3.335 \(\int \cot ^4(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=45 \[ \frac {\left (a^2-b^2\right ) \cot (e+f x)}{f}+a^2 x-\frac {(a+b)^2 \cot ^3(e+f x)}{3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4141, 1802, 203} \[ \frac {\left (a^2-b^2\right ) \cot (e+f x)}{f}+a^2 x-\frac {(a+b)^2 \cot ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.94, size = 160, normalized size = 3.56 \[ \frac {\csc (e) \csc ^3(e+f x) \left (-12 a^2 \sin (2 e+f x)+8 a^2 \sin (2 e+3 f x)-9 a^2 f x \cos (2 e+f x)-3 a^2 f x \cos (2 e+3 f x)+3 a^2 f x \cos (4 e+3 f x)-12 a^2 \sin (f x)+9 a^2 f x \cos (f x)-12 a b \sin (2 e+f x)+4 a b \sin (2 e+3 f x)-4 b^2 \sin (2 e+3 f x)+12 b^2 \sin (f x)\right )}{24 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e]*Csc[e + f*x]^3*(9*a^2*f*x*Cos[f*x] - 9*a^2*f*x*Cos[2*e + 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] + 12*b^2*Sin[f*x] - 12*a^2*Sin[2*e + f*x] - 12*a*b*Sin[2*e + f*x] + 8*a^2*
Sin[2*e + 3*f*x] + 4*a*b*Sin[2*e + 3*f*x] - 4*b^2*Sin[2*e + 3*f*x]))/(24*f)

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 98, normalized size = 2.18 \[ \frac {2 \, {\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right ) + 3 \, {\left (a^{2} f x \cos \left (f x + e\right )^{2} - a^{2} f x\right )} \sin \left (f x + e\right )}{3 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*(2*a^2 + a*b - b^2)*cos(f*x + e)^3 - 3*(a^2 - b^2)*cos(f*x + e) + 3*(a^2*f*x*cos(f*x + e)^2 - a^2*f*x)*
sin(f*x + e))/((f*cos(f*x + e)^2 - f)*sin(f*x + e))

________________________________________________________________________________________

giac [B]  time = 0.57, size = 187, normalized size = 4.16 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 24 \, {\left (f x + e\right )} a^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^2*tan(1/2*f*x + 1/2*e)^3 + 2*a*b*tan(1/2*f*x + 1/2*e)^3 + b^2*tan(1/2*f*x + 1/2*e)^3 + 24*(f*x + e)*a^
2 - 15*a^2*tan(1/2*f*x + 1/2*e) - 6*a*b*tan(1/2*f*x + 1/2*e) + 9*b^2*tan(1/2*f*x + 1/2*e) + (15*a^2*tan(1/2*f*
x + 1/2*e)^2 + 6*a*b*tan(1/2*f*x + 1/2*e)^2 - 9*b^2*tan(1/2*f*x + 1/2*e)^2 - a^2 - 2*a*b - b^2)/tan(1/2*f*x +
1/2*e)^3)/f

________________________________________________________________________________________

maple [A]  time = 1.28, size = 73, normalized size = 1.62 \[ \frac {a^{2} \left (-\frac {\left (\cot ^{3}\left (f x +e \right )\right )}{3}+\cot \left (f x +e \right )+f x +e \right )-\frac {2 a b \left (\cos ^{3}\left (f x +e \right )\right )}{3 \sin \left (f x +e \right )^{3}}+b^{2} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a^2*(-1/3*cot(f*x+e)^3+cot(f*x+e)+f*x+e)-2/3*a*b/sin(f*x+e)^3*cos(f*x+e)^3+b^2*(-2/3-1/3*csc(f*x+e)^2)*co
t(f*x+e))

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 59, normalized size = 1.31 \[ \frac {3 \, {\left (f x + e\right )} a^{2} + \frac {3 \, {\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (f x + e\right )^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.61, size = 53, normalized size = 1.18 \[ a^2\,x-\frac {\frac {2\,a\,b}{3}-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a^2-b^2\right )+\frac {a^2}{3}+\frac {b^2}{3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cot ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________