3.158 \(\int \cos ^4(a+b x) \cot ^4(a+b x) \, dx\)

Optimal. Leaf size=80 \[ -\frac{35 \cot ^3(a+b x)}{24 b}+\frac{35 \cot (a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}+\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{35 x}{8} \]

[Out]

(35*x)/8 + (35*Cot[a + b*x])/(8*b) - (35*Cot[a + b*x]^3)/(24*b) + (7*Cos[a + b*x]^2*Cot[a + b*x]^3)/(8*b) + (C
os[a + b*x]^4*Cot[a + b*x]^3)/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0484905, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2591, 288, 302, 203} \[ -\frac{35 \cot ^3(a+b x)}{24 b}+\frac{35 \cot (a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}+\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{35 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^4*Cot[a + b*x]^4,x]

[Out]

(35*x)/8 + (35*Cot[a + b*x])/(8*b) - (35*Cot[a + b*x]^3)/(24*b) + (7*Cos[a + b*x]^2*Cot[a + b*x]^3)/(8*b) + (C
os[a + b*x]^4*Cot[a + b*x]^3)/(4*b)

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^4(a+b x) \cot ^4(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^2\right )^3} \, dx,x,\cot (a+b x)\right )}{b}\\ &=\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}-\frac{7 \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (a+b x)\right )}{4 b}\\ &=\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (a+b x)\right )}{8 b}\\ &=\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (a+b x)\right )}{8 b}\\ &=\frac{35 \cot (a+b x)}{8 b}-\frac{35 \cot ^3(a+b x)}{24 b}+\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (a+b x)\right )}{8 b}\\ &=\frac{35 x}{8}+\frac{35 \cot (a+b x)}{8 b}-\frac{35 \cot ^3(a+b x)}{24 b}+\frac{7 \cos ^2(a+b x) \cot ^3(a+b x)}{8 b}+\frac{\cos ^4(a+b x) \cot ^3(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.280761, size = 53, normalized size = 0.66 \[ \frac{420 (a+b x)+72 \sin (2 (a+b x))+3 \sin (4 (a+b x))-32 \cot (a+b x) \left (\csc ^2(a+b x)-10\right )}{96 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^4*Cot[a + b*x]^4,x]

[Out]

(420*(a + b*x) - 32*Cot[a + b*x]*(-10 + Csc[a + b*x]^2) + 72*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)])/(96*b)

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 94, normalized size = 1.2 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{9}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}}+2\,{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{9}}{\sin \left ( bx+a \right ) }}+2\, \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{7}+7/6\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}+{\frac{35\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{24}}+{\frac{35\,\cos \left ( bx+a \right ) }{16}} \right ) \sin \left ( bx+a \right ) +{\frac{35\,bx}{8}}+{\frac{35\,a}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^8/sin(b*x+a)^4,x)

[Out]

1/b*(-1/3/sin(b*x+a)^3*cos(b*x+a)^9+2/sin(b*x+a)*cos(b*x+a)^9+2*(cos(b*x+a)^7+7/6*cos(b*x+a)^5+35/24*cos(b*x+a
)^3+35/16*cos(b*x+a))*sin(b*x+a)+35/8*b*x+35/8*a)

________________________________________________________________________________________

Maxima [A]  time = 1.49196, size = 101, normalized size = 1.26 \begin{align*} \frac{105 \, b x + 105 \, a + \frac{105 \, \tan \left (b x + a\right )^{6} + 175 \, \tan \left (b x + a\right )^{4} + 56 \, \tan \left (b x + a\right )^{2} - 8}{\tan \left (b x + a\right )^{7} + 2 \, \tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/24*(105*b*x + 105*a + (105*tan(b*x + a)^6 + 175*tan(b*x + a)^4 + 56*tan(b*x + a)^2 - 8)/(tan(b*x + a)^7 + 2*
tan(b*x + a)^5 + tan(b*x + a)^3))/b

________________________________________________________________________________________

Fricas [A]  time = 2.32568, size = 230, normalized size = 2.88 \begin{align*} -\frac{6 \, \cos \left (b x + a\right )^{7} + 21 \, \cos \left (b x + a\right )^{5} - 140 \, \cos \left (b x + a\right )^{3} - 105 \,{\left (b x \cos \left (b x + a\right )^{2} - b x\right )} \sin \left (b x + a\right ) + 105 \, \cos \left (b x + a\right )}{24 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/24*(6*cos(b*x + a)^7 + 21*cos(b*x + a)^5 - 140*cos(b*x + a)^3 - 105*(b*x*cos(b*x + a)^2 - b*x)*sin(b*x + a)
+ 105*cos(b*x + a))/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

________________________________________________________________________________________

Sympy [A]  time = 15.6449, size = 141, normalized size = 1.76 \begin{align*} \begin{cases} \frac{35 x \sin ^{4}{\left (a + b x \right )}}{8} + \frac{35 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{35 x \cos ^{4}{\left (a + b x \right )}}{8} + \frac{35 \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} + \frac{175 \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{24 b} + \frac{7 \cos ^{5}{\left (a + b x \right )}}{3 b \sin{\left (a + b x \right )}} - \frac{\cos ^{7}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{8}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**8/sin(b*x+a)**4,x)

[Out]

Piecewise((35*x*sin(a + b*x)**4/8 + 35*x*sin(a + b*x)**2*cos(a + b*x)**2/4 + 35*x*cos(a + b*x)**4/8 + 35*sin(a
+ b*x)**3*cos(a + b*x)/(8*b) + 175*sin(a + b*x)*cos(a + b*x)**3/(24*b) + 7*cos(a + b*x)**5/(3*b*sin(a + b*x))
- cos(a + b*x)**7/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**8/sin(a)**4, True))

________________________________________________________________________________________

Giac [A]  time = 1.18604, size = 92, normalized size = 1.15 \begin{align*} \frac{105 \, b x + 105 \, a + \frac{3 \,{\left (11 \, \tan \left (b x + a\right )^{3} + 13 \, \tan \left (b x + a\right )\right )}}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{2}} + \frac{8 \,{\left (9 \, \tan \left (b x + a\right )^{2} - 1\right )}}{\tan \left (b x + a\right )^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^8/sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/24*(105*b*x + 105*a + 3*(11*tan(b*x + a)^3 + 13*tan(b*x + a))/(tan(b*x + a)^2 + 1)^2 + 8*(9*tan(b*x + a)^2 -
1)/tan(b*x + a)^3)/b