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(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a/(a+b)^(5/2)/f+(a+2*b)*cot(f*x+e)/(a+b)^2/f-1/3*cot(f*x+e)
^3/(a+b)/f

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Rubi [A]  time = 0.25, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 \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*}

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Mathematica [C]  time = 3.38, 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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fricas [B]  time = 0.53, size = 533, normalized size = 6.20 \[ \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 ] \]

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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giac [A]  time = 0.33, size = 140, normalized size = 1.63 \[ -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b + b^{2}}} - \frac {3 \, {\left (f x + e\right )}}{a} - \frac {3 \, a \tan \left (f x + e\right )^{2} + 6 \, b \tan \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

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

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maple [A]  time = 1.06, size = 110, normalized size = 1.28 \[ -\frac {b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{2} a \sqrt {\left (a +b \right ) b}}-\frac {1}{3 f \left (a +b \right ) \tan \left (f x +e \right )^{3}}+\frac {a}{f \left (a +b \right )^{2} \tan \left (f x +e \right )}+\frac {2 b}{f \left (a +b \right )^{2} \tan \left (f x +e \right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a} \]

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

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maxima [A]  time = 0.46, size = 106, normalized size = 1.23 \[ -\frac {\frac {3 \, b^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {3 \, {\left (f x + e\right )}}{a} - \frac {3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]

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]

