\(\int \csc ^5(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [85]

Optimal result

Integrand size = 25, antiderivative size = 218 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {3 \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 \sqrt {a+b} f}+\frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \]

[Out]

-1/4*cot(f*x+e)*csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2)/f+3/2*(a+2*b)*arctanh(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)
^2)^(1/2))*b^(1/2)/f-3/8*(a^2+8*a*b+8*b^2)*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/f/(a+b)^(1
/2)+3/8*(a+4*b)*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f-3/8*(a+2*b)*csc(f*x+e)^2*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(
1/2)/f

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4219, 478, 591, 596, 537, 223, 212, 385, 213} \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 f \sqrt {a+b}}+\frac {3 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \]

[In]

Int[Csc[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(3*Sqrt[b]*(a + 2*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) - (3*(a^2 + 8*a*b + 8*b
^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*Sqrt[a + b]*f) + (3*(a + 4*b)*Sec[e + f
*x]*Sqrt[a + b*Sec[e + f*x]^2])/(8*f) - (3*(a + 2*b)*Csc[e + f*x]^2*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(
8*f) - (Cot[e + f*x]*Csc[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(4*f)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 591

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 596

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q +
1) + 1))), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 4219

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^
n)^p/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^{3/2}}{\left (-1+x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\text {Subst}\left (\int \frac {x^2 \sqrt {a+b x^2} \left (3 a+6 b x^2\right )}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 f} \\ & = -\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 a (a+6 b)+6 b (a+4 b) x^2\right )}{\left (-1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 f} \\ & = \frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f}-\frac {\text {Subst}\left (\int \frac {-6 a b (a+4 b)-24 b^2 (a+2 b) x^2}{\left (-1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{16 b f} \\ & = \frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 b (a+2 b)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}+\frac {\left (3 \left (a^2+8 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 f} \\ & = \frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 b (a+2 b)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {\left (3 \left (a^2+8 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-(-a-b) x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 f} \\ & = \frac {3 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {3 \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 \sqrt {a+b} f}+\frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.60 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.20 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\left (b+a \cos ^2(e+f x)\right ) \left (-12 b^{3/2} \left (a^2+3 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right ) \cos ^2(e+f x)+3 b \sqrt {a+b} \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \cos ^2(e+f x)+\frac {b (a+b) \sqrt {a+2 b+a \cos (2 (e+f x))} (11 a+4 b+8 (a+3 b) \cos (2 (e+f x))-3 (a+4 b) \cos (4 (e+f x))) \csc ^4(e+f x)}{8 \sqrt {2}}\right ) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} b (a+b) f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \]

[In]

Integrate[Csc[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-1/2*((b + a*Cos[e + f*x]^2)*(-12*b^(3/2)*(a^2 + 3*a*b + 2*b^2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]
]*Cos[e + f*x]^2 + 3*b*Sqrt[a + b]*(a^2 + 8*a*b + 8*b^2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*C
os[e + f*x]^2 + (b*(a + b)*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(11*a + 4*b + 8*(a + 3*b)*Cos[2*(e + f*x)] - 3*(
a + 4*b)*Cos[4*(e + f*x)])*Csc[e + f*x]^4)/(8*Sqrt[2]))*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b*(a
+ b)*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5130\) vs. \(2(190)=380\).

Time = 6.61 (sec) , antiderivative size = 5131, normalized size of antiderivative = 23.54

method	result	size
default	\(\text {Expression too large to display}\)	\(5131\)

[In]

int(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 1511, normalized size of antiderivative = 6.93 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2
)*cos(f*x + e))*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2
)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + 12*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^5 - 2*(a^2 + 3*a*b +
2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2*b^2)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*
cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*(3*(a^2 + 5*a*b + 4*b^2)*cos(f*x +
e)^4 - (5*a^2 + 23*a*b + 18*b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))
/((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e)), 1/8*(3*((a^2 + 8*a*b + 8*b^
2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(-a - b)*
arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + 6*((a^2 + 3*a*b + 2*b^
2)*cos(f*x + e)^5 - 2*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2*b^2)*cos(f*x + e))*sqrt(b)*log((
a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) +
(3*(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^2 + 23*a*b + 18*b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x
+ e)), -1/16*(24*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^5 - 2*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*
b + 2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)
- 3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos
(f*x + e))*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos
(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) - 2*(3*(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^2 + 23*a*b + 18*
b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 -
2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e)), 1/8*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 +
8*a*b + 8*b^2)*cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) - 12*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^5 - 2*(a^2 +
3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^2 + 23*a*b + 18*b
^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 - 2
*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(csc(f*x+e)**5*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{5} \,d x } \]

[In]

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^5, x)

Giac [F(-2)]

Exception generated. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{262144,[6,6]%%%},[12]%%%}+%%%{%%{[%%%{1572864,[6,6]%
%%},0]:[1,0

Mupad [F(-1)]

Timed out. \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^5} \,d x \]

[In]

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^5,x)

[Out]

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^5, x)