3.348 \(\int \frac{\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=86 \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a f (a+b)^{5/2}}-\frac{\cot ^3(e+f x)}{3 f (a+b)}+\frac{(a+2 b) \cot (e+f x)}{f (a+b)^2}+\frac{x}{a} \]

[Out]

x/a - (b^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*(a + b)^(5/2)*f) + ((a + 2*b)*Cot[e + f*x])/((a
+ b)^2*f) - Cot[e + f*x]^3/(3*(a + b)*f)

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Rubi [A]  time = 0.248174, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4141, 1975, 480, 583, 522, 203, 205} \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a f (a+b)^{5/2}}-\frac{\cot ^3(e+f x)}{3 f (a+b)}+\frac{(a+2 b) \cot (e+f x)}{f (a+b)^2}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

x/a - (b^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*(a + b)^(5/2)*f) + ((a + 2*b)*Cot[e + f*x])/((a
+ b)^2*f) - Cot[e + f*x]^3/(3*(a + b)*f)

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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])

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x)}{3 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{-3 (a+2 b)-3 b x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=\frac{(a+2 b) \cot (e+f x)}{(a+b)^2 f}-\frac{\cot ^3(e+f x)}{3 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (a^2+3 a b+3 b^2\right )-3 b (a+2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac{(a+2 b) \cot (e+f x)}{(a+b)^2 f}-\frac{\cot ^3(e+f x)}{3 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a f}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a (a+b)^2 f}\\ &=\frac{x}{a}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a (a+b)^{5/2} f}+\frac{(a+2 b) \cot (e+f x)}{(a+b)^2 f}-\frac{\cot ^3(e+f x)}{3 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 3.70735, size = 390, normalized size = 4.53 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{1}{8} \sqrt{a+b} \csc (e) \sqrt{b (\cos (e)-i \sin (e))^4} \csc ^3(e+f x) \left (-12 a^2 \sin (2 e+f x)+8 a^2 \sin (2 e+3 f x)-3 a^2 f x \cos (2 e+3 f x)+3 a^2 f x \cos (4 e+3 f x)-12 a^2 \sin (f x)-18 a b \sin (2 e+f x)+14 a b \sin (2 e+3 f x)-6 a b f x \cos (2 e+3 f x)+6 a b f x \cos (4 e+3 f x)-9 f x (a+b)^2 \cos (2 e+f x)-24 a b \sin (f x)+9 f x (a+b)^2 \cos (f x)-3 b^2 f x \cos (2 e+3 f x)+3 b^2 f x \cos (4 e+3 f x)\right )+3 b^3 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{6 a f (a+b)^{5/2} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(3*b^3*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Si
n[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + (Sqrt[a
+ b]*Csc[e]*Csc[e + f*x]^3*Sqrt[b*(Cos[e] - I*Sin[e])^4]*(9*(a + b)^2*f*x*Cos[f*x] - 9*(a + b)^2*f*x*Cos[2*e +
f*x] - 3*a^2*f*x*Cos[2*e + 3*f*x] - 6*a*b*f*x*Cos[2*e + 3*f*x] - 3*b^2*f*x*Cos[2*e + 3*f*x] + 3*a^2*f*x*Cos[4
*e + 3*f*x] + 6*a*b*f*x*Cos[4*e + 3*f*x] + 3*b^2*f*x*Cos[4*e + 3*f*x] - 12*a^2*Sin[f*x] - 24*a*b*Sin[f*x] - 12
*a^2*Sin[2*e + f*x] - 18*a*b*Sin[2*e + f*x] + 8*a^2*Sin[2*e + 3*f*x] + 14*a*b*Sin[2*e + 3*f*x]))/8))/(6*a*(a +
b)^(5/2)*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

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Maple [A]  time = 0.099, size = 110, normalized size = 1.3 \begin{align*}{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{fa}}-{\frac{1}{3\,f \left ( a+b \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{a}{f \left ( a+b \right ) ^{2}\tan \left ( fx+e \right ) }}+2\,{\frac{b}{f \left ( a+b \right ) ^{2}\tan \left ( fx+e \right ) }}-{\frac{{b}^{3}}{f \left ( a+b \right ) ^{2}a}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2),x)

[Out]

1/f/a*arctan(tan(f*x+e))-1/3/f/(a+b)/tan(f*x+e)^3+1/f/(a+b)^2/tan(f*x+e)*a+2/f/(a+b)^2/tan(f*x+e)*b-1/f/(a+b)^
2*b^3/a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.621233, size = 1261, normalized size = 14.66 \begin{align*} \left [\frac{4 \,{\left (4 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \sqrt{-\frac{b}{a + b}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \,{\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right ) + 12 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f x \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\right )} f x\right )} \sin \left (f x + e\right )}{12 \,{\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac{2 \,{\left (4 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \sqrt{\frac{b}{a + b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \,{\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right ) + 6 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f x \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\right )} f x\right )} \sin \left (f x + e\right )}{6 \,{\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/12*(4*(4*a^2 + 7*a*b)*cos(f*x + e)^3 + 3*(b^2*cos(f*x + e)^2 - b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*
b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)
*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x
+ e) - 12*(a^2 + 2*a*b)*cos(f*x + e) + 12*((a^2 + 2*a*b + b^2)*f*x*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f*x)*
sin(f*x + e))/(((a^3 + 2*a^2*b + a*b^2)*f*cos(f*x + e)^2 - (a^3 + 2*a^2*b + a*b^2)*f)*sin(f*x + e)), 1/6*(2*(4
*a^2 + 7*a*b)*cos(f*x + e)^3 + 3*(b^2*cos(f*x + e)^2 - b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)
^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(a^2 + 2*a*b)*cos(f*x + e) + 6*((a^2 +
2*a*b + b^2)*f*x*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f*x)*sin(f*x + e))/(((a^3 + 2*a^2*b + a*b^2)*f*cos(f*x
+ e)^2 - (a^3 + 2*a^2*b + a*b^2)*f)*sin(f*x + e))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(cot(e + f*x)**4/(a + b*sec(e + f*x)**2), x)

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Giac [A]  time = 1.43611, size = 189, normalized size = 2.2 \begin{align*} -\frac{\frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{a b + b^{2}}} - \frac{3 \,{\left (f x + e\right )}}{a} - \frac{3 \, a \tan \left (f x + e\right )^{2} + 6 \, b \tan \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/3*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^3/((a^3 + 2*a^2*b + a
*b^2)*sqrt(a*b + b^2)) - 3*(f*x + e)/a - (3*a*tan(f*x + e)^2 + 6*b*tan(f*x + e)^2 - a - b)/((a^2 + 2*a*b + b^2
)*tan(f*x + e)^3))/f