3.4.32 \(\int \frac {\cot ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [332]
Optimal. Leaf size=170 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \]
[Out]
-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)-1/15*(15*a^2+10*a*b+8*b^2)*cot(f*x+e)*(
a+b*tan(f*x+e)^2)^(1/2)/a^3/f+1/15*(5*a+4*b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a^2/f-1/5*cot(f*x+e)^5*(a+b
*tan(f*x+e)^2)^(1/2)/a/f
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Rubi [A]
time = 0.16, antiderivative size = 170, 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 = {3751, 491, 597,
12, 385, 209} \begin {gather*} \frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^6/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)) - ((15*a^2 + 10*a*b + 8*b^2)*
Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) + ((5*a + 4*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/
(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*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 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 491
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 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 ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}+\frac {\text {Subst}\left (\int \frac {-5 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 )}{5 a f}\\ &=\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}-\frac {\text {Subst}\left (\int \frac {-15 a^2-10 a b-8 b^2-2 b (5 a+4 b) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}+\frac {\text {Subst}\left (\int -\frac {15 a^3}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a 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 )}{f}\\ &=-\frac {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a 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 )}{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 {\left (15 a^2+10 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}+\frac {(5 a+4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a 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 = 12.24, size = 1374, normalized size = 8.08 \begin {gather*} -\frac {\cos ^4(e+f x) \cot ^5(e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (9 a^4 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right )+9 a^4 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)-18 a^3 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)-18 a^3 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)+72 a^2 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)+72 a^2 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^6(e+f x)+144 a b^3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^6(e+f x)+144 a b^3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^8(e+f x)-4 a^4 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}+4 a^3 b \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}-12 a^3 b \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^4(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}+12 a^2 b^2 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^4(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}+168 a^2 b^2 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^6(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}-168 a b^3 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^6(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}+176 a b^3 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^8(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}-176 b^4 \, _2F_1\left (2,2;\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^8(e+f x) \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}+16 (a-b) \, _4F_3\left (2,2,2,2;1,1,\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^3 \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}-24 \, _3F_2\left (2,2,2;1,\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}} \left (a^2+4 b^2 \tan ^2(e+f x)-a \left (b+4 b \tan ^2(e+f x)\right )\right )\right )}{45 a^4 f \sqrt {a+b \tan ^2(e+f x)} \sqrt {\frac {\cos ^2(e+f x) \sin ^2(e+f x) \left (a^2-b^2 \tan ^2(e+f x)+a b \left (-1+\tan ^2(e+f x)\right )\right )}{a^2}}} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^6/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
-1/45*(Cos[e + f*x]^4*Cot[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*(9*a^4*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]
] + 9*a^4*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2 - 18*a^3*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]
^2)/a]]*Tan[e + f*x]^2 - 18*a^3*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4 + 72*a^2*b^2*ArcSin[
Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^4 + 72*a^2*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e +
f*x]^6 + 144*a*b^3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^6 + 144*a*b^3*ArcSin[Sqrt[((a - b)*S
in[e + f*x]^2)/a]]*Tan[e + f*x]^8 - 4*a^4*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x
]^2*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] + 4*a^3*b
*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(
a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] - 12*a^3*b*Hypergeometric2F1[2, 2, 5/2, ((a - b)*S
in[e + f*x]^2)/a]*Tan[e + f*x]^4*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan
[e + f*x]^2)))/a^2] + 12*a^2*b^2*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^4*Sqrt[
(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] + 168*a^2*b^2*Hype
rgeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^6*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 -
b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] - 168*a*b^3*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e
+ f*x]^2)/a]*Tan[e + f*x]^6*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e +
f*x]^2)))/a^2] + 176*a*b^3*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^8*Sqrt[(Cos[
e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] - 176*b^4*Hypergeometri
c2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^8*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[
e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] + 16*(a - b)*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 5/2}, ((a -
b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^3*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*T
an[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2] - 24*HypergeometricPFQ[{2, 2, 2}, {1, 5/2}, ((a - b)*Sin[e +
f*x]^2)/a]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^2*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]
^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2]*(a^2 + 4*b^2*Tan[e + f*x]^2 - a*(b + 4*b*Tan[e + f*x]^2))))/(a^4*f*Sqrt[
a + b*Tan[e + f*x]^2]*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2
)))/a^2])
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.42, size = 3741, normalized size = 22.01
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| method |
result |
size |
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\(\text {Expression too large to display}\) |
\(3741\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-1/15/f*(-30*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)/(co
s(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))*sin(f*x+e)*a^3-30*cos(f*x+e)^2*sin(f*x+e)*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^
3-30*cos(f*x+e)*sin(f*x+e)*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^3+15*cos(f*x+e)*sin(f*x+e)*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
^3+15*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+10*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2+24*co
s(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3-24*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(
1/2)*b^3+23*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3-35*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/
2)+a-2*b)/a)^(1/2)*a^3+15*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3-9*cos(f*x+e)^6*((2*I*b^(1
/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b-6*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2+34*cos(f*
x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+22*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2)*a*b^2-40*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b-26*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(
1/2)+a-2*b)/a)^(1/2)*a*b^2+8*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3-8*cos(f*x+e)^6*((2*I*b^(1/2)*(a-b)^
(1/2)+a-2*b)/a)^(1/2)*b^3-30*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*c
os(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)/si
n(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)^5*sin(f*x+e)*a^3+15*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)^5*sin(f*x+e)*a^3-30*cos(f*x+e)^4*sin(f*x+e)*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^3+15*cos(f*x+e)^4*sin(f*x+e)*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^3+60*cos(f*x+e)^3*sin(f*x+e)*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^3-30*cos(f*x+e)^3*sin(f*x+e)*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...
________________________________________________________________________________________
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)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A]
time = 4.79, size = 457, normalized size = 2.69 \begin {gather*} \left [-\frac {15 \, a^{3} \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 )^{5} + 4 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}, -\frac {15 \, \sqrt {a - b} a^{3} \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 )^{5} + 2 \, {\left ({\left (15 \, a^{3} - 5 \, a^{2} b - 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 3 \, a^{3} - 3 \, a^{2} b - {\left (5 \, a^{3} - a^{2} b - 4 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, {\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[-1/60*(15*a^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))*tan(f*x + e)^5 + 4*((15*a^3 - 5*a^2*b - 2*a*b^2 - 8*b^3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*
b - (5*a^3 - a^2*b - 4*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^5), -1
/30*(15*sqrt(a - b)*a^3*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)^5 + 2*((15*a^3 - 5*a^2*b - 2*a*b^2 - 8*b^3)*tan(f*x + e)^4 + 3*a^3 - 3*a^2*b - (5*a^3 - a
^2*b - 4*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^5)]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{6}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**6/(a+b*tan(f*x+e)**2)**(1/2),x)
[Out]
Integral(cot(e + f*x)**6/sqrt(a + b*tan(e + f*x)**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)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(cot(f*x + e)^6/sqrt(b*tan(f*x + e)^2 + a), 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 )}^6}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(e + f*x)^6/(a + b*tan(e + f*x)^2)^(1/2),x)
[Out]
int(cot(e + f*x)^6/(a + b*tan(e + f*x)^2)^(1/2), x)
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