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

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mupad [B]  time = 8.97, size = 2644, normalized size = 30.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((10*b^12*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 66
0*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (80*a*b^11*tan(e + f*x))/(80*a*b^11 + 10*b^
12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9
*b^3 + 2*a^10*b^2) + (290*a^2*b^10*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b
^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (630*a^3*b^9*tan(e + f
*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5
+ 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (912*a^4*b^8*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 6
30*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) +
(922*a^5*b^7*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660
*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (660*a^6*b^6*tan(e + f*x))/(80*a*b^11 + 10*b
^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^
9*b^3 + 2*a^10*b^2) + (330*a^7*b^5*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b
^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (110*a^8*b^4*tan(e + f
*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5
+ 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (22*a^9*b^3*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 63
0*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (
2*a^10*b^2*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a
^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2))/(a*f) - (1/(3*(a + b)) - (tan(e + f*x)^2*(a + 2
*b))/(a + b)^2)/(f*tan(e + f*x)^3) - (atan((((-b^5*(a + b)^5)^(1/2)*((tan(e + f*x)*(32*a*b^12 + 4*b^13 + 120*a
^2*b^11 + 280*a^3*b^10 + 450*a^4*b^9 + 516*a^5*b^8 + 422*a^6*b^7 + 240*a^7*b^6 + 90*a^8*b^5 + 20*a^9*b^4 + 2*a
^10*b^3))/2 - ((-b^5*(a + b)^5)^(1/2)*(6*a^2*b^12 + 54*a^3*b^11 + 218*a^4*b^10 + 520*a^5*b^9 + 812*a^6*b^8 + 8
68*a^7*b^7 + 644*a^8*b^6 + 328*a^9*b^5 + 110*a^10*b^4 + 22*a^11*b^3 + 2*a^12*b^2 - (tan(e + f*x)*(-b^5*(a + b)
^5)^(1/2)*(16*a^2*b^13 + 168*a^3*b^12 + 800*a^4*b^11 + 2280*a^5*b^10 + 4320*a^6*b^9 + 5712*a^7*b^8 + 5376*a^8*
b^7 + 3600*a^9*b^6 + 1680*a^10*b^5 + 520*a^11*b^4 + 96*a^12*b^3 + 8*a^13*b^2))/(4*a*(a + b)^5)))/(2*a*(a + b)^
5))*1i)/(a*(a + b)^5) + ((-b^5*(a + b)^5)^(1/2)*((tan(e + f*x)*(32*a*b^12 + 4*b^13 + 120*a^2*b^11 + 280*a^3*b^
10 + 450*a^4*b^9 + 516*a^5*b^8 + 422*a^6*b^7 + 240*a^7*b^6 + 90*a^8*b^5 + 20*a^9*b^4 + 2*a^10*b^3))/2 + ((-b^5
*(a + b)^5)^(1/2)*(6*a^2*b^12 + 54*a^3*b^11 + 218*a^4*b^10 + 520*a^5*b^9 + 812*a^6*b^8 + 868*a^7*b^7 + 644*a^8
*b^6 + 328*a^9*b^5 + 110*a^10*b^4 + 22*a^11*b^3 + 2*a^12*b^2 + (tan(e + f*x)*(-b^5*(a + b)^5)^(1/2)*(16*a^2*b^
13 + 168*a^3*b^12 + 800*a^4*b^11 + 2280*a^5*b^10 + 4320*a^6*b^9 + 5712*a^7*b^8 + 5376*a^8*b^7 + 3600*a^9*b^6 +
1680*a^10*b^5 + 520*a^11*b^4 + 96*a^12*b^3 + 8*a^13*b^2))/(4*a*(a + b)^5)))/(2*a*(a + b)^5))*1i)/(a*(a + b)^5
))/(26*a*b^11 + 4*b^12 + 72*a^2*b^10 + 110*a^3*b^9 + 100*a^4*b^8 + 54*a^5*b^7 + 16*a^6*b^6 + 2*a^7*b^5 + ((-b^
5*(a + b)^5)^(1/2)*((tan(e + f*x)*(32*a*b^12 + 4*b^13 + 120*a^2*b^11 + 280*a^3*b^10 + 450*a^4*b^9 + 516*a^5*b^
8 + 422*a^6*b^7 + 240*a^7*b^6 + 90*a^8*b^5 + 20*a^9*b^4 + 2*a^10*b^3))/2 - ((-b^5*(a + b)^5)^(1/2)*(6*a^2*b^12
+ 54*a^3*b^11 + 218*a^4*b^10 + 520*a^5*b^9 + 812*a^6*b^8 + 868*a^7*b^7 + 644*a^8*b^6 + 328*a^9*b^5 + 110*a^10
*b^4 + 22*a^11*b^3 + 2*a^12*b^2 - (tan(e + f*x)*(-b^5*(a + b)^5)^(1/2)*(16*a^2*b^13 + 168*a^3*b^12 + 800*a^4*b
^11 + 2280*a^5*b^10 + 4320*a^6*b^9 + 5712*a^7*b^8 + 5376*a^8*b^7 + 3600*a^9*b^6 + 1680*a^10*b^5 + 520*a^11*b^4
+ 96*a^12*b^3 + 8*a^13*b^2))/(4*a*(a + b)^5)))/(2*a*(a + b)^5)))/(a*(a + b)^5) - ((-b^5*(a + b)^5)^(1/2)*((ta
n(e + f*x)*(32*a*b^12 + 4*b^13 + 120*a^2*b^11 + 280*a^3*b^10 + 450*a^4*b^9 + 516*a^5*b^8 + 422*a^6*b^7 + 240*a
^7*b^6 + 90*a^8*b^5 + 20*a^9*b^4 + 2*a^10*b^3))/2 + ((-b^5*(a + b)^5)^(1/2)*(6*a^2*b^12 + 54*a^3*b^11 + 218*a^
4*b^10 + 520*a^5*b^9 + 812*a^6*b^8 + 868*a^7*b^7 + 644*a^8*b^6 + 328*a^9*b^5 + 110*a^10*b^4 + 22*a^11*b^3 + 2*
a^12*b^2 + (tan(e + f*x)*(-b^5*(a + b)^5)^(1/2)*(16*a^2*b^13 + 168*a^3*b^12 + 800*a^4*b^11 + 2280*a^5*b^10 + 4
320*a^6*b^9 + 5712*a^7*b^8 + 5376*a^8*b^7 + 3600*a^9*b^6 + 1680*a^10*b^5 + 520*a^11*b^4 + 96*a^12*b^3 + 8*a^13
*b^2))/(4*a*(a + b)^5)))/(2*a*(a + b)^5)))/(a*(a + b)^5)))*(-b^5*(a + b)^5)^(1/2)*1i)/(a*f*(a + b)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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