3.5.26 \(\int \frac {\cot ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [426]

Optimal. Leaf size=174 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 483, 597, 12, 385, 209} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 a f (a+b)^2}-\frac {b \cot ^3(e+f x)}{a f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 a f (a+b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f) - (b*Cot[e + f*x]^3)/(a*(a + b)*f*Sq
rt[a + b + b*Tan[e + f*x]^2]) + ((3*a - b)*(a + 3*b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(3*a*(a + b)
^3*f) - ((a - 3*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(3*a*(a + b)^2*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 2000

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4226

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

\begin {align*} \int \frac {\cot ^4(e+f x)}{\left (a+b \sec ^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 \left (1+x^2\right )\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right ) \left (a+b+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+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {a-3 b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+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+b \tan ^2(e+f x)}}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f}-\frac {\text {Subst}\left (\int \frac {(3 a-b) (a+3 b)+2 (a-3 b) b x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a+b)^2 f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f}+\frac {\text {Subst}\left (\int \frac {3 (a+b)^3}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a+b)^3 f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f}+\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f}+\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {b \cot ^3(e+f x)}{a (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a-b) (a+3 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^3 f}-\frac {(a-3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 a (a+b)^2 f}\\ \end {align*}

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Mathematica [A]
time = 5.83, size = 224, normalized size = 1.29 \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{2 \sqrt {2} a^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 e+2 f x))^2 \sec ^3(e+f x) \left (\frac {(4 a+9 b) \csc (e+f x)}{12 (a+b)^3 f}-\frac {\csc ^3(e+f x)}{12 (a+b)^2 f}-\frac {b^3 \sin (e+f x)}{2 a (a+b)^3 f (a+2 b+a \cos (2 e+2 f x))}\right )}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*
x]^3)/(2*Sqrt[2]*a^(3/2)*f*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^3*((
(4*a + 9*b)*Csc[e + f*x])/(12*(a + b)^3*f) - Csc[e + f*x]^3/(12*(a + b)^2*f) - (b^3*Sin[e + f*x])/(2*a*(a + b)
^3*f*(a + 2*b + a*Cos[2*e + 2*f*x]))))/(a + b*Sec[e + f*x]^2)^(3/2)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.18, size = 7541, normalized size = 43.34

method result size
default \(\text {Expression too large to display}\) \(7541\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 489 vs. \(2 (166) = 332\).
time = 9.49, size = 1102, normalized size = 6.33 \begin {gather*} \left [-\frac {3 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} b - 3 \, a^{2} b^{2} - 3 \, a b^{3} - b^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 8 \, {\left ({\left (4 \, a^{4} + 9 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{4} + 4 \, a^{3} b - 9 \, a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{6} + 2 \, a^{5} b - 2 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {3 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} b - 3 \, a^{2} b^{2} - 3 \, a b^{3} - b^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (4 \, a^{4} + 9 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{4} + 4 \, a^{3} b - 9 \, a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{6} + 2 \, a^{5} b - 2 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(3*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a
^3*b - 2*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 +
 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7
*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5
*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos
(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f*x + e) - 8*((4*a^4 + 9*a^3*b + 3*a*b^3)*cos(f*x + e)^5 -
(3*a^4 + 4*a^3*b - 9*a^2*b^2 + 6*a*b^3)*cos(f*x + e)^3 - (3*a^3*b + 8*a^2*b^2 - 3*a*b^3)*cos(f*x + e))*sqrt((a
*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (a^6 + 2*a^5*
b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f)*sin(f*x + e)), -1/12*
(3*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - 3*a*b^3 - b^4 - (a^4 + 2*a^3*b -
2*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2
- 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*
b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)))*sin(f*x + e) - 4*((4*a^4 + 9*a^3*b + 3*a*b^3)*cos(f
*x + e)^5 - (3*a^4 + 4*a^3*b - 9*a^2*b^2 + 6*a*b^3)*cos(f*x + e)^3 - (3*a^3*b + 8*a^2*b^2 - 3*a*b^3)*cos(f*x +
 e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 - (
a^6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 - (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f)*sin(f*x +
 e))]

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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 \sec ^{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*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**4/(a + b*sec(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)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^4/(b*sec(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 )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\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/cos(e + f*x)^2)^(3/2),x)

[Out]

int(cot(e + f*x)^4/(a + b/cos(e + f*x)^2)^(3/2), x)

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