\(\int \cot ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [401]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \]

[Out]

-a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f-1/15*(15*a^2+10*a*b-2*b^2)*cot(f*x+e)*(a+b+b*
tan(f*x+e)^2)^(1/2)/(a+b)/f+1/15*(5*a-b)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/f-1/5*(a+b)*cot(f*x+e)^5*(a+b
+b*tan(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 485, 597, 12, 385, 209} \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f} \]

[In]

Int[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f) - ((15*a^2 + 10*a*b - 2*b^2)*Cot[
e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)*f) + ((5*a - b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x
]^2])/(15*f) - ((a + b)*Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(5*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 485

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[c*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b \left (1+x^2\right )\right )^{3/2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}+\frac {\text {Subst}\left (\int \frac {-((5 a-b) (a+b))-(4 a-b) b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = \frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {\text {Subst}\left (\int \frac {-\left ((a+b) \left (15 a^2+10 a b-2 b^2\right )\right )-2 (5 a-b) b (a+b) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b) f} \\ & = -\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}+\frac {\text {Subst}\left (\int -\frac {15 a^2 (a+b)^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {a^2 \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 )}{f} \\ & = -\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.84 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {2 \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-\frac {3}{4} (a+2 b+a \cos (2 (e+f x)))^2 \csc ^2(e+f x)+\frac {5 (a+b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {a \sin ^2(e+f x)}{a+b}\right )}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{15 (a+b) f (a+2 b+a \cos (2 (e+f x)))} \]

[In]

Integrate[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(2*Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*((-3*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e + f*x]^2)/4 + (5*(a
+ b)^2*Hypergeometric2F1[-3/2, -3/2, -1/2, (a*Sin[e + f*x]^2)/(a + b)])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a +
b)]))/(15*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(792\) vs. \(2(147)=294\).

Time = 5.02 (sec) , antiderivative size = 793, normalized size of antiderivative = 4.81

method	result	size
default	\(-\frac {\left (23 \cos \left (f x +e \right )^{4} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+20 \cos \left (f x +e \right )^{4} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -15 \sin \left (f x +e \right )^{3} a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \cos \left (f x +e \right )-15 \sin \left (f x +e \right )^{3} a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \cos \left (f x +e \right )+15 \sin \left (f x +e \right )^{3} a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right )+15 \sin \left (f x +e \right )^{3} a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right )-35 \cos \left (f x +e \right )^{2} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}-24 \cos \left (f x +e \right )^{2} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +5 \cos \left (f x +e \right )^{2} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+15 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+10 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -2 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \cot \left (f x +e \right )^{3} \csc \left (f x +e \right )^{2}}{15 f \left (a +b \right ) \sqrt {-a}\, \left (b +a \cos \left (f x +e \right )^{2}\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(793\)

[In]

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/f/(a+b)/(-a)^(1/2)*(23*cos(f*x+e)^4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+20*cos(f*
x+e)^4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-15*sin(f*x+e)^3*a^3*ln(4*(-a)^(1/2)*((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(
f*x+e)*a)*cos(f*x+e)-15*sin(f*x+e)^3*a^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x
+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*cos(f*x+e)+15*sin(f*x+e)^3*a^3*ln
(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f
*x+e))^2)^(1/2)-4*sin(f*x+e)*a)+15*sin(f*x+e)^3*a^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)-35*cos(f*x+e)^2*(-a)^(
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-24*cos(f*x+e)^2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*a*b+5*cos(f*x+e)^2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+15*(-a)^(1/2)*((b+
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+10*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-2*(-
a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b*sec(f*x+e)^2)^(3/2)/(b+a*cos(f*x+e)^2)/((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cot(f*x+e)^3*csc(f*x+e)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (147) = 294\).

Time = 3.06 (sec) , antiderivative size = 767, normalized size of antiderivative = 4.65 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {15 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a^{2} + a b\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 (23 \, a^{2} + 20 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 24 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, b^{2}\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}}}}{120 \, {\left ({\left (a + b\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2} + {\left (a + b\right )} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a^{2} + a b\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 (23 \, a^{2} + 20 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 24 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b - 2 \, b^{2}\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}}}}{60 \, {\left ({\left (a + b\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2} + {\left (a + b\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*((a^2 + a*b)*cos(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^2 + a*b)*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*(
(23*a^2 + 20*a*b)*cos(f*x + e)^5 - (35*a^2 + 24*a*b - 5*b^2)*cos(f*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*
x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^4 - 2*(a + b)*f*cos(f*x + e)^2 +
(a + b)*f)*sin(f*x + e)), 1/60*(15*((a^2 + a*b)*cos(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^2 + a*b)*sq
rt(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*((23*a^2 + 20*a*b)*cos(f*x + e)^5 - (35*a^2 + 24*a*b - 5*b^2)*cos(f
*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*c
os(f*x + e)^4 - 2*(a + b)*f*cos(f*x + e)^2 + (a + b)*f)*sin(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)

Giac [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]

[In]

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2),x)

[Out]

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2), x)