3.4.43 \(\int \frac {\cot ^2(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [343]

Optimal. Leaf size=128 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) f} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 483, 597, 12, 385, 209} \begin {gather*} -\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 f (a-b)}-\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac {b \cot (e+f x)}{a f (a-b) \sqrt {a+b \tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f)) - (b*Cot[e + f*x])/(a*(a -
b)*f*Sqrt[a + b*Tan[e + f*x]^2]) - ((a - 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(a^2*(a - b)*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 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 )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {a-2 b-2 b x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{a (a-b) f}\\ &=-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) f}-\frac {\text {Subst}\left (\int \frac {a^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{a^2 (a-b) f}\\ &=-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) 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) f}\\ &=-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) 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) f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot (e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a^2 (a-b) f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.01, size = 882, normalized size = 6.89 \begin {gather*} -\frac {\cos ^2(e+f x) \cot (e+f x) \left (\frac {3 a \csc ^2(e+f x)}{a-b}+\frac {12 b \sec ^2(e+f x)}{a-b}+\frac {16 (a-b) \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{15 a}+\frac {8 (a-b) \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{15 a}+\frac {8 b^2 \sec ^2(e+f x) \tan ^2(e+f x)}{a (a-b)}+\frac {8 (a-b) b \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{3 a^2}+\frac {16 (a-b) b \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{15 a^2}+\frac {8 (a-b) b^2 \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{5 a^3}+\frac {8 (a-b) b^2 \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{15 a^3}-\frac {3 \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 )^{3/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}-\frac {12 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 )^{3/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}-\frac {8 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 )^{3/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {3 \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 {12 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 {8 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}}}\right )}{a f \sqrt {a+b \tan ^2(e+f x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((Cos[e + f*x]^2*Cot[e + f*x]*((3*a*Csc[e + f*x]^2)/(a - b) + (12*b*Sec[e + f*x]^2)/(a - b) + (16*(a - b)*Hyp
ergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(15*a) + (8*(a - b)*HypergeometricPFQ[{2
, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/(15*a) + (8*b^2*Sec[e + f*x]^2*Tan[e + f*x]^2)/
(a*(a - b)) + (8*(a - b)*b*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x
]^2)/(3*a^2) + (16*(a - b)*b*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2
*Tan[e + f*x]^2)/(15*a^2) + (8*(a - b)*b^2*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*
x]^2*Tan[e + f*x]^4)/(5*a^3) + (8*(a - b)*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/
a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/(15*a^3) - (3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]])/((((a - b)*Sin[e + f
*x]^2)/a)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) - (12*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)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) - (
8*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4)/(a^2*(((a - b)*Sin[e + f*x]^2)/a)^(3/2)*Sqrt[(C
os[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (3*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] + (12*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]) + (8*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])))/(a*f*Sqrt[a + b*Tan[e + f*x]^2]))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.44, size = 1305, normalized size = 10.20

method	result	size
default	\(\text {Expression too large to display}\)	\(1305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos
(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-
cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)
/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+
e)*sin(f*x+e)*a^2-2*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)
+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x
+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),
-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*
b)/a)^(1/2))*cos(f*x+e)*sin(f*x+e)*a^2+2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*
x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos
(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)
^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a^2*sin(f*x
+e)-2*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e
)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*
x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2
)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a
^2*sin(f*x+e)+((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a^2-2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)
^(1/2)*cos(f*x+e)^2*a*b+2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*b^2+((2*I*b^(1/2)*(a-b)^(1/2)
+a-2*b)/a)^(1/2)*a*b-2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2)*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/cos(f*x+e)^2)^(3/2)/sin(f*x+e)/a^2/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/(a-b)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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)^(3/2),x, algorithm="maxima")

[Out]

integrate(cot(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(3/2), x)

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Fricas [A]
time = 3.21, size = 493, normalized size = 3.85 \begin {gather*} \left [\frac {{\left (a^{2} b \tan \left (f x + e\right )^{3} + a^{3} \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 (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{4 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {{\left (a^{2} b \tan \left (f x + e\right )^{3} + a^{3} \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 (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{2 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/4*((a^2*b*tan(f*x + e)^3 + a^3*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*(a^3 - 2*a^2*b + a*b^2 + (a^2*b - 3*a*b^2 + 2*b^3)*
tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b - 2*a^3*b^2 + a^2*b^3)*f*tan(f*x + e)^3 + (a^5 - 2*a^4*b +
a^3*b^2)*f*tan(f*x + e)), -1/2*((a^2*b*tan(f*x + e)^3 + a^3*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*(a^3 - 2*a^2*b + a*b^2 + (a^2*b - 3
*a*b^2 + 2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b - 2*a^3*b^2 + a^2*b^3)*f*tan(f*x + e)^3 +
(a^5 - 2*a^4*b + a^3*b^2)*f*tan(f*x + e))]

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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 {3}{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)**(3/2),x)

[Out]

Integral(cot(e + f*x)**2/(a + b*tan(e + f*x)**2)**(3/2), x)

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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)^(3/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(3/2), x)

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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 )}^{3/2}} \,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)^(3/2),x)

[Out]

int(cot(e + f*x)^2/(a + b*tan(e + f*x)^2)^(3/2), x)

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