3.331 \(\int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)
Optimal. Leaf size=120 \[ \frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f} \]
[Out]
arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)+1/3*(3*a+2*b)*cot(f*x+e)*(a+b*tan(f*x+e)
^2)^(1/2)/a^2/f-1/3*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a/f
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Rubi [A] time = 0.16, antiderivative size = 120, 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 =
{3670, 480, 583, 12, 377, 203} \[ \frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(Sqrt[a - b]*f) + ((3*a + 2*b)*Cot[e + f*x]*Sqrt
[a + b*Tan[e + f*x]^2])/(3*a^2*f) - (Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*a*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 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 377
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 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 3670
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 ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f}+\frac {\operatorname {Subst}\left (\int \frac {-3 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 )}{3 a f}\\ &=\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f}-\frac {\operatorname {Subst}\left (\int -\frac {3 a^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 f}\\ &=\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f}+\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f}+\frac {\operatorname {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 )}{f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}+\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f}\\ \end {align*}
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Mathematica [C] time = 11.07, size = 263, normalized size = 2.19 \[ -\frac {\cos ^2(e+f x) \cot ^3(e+f x) \left (\frac {b \tan ^2(e+f x)}{a}+1\right ) \left (-8 (a-b) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, _3F_2\left (2,2,2;1,\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )+\frac {6 a \sin ^{-1}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a^2-4 a b \tan ^2(e+f x)-8 b^2 \tan ^4(e+f x)\right )}{\sqrt {\frac {(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}-12 b (b-a) \tan ^4(e+f x) ((b-a) \cos (2 (e+f x))-a-b) \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )\right )}{9 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
-1/9*(Cos[e + f*x]^2*Cot[e + f*x]^3*(1 + (b*Tan[e + f*x]^2)/a)*(-12*b*(-a + b)*(-a - b + (-a + b)*Cos[2*(e + f
*x)])*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^4 - 8*(a - b)*HypergeometricPFQ[{2
, 2, 2}, {1, 5/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2)^2 + (6*a*ArcSin[Sqrt[((a
- b)*Sin[e + f*x]^2)/a]]*(a^2 - 4*a*b*Tan[e + f*x]^2 - 8*b^2*Tan[e + f*x]^4))/Sqrt[((a - b)*Sin[2*(e + f*x)]^2
*(a + b*Tan[e + f*x]^2))/a^2]))/(a^3*f*Sqrt[a + b*Tan[e + f*x]^2])
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fricas [A] time = 0.56, size = 359, normalized size = 2.99 \[ \left [-\frac {3 \, a^{2} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 4 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + a b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{3}}, \frac {3 \, \sqrt {a - b} a^{2} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{3} + 2 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + a b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[-1/12*(3*a^2*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))*tan(f*x + e)^3 - 4*((3*a^2 - a*b - 2*b^2)*tan(f*x + e)^2 - a^2 + a*b)*sqrt(b*tan(f*x + e)
^2 + a))/((a^3 - a^2*b)*f*tan(f*x + e)^3), 1/6*(3*sqrt(a - b)*a^2*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))*tan(f*x + e)^3 + 2*((3*a^2 - a*b - 2*b^2)*tan(f*x + e)^2 - a
^2 + a*b)*sqrt(b*tan(f*x + e)^2 + a))/((a^3 - a^2*b)*f*tan(f*x + e)^3)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^4/sqrt(b*tan(f*x + e)^2 + a), x)
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maple [C] time = 1.77, size = 2433, normalized size = 20.28 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x)
[Out]
-1/3/f*(-6*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+co
s(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1
+cos(f*x+e))/a)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*
(a-b)^(1/2)*b^(1/2)+a-2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1
/2))*cos(f*x+e)^3*sin(f*x+e)*a^2+3*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+
e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f
*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1
/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^3*s
in(f*x+e)*a^2-6*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/
(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-
b)/(1+cos(f*x+e))/a)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/
(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/
a)^(1/2))*cos(f*x+e)^2*sin(f*x+e)*a^2+3*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos
(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*
cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/
a)^(1/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e
)^2*sin(f*x+e)*a^2+6*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e
)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*
x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e
),-1/(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)*b^(1/2)+a-
2*b)/a)^(1/2))*cos(f*x+e)*sin(f*x+e)*a^2-3*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*
cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)
-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*
b)/a)^(1/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*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+4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*a^2-2*((2*I*(a-b)^(1/2)*b^(1/2)+a
-2*b)/a)^(1/2)*cos(f*x+e)^4*a*b-2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*b^2+6*a^2*2^(1/2)*((I
*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2
*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*
EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*(a-b)^(1/2)*b^(1/2)+a-
2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2))*sin(f*x+e)-3*2^(1
/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1
/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)
^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2
)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*sin(f*x+e)*a^2-3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^
(1/2)*cos(f*x+e)^2*a^2+5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a*b+4*((2*I*(a-b)^(1/2)*b^(1/2
)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*b^2-3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b-2*((2*I*(a-b)^(1/2)*b^(1/2)
+a-2*b)/a)^(1/2)*b^2)/cos(f*x+e)/sin(f*x+e)^3/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/((2*I*(a-
b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/a^2
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(1/2),x)
[Out]
int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**4/(a+b*tan(f*x+e)**2)**(1/2),x)
[Out]
Integral(cot(e + f*x)**4/sqrt(a + b*tan(e + f*x)**2), x)
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