3.4.44 \(\int \frac {\cot ^4(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [344]
Optimal. Leaf size=184 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2} f}-\frac {b \cot ^3(e+f x)}{a (a-b) f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-4 b) (a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^3 (a-b) f}-\frac {(a-4 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 (a-b) f} \]
[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
________________________________________________________________________________________
Rubi [A]
time = 0.17, 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 = {3751, 483, 597,
12, 385, 209} \begin {gather*} \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 {\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 ^3(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]^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 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 ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\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) 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 {\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 ^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 {\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 ^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*}
________________________________________________________________________________________
Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 11.82, size = 1398, normalized size = 7.60 \begin {gather*} -\frac {\cos ^2(e+f x) \cot ^3(e+f x) \left (\frac {45 a \csc ^2(e+f x)}{a-b}-\frac {270 b \sec ^2(e+f x)}{a-b}+\frac {4 (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)}{a}-\frac {24 (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)}{a}-\frac {16 (a-b) \, _4F_3\left (2,2,2,2;1,1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x)}{a}-\frac {1080 b^2 \sec ^2(e+f x) \tan ^2(e+f x)}{a (a-b)}-\frac {132 (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)}{a^2}-\frac {144 (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)}{a^2}-\frac {48 (a-b) b \, _4F_3\left (2,2,2,2;1,1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a^2}-\frac {720 b^3 \sec ^2(e+f x) \tan ^4(e+f x)}{a^2 (a-b)}-\frac {312 (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)}{a^3}-\frac {216 (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)}{a^3}-\frac {48 (a-b) b^2 \, _4F_3\left (2,2,2,2;1,1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{a^3}-\frac {176 (a-b) b^3 \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^6(e+f x)}{a^4}-\frac {96 (a-b) b^3 \, _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 ^6(e+f x)}{a^4}-\frac {16 (a-b) b^3 \, _4F_3\left (2,2,2,2;1,1,\frac {7}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^6(e+f x)}{a^4}-\frac {45 \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 {270 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 {1080 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 {720 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 )^{3/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}+\frac {45 \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 {270 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 {1080 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 {720 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}}}\right )}{45 a f \sqrt {a+b \tan ^2(e+f x)}} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
-1/45*(Cos[e + f*x]^2*Cot[e + f*x]^3*((45*a*Csc[e + f*x]^2)/(a - b) - (270*b*Sec[e + f*x]^2)/(a - b) + (4*(a -
b)*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/a - (24*(a - b)*HypergeometricPFQ
[{2, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/a - (16*(a - b)*HypergeometricPFQ[{2, 2, 2,
2}, {1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2)/a - (1080*b^2*Sec[e + f*x]^2*Tan[e + f*x]^2)/(a*(
a - b)) - (132*(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)/a^2 - (144*(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)/a^2 - (48*(a - b)*b*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e
+ f*x]^2*Tan[e + f*x]^2)/a^2 - (720*b^3*Sec[e + f*x]^2*Tan[e + f*x]^4)/(a^2*(a - b)) - (312*(a - b)*b^2*Hyper
geometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/a^3 - (216*(a - b)*b^2*Hype
rgeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/a^3 - (48*(a - b
)*b^2*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^4)/
a^3 - (176*(a - b)*b^3*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e + f*x]^6)
/a^4 - (96*(a - b)*b^3*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*Tan[e
+ f*x]^6)/a^4 - (16*(a - b)*b^3*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 7/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[
e + f*x]^2*Tan[e + f*x]^6)/a^4 - (45*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]) + (270*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]) + (1080*b^2*A
rcSin[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[(Cos[e +
f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (720*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)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]) + (45*ArcSin[Sqrt[((a - b)*S
in[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] - (270*b*ArcSin[S
qrt[((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]) - (1080*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]) - (720*b^3*ArcSin[Sqrt[((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])))/(a*f*Sqrt[a + b*T
an[e + f*x]^2])
________________________________________________________________________________________
Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.37, size = 2577, normalized size = 14.01
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(2577\) |
| | |
|
|
|
|
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,method=_RETURNVERBOSE)
[Out]
1/3/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(6*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^3-3*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)*Ellipti
cF((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+6*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)*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^3-3*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)*Elli
pticF((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^3-6*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+3*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*c
os(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)*Ellip
ticF((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^3-6*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))*sin(f*x+e)*a^3+3*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))*sin(f*x+e)*a^3-4*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^3+3*cos(f*x+e)^4*((2*I*b^(
1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b+6*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b^2-8*cos(f*
x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3+3*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*
a^3-5*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2*b-8*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-
2*b)/a)^(1/2)*a*b^2+16*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^3+3*((2*I*b^(1/2)*(a-b)^(1/2)+
a-2*b)/a)^(1/2)*a^2*b+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)*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*b^(1/2)*
(a-b)^(1/2)+a-2*b)/a)^(1/2)/(a-b)
________________________________________________________________________________________
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)^4/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A]
time = 4.62, size = 603, normalized size = 3.28 \begin {gather*} \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 ] \end {gather*}
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)]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{4}{\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)**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)
________________________________________________________________________________________
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)^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)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}
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)
________________________________________________________________________________________