3.4.56 \(\int \frac {\cot ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [356]

Optimal. Leaf size=186 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f} \]

[Out]

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/3*(7*a-4*b)*b*cot(f*x+e)/a^2/(a-b)^2/
f/(a+b*tan(f*x+e)^2)^(1/2)-1/3*(a-4*b)*(3*a-2*b)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^3/(a-b)^2/f-1/3*b*cot(f
*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3751, 483, 593, 597, 12, 385, 209} \begin {gather*} -\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)^2}-\frac {b (7 a-4 b) \cot (e+f x)}{3 a^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac {b \cot (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) - (b*Cot[e + f*x])/(3*a*(a
- b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((7*a - 4*b)*b*Cot[e + f*x])/(3*a^2*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]
^2]) - ((a - 4*b)*(3*a - 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^3*(a - b)^2*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 483

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

Rule 593

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

Rule 597

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 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {3 a-4 b-4 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b) f}\\ &=-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {(a-4 b) (3 a-2 b)-2 (7 a-4 b) b x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 (a-b)^2 f}\\ &=-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac {\text {Subst}\left (\int \frac {3 a^3}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^3 (a-b)^2 f}\\ &=-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}\\ &=-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}-\frac {\text {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^2 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {b \cot (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(7 a-4 b) b \cot (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) (3 a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b)^2 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 12.08, size = 1890, normalized size = 10.16 \begin {gather*} -\frac {\cos ^2(e+f x) \cot (e+f x) \left (\frac {20 a \csc ^2(e+f x)}{3 (a-b)}-\frac {5 a^2 \csc ^4(e+f x)}{(a-b)^2}+\frac {40 b \sec ^2(e+f x)}{a-b}-\frac {30 a b \csc ^2(e+f x) \sec ^2(e+f x)}{(a-b)^2}-\frac {40 b^2 \sec ^4(e+f x)}{(a-b)^2}+\frac {92 (a-b) \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{105 a}+\frac {24 (a-b) \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{35 a}+\frac {16 (a-b) \, _4F_3\left (2,2,2,2;1,1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{105 a}+\frac {160 b^2 \sec ^2(e+f x) \tan ^2(e+f x)}{3 a (a-b)}+\frac {124 (a-b) b \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{35 a^2}+\frac {16 (a-b) b \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{7 a^2}+\frac {16 (a-b) b \, _4F_3\left (2,2,2,2;1,1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{35 a^2}+\frac {64 b^3 \sec ^2(e+f x) \tan ^4(e+f x)}{3 a^2 (a-b)}+\frac {152 (a-b) b^2 \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{35 a^3}+\frac {88 (a-b) b^2 \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{35 a^3}+\frac {16 (a-b) b^2 \, _4F_3\left (2,2,2,2;1,1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{35 a^3}+\frac {176 (a-b) b^3 \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^6(e+f x)}{105 a^4}+\frac {32 (a-b) b^3 \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^6(e+f x)}{35 a^4}+\frac {16 (a-b) b^3 \, _4F_3\left (2,2,2,2;1,1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^6(e+f x)}{105 a^4}+\frac {5 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right )}{\left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{5/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {30 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{5/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {40 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)}{a^2 \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{5/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {16 b^3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^6(e+f x)}{a^3 \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{5/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {5 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right )}{\sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-\frac {10 a \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \csc ^2(e+f x)}{(a-b) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-\frac {60 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sec ^2(e+f x)}{(a-b) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}+\frac {30 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-\frac {80 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sec ^2(e+f x) \tan ^2(e+f x)}{a (a-b) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}+\frac {40 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)}{a^2 \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-\frac {32 b^3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sec ^2(e+f x) \tan ^4(e+f x)}{a^2 (a-b) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}+\frac {16 b^3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^6(e+f x)}{a^3 \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-\frac {16 b^3 \left (\tan (e+f x)+\tan ^3(e+f x)\right )^2}{a (a-b)^2}\right )}{a^2 f \sqrt {a+b \tan ^2(e+f x)} \left (1+\frac {b \tan ^2(e+f x)}{a}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-((Cos[e + f*x]^2*Cot[e + f*x]*((20*a*Csc[e + f*x]^2)/(3*(a - b)) - (5*a^2*Csc[e + f*x]^4)/(a - b)^2 + (40*b*S
ec[e + f*x]^2)/(a - b) - (30*a*b*Csc[e + f*x]^2*Sec[e + f*x]^2)/(a - b)^2 - (40*b^2*Sec[e + f*x]^4)/(a - b)^2
+ (92*(a - b)*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(105*a) + (24*(a - b)*H
ypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(35*a) + (16*(a - b)*Hyperge
ometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(105*a) + (160*b^2*Sec[e + f
*x]^2*Tan[e + f*x]^2)/(3*a*(a - b)) + (124*(a - b)*b*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*
Sin[e + f*x]^2*Tan[e + f*x]^2)/(35*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e
+ f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(7*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2
}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(35*a^2) + (64*b^3*Sec[e + f*x]^2*Tan[e + f*x]^4
)/(3*a^2*(a - b)) + (152*(a - b)*b^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*T
an[e + f*x]^4)/(35*a^3) + (88*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*S
in[e + f*x]^2*Tan[e + f*x]^4)/(35*a^3) + (16*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)
*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/(35*a^3) + (176*(a - b)*b^3*Hypergeometric2F1[2, 2, 9/2, ((
a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(105*a^4) + (32*(a - b)*b^3*HypergeometricPFQ[{2, 2,
2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(35*a^4) + (16*(a - b)*b^3*Hypergeome
tricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)/(105*a^4) + (5*A
rcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/((((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e
+ f*x]^2))/a]) + (30*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*(((a - b)*Sin[e + f*x]^2)/
a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (40*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*T
an[e + f*x]^4)/(a^2*(((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (16*
b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6)/(a^3*(((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[(Cos
[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (5*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/Sqrt[((a - b)*Cos[e + f
*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2] - (10*a*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Csc[e + f*x
]^2)/((a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) - (60*b*ArcSin[Sqrt[((
a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x]^2)/((a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e +
f*x]^2))/a^2]) + (30*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/(a*Sqrt[((a - b)*Cos[e + f*x]
^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) - (80*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x
]^2*Tan[e + f*x]^2)/(a*(a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) + (40
*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4)/(a^2*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2
*(a + b*Tan[e + f*x]^2))/a^2]) - (32*b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sec[e + f*x]^2*Tan[e + f*x]^
4)/(a^2*(a - b)*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]) + (16*b^3*ArcSin[Sqr
t[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6)/(a^3*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e +
f*x]^2))/a^2]) - (16*b^3*(Tan[e + f*x] + Tan[e + f*x]^3)^2)/(a*(a - b)^2)))/(a^2*f*Sqrt[a + b*Tan[e + f*x]^2]
*(1 + (b*Tan[e + f*x]^2)/a)))

