + 1/2*c)^3 - 255*a^2*tan(1/2*d*x + 1/2*c)^2 + 840*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 240*a^2*ta
n(1/2*d*x + 1/2*c) - (2058*a^2*tan(1/2*d*x + 1/2*c)^6 + 240*a^2*tan(1/2*d*x + 1/2*c)^5 - 255*a^2*tan(1/2*d*x +
1/2*c)^4 - 120*a^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*a^2*tan(1/2*d*x + 1/2*c) + 5*a
^2)/tan(1/2*d*x + 1/2*c)^6)/d