3.344 \(\int \frac {\cot ^4(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=184 \[ \frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f (a-b)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac {b \cot ^3(e+f x)}{a f (a-b) \sqrt {a+b \tan ^2(e+f x)}} \]
[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)^3/a/(a-b)/f/(a+b*tan(f*x+e)
^2)^(1/2)+1/3*(3*a-4*b)*(a+2*b)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^3/(a-b)/f-1/3*(a-4*b)*cot(f*x+e)^3*(a+b*
tan(f*x+e)^2)^(1/2)/a^2/(a-b)/f
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Rubi [A] time = 0.26, antiderivative size = 184, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used =
{3670, 472, 583, 12, 377, 203} \[ -\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f (a-b)}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 f (a-b)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac {b \cot ^3(e+f x)}{a f (a-b) \sqrt {a+b \tan ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^4/(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]^3)/(a*(a - b
)*f*Sqrt[a + b*Tan[e + f*x]^2]) + ((3*a - 4*b)*(a + 2*b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(3*a^3*(a -
b)*f) - ((a - 4*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*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 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 472
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 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)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {a-4 b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{a (a-b) f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) f}-\frac {\operatorname {Subst}\left (\int \frac {(3 a-4 b) (a+2 b)+2 (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) f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) f}+\frac {\operatorname {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) f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) 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 )}{(a-b) f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) 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 )}{(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 ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) f}\\ \end {align*}
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Mathematica [C] time = 16.45, size = 802, normalized size = 4.36 \[ \frac {-\frac {b \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{(a-b) f}+\frac {\sqrt {\frac {\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (-\frac {\sin (2 (e+f x)) b^3}{a^3 (a-b) (\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x)))}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2}+\frac {(4 a \cos (e+f x)+5 b \cos (e+f x)) \csc (e+f x)}{3 a^3}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
(-((b*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*
(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(
e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*
x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a + b + (a - b)*Cos[2*(e
+ f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x
]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((
a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x
)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*
Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(
b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(2
*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])))/Sqrt[a + b + (a - b)*Cos[2*(e +
f*x)]])/((a - b)*f) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(((4*a*C
os[e + f*x] + 5*b*Cos[e + f*x])*Csc[e + f*x])/(3*a^3) - (Cot[e + f*x]*Csc[e + f*x]^2)/(3*a^2) - (b^3*Sin[2*(e
+ f*x)])/(a^3*(a - b)*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)]))))/f
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fricas [A] time = 0.58, size = 579, normalized size = 3.15 \[ \left [\frac {3 \, {\left (a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (3 \, a^{3} b - a^{2} b^{2} - 10 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} - a^{4} + 2 \, a^{3} b - a^{2} b^{2} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \tan \left (f x + e\right )^{5} + {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \tan \left (f x + e\right )^{3}\right )}}, \frac {3 \, {\left (a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) + 2 \, {\left ({\left (3 \, a^{3} b - a^{2} b^{2} - 10 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} - a^{4} + 2 \, a^{3} b - a^{2} b^{2} + {\left (3 \, a^{4} - 2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \tan \left (f x + e\right )^{5} + {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} f \tan \left (f x + e\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/12*(3*(a^3*b*tan(f*x + e)^5 + a^4*tan(f*x + e)^3)*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^3*b - a^2*b^2 - 10*a*b^3 + 8*b^4)*tan(f*
x + e)^4 - a^4 + 2*a^3*b - a^2*b^2 + (3*a^4 - 2*a^3*b - 5*a^2*b^2 + 4*a*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x +
e)^2 + a))/((a^5*b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^5 + (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^3), 1/6*
(3*(a^3*b*tan(f*x + e)^5 + a^4*tan(f*x + e)^3)*sqrt(a - b)*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*ta
n(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a)) + 2*((3*a^3*b - a^2*b^2 - 10*a*b^3 + 8*b^4)*tan(f*x + e)^4 - a^4 +
2*a^3*b - a^2*b^2 + (3*a^4 - 2*a^3*b - 5*a^2*b^2 + 4*a*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^5*
b - 2*a^4*b^2 + a^3*b^3)*f*tan(f*x + e)^5 + (a^6 - 2*a^5*b + a^4*b^2)*f*tan(f*x + e)^3)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(3/2), x)
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maple [C] time = 1.21, size = 2577, normalized size = 14.01 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^4/(a+b*tan(f*x+e)^2)^(3/2),x)
[Out]
-1/3/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^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)^3*sin(f*x+e)*a^3+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*sin(f*x+e)*a^3-6*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^3+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^3+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^3-3*cos(f*x+e)*sin(f*x+e)*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))*a^3+4*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)
^4*a^3-3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*a^2*b-6*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1
/2)*cos(f*x+e)^4*a*b^2+8*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*b^3+6*sin(f*x+e)*2^(1/2)*((I*c
os(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)*El
lipticPi((-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))*a^3-3*sin(f*x+e)*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))*a^3-3*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos
(f*x+e)^2*a^3+5*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*a^2*b+8*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b
)/a)^(1/2)*cos(f*x+e)^2*a*b^2-16*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^2*b^3-3*((2*I*(a-b)^(1/2
)*b^(1/2)+a-2*b)/a)^(1/2)*a^2*b-2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a*b^2+8*((2*I*(a-b)^(1/2)*b^(1/2)+
a-2*b)/a)^(1/2)*b^3)*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)^3/a^3/((2*
I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/(a-b)
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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)^(3/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2),x)
[Out]
int(cot(e + f*x)^4/(a + b*tan(e + f*x)^2)^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**4/(a+b*tan(f*x+e)**2)**(3/2),x)
[Out]
Integral(cot(e + f*x)**4/(a + b*tan(e + f*x)**2)**(3/2), x)
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