________________________________________________________________________________________

Maple [F]
time = 0.33, size = 0, normalized size = 0.00 \[\int \frac {\cot ^{2}\left (f x +e \right )}{\left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (176) = 352\).
time = 3.57, size = 781, normalized size = 4.20 \begin {gather*} \left [-\frac {3 \, {\left (a^{3} b^{2} \tan \left (f x + e\right )^{5} + 2 \, a^{4} b \tan \left (f x + e\right )^{3} + a^{5} \tan \left (f x + e\right )\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, a^{5} - 9 \, a^{4} b + 9 \, a^{3} b^{2} - 3 \, a^{2} b^{3} + {\left (3 \, a^{3} b^{2} - 17 \, a^{2} b^{3} + 22 \, a b^{4} - 8 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 3 \, {\left (2 \, a^{4} b - 9 \, a^{3} b^{2} + 11 \, a^{2} b^{3} - 4 \, a b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {3 \, {\left (a^{3} b^{2} \tan \left (f x + e\right )^{5} + 2 \, a^{4} b \tan \left (f x + e\right )^{3} + a^{5} \tan \left (f x + e\right )\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) + 2 \, {\left (3 \, a^{5} - 9 \, a^{4} b + 9 \, a^{3} b^{2} - 3 \, a^{2} b^{3} + {\left (3 \, a^{3} b^{2} - 17 \, a^{2} b^{3} + 22 \, a b^{4} - 8 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 3 \, {\left (2 \, a^{4} b - 9 \, a^{3} b^{2} + 11 \, a^{2} b^{3} - 4 \, a b^{4}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^3*b^2*tan(f*x + e)^5 + 2*a^4*b*tan(f*x + e)^3 + a^5*tan(f*x + e))*sqrt(-a + b)*log(-((a^2 - 8*a*b
+ 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2 + 4*((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x +
e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*(3*a^5 - 9*a^4*b + 9
*a^3*b^2 - 3*a^2*b^3 + (3*a^3*b^2 - 17*a^2*b^3 + 22*a*b^4 - 8*b^5)*tan(f*x + e)^4 + 3*(2*a^4*b - 9*a^3*b^2 + 1
1*a^2*b^3 - 4*a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*
f*tan(f*x + e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b + 3*a^6*b^2 -
a^5*b^3)*f*tan(f*x + e)), -1/6*(3*(a^3*b^2*tan(f*x + e)^5 + 2*a^4*b*tan(f*x + e)^3 + a^5*tan(f*x + e))*sqrt(a
- b)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a)) + 2*(3*a^5
- 9*a^4*b + 9*a^3*b^2 - 3*a^2*b^3 + (3*a^3*b^2 - 17*a^2*b^3 + 22*a*b^4 - 8*b^5)*tan(f*x + e)^4 + 3*(2*a^4*b -
9*a^3*b^2 + 11*a^2*b^3 - 4*a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b
^4 - a^3*b^5)*f*tan(f*x + e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b
+ 3*a^6*b^2 - a^5*b^3)*f*tan(f*x + e))]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(cot(e + f*x)**2/(a + b*tan(e + f*x)**2)**(5/2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^2}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^2/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(cot(e + f*x)^2/(a + b*tan(e + f*x)^2)^(5/2), x)

________________________________________________________________________________________