\(\int \frac {\csc ^5(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [124]

Optimal result

Integrand size = 25, antiderivative size = 234 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (3 a^2-24 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{9/2} f}-\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

-1/8*(3*a^2-24*a*b+8*b^2)*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(9/2)/f-1/8*(5*a-2*b)
*cot(f*x+e)*csc(f*x+e)/(a+b)^2/f/(a+b*sec(f*x+e)^2)^(3/2)-1/4*cot(f*x+e)^3*csc(f*x+e)/(a+b)/f/(a+b*sec(f*x+e)^
2)^(3/2)-1/24*(23*a-12*b)*b*sec(f*x+e)/(a+b)^3/f/(a+b*sec(f*x+e)^2)^(3/2)-5/24*(11*a-10*b)*b*sec(f*x+e)/(a+b)^
4/f/(a+b*sec(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4219, 481, 541, 12, 385, 213} \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (3 a^2-24 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{9/2}}-\frac {5 b (11 a-10 b) \sec (e+f x)}{24 f (a+b)^4 \sqrt {a+b \sec ^2(e+f x)}}-\frac {b (23 a-12 b) \sec (e+f x)}{24 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-1/8*((3*a^2 - 24*a*b + 8*b^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/((a + b)^(9/2)*
f) - ((5*a - 2*b)*Cot[e + f*x]*Csc[e + f*x])/(8*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (Cot[e + f*x]^3*Cs
c[e + f*x])/(4*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - ((23*a - 12*b)*b*Sec[e + f*x])/(24*(a + b)^3*f*(a + b
*Sec[e + f*x]^2)^(3/2)) - (5*(11*a - 10*b)*b*Sec[e + f*x])/(24*(a + b)^4*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -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 (-1+x^2\right )^3 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-a-2 (2 a-b) x^2}{\left (-1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{4 (a+b) f} \\ & = -\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-a (3 a-4 b)+4 (5 a-2 b) b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{8 (a+b)^2 f} \\ & = -\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-a^2 (9 a-26 b)+2 a (23 a-12 b) b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{24 a (a+b)^3 f} \\ & = -\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\text {Subst}\left (\int -\frac {3 a^2 \left (3 a^2-24 a b+8 b^2\right )}{\left (-1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{24 a^2 (a+b)^4 f} \\ & = -\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (3 a^2-24 a b+8 b^2\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 (a+b)^4 f} \\ & = -\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\left (3 a^2-24 a b+8 b^2\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 (a+b)^4 f} \\ & = -\frac {\left (3 a^2-24 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 (a+b)^{9/2} f}-\frac {(5 a-2 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-12 b) b \sec (e+f x)}{24 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-10 b) b \sec (e+f x)}{24 (a+b)^4 f \sqrt {a+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.55 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 (a+b) (3 a-4 b+(a+8 b) \cos (2 (e+f x))) \csc ^4(e+f x)-2 \left (3 a^2-24 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1-\frac {a \sin ^2(e+f x)}{a+b}\right )\right ) \sec ^5(e+f x)}{96 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

[In]

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

[Out]

-1/96*((a + 2*b + a*Cos[2*(e + f*x)])*(3*(a + b)*(3*a - 4*b + (a + 8*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^4 - 2*(
3*a^2 - 24*a*b + 8*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, 1 - (a*Sin[e + f*x]^2)/(a + b)])*Sec[e + f*x]^5)/((a
+ b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(8114\) vs. \(2(210)=420\).

Time = 6.06 (sec) , antiderivative size = 8115, normalized size of antiderivative = 34.68

method	result	size
default	\(\text {Expression too large to display}\)	\(8115\)

[In]

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

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 644 vs. \(2 (210) = 420\).

Time = 0.56 (sec) , antiderivative size = 1307, normalized size of antiderivative = 5.59 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/48*(3*((3*a^4 - 24*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(3*a^4 - 27*a^3*b + 32*a^2*b^2 - 8*a*b^3)*cos(f*x
+ e)^6 + (3*a^4 - 36*a^3*b + 107*a^2*b^2 - 56*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 24*a*b^3 + 8*b^4 + 2
*(3*a^3*b - 27*a^2*b^2 + 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)*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*(3*a^4 - 21*
a^3*b - 16*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 117*a^3*b + 4*a^2*b^2 + 104*a*b^3 - 32*b^4)*cos(f*x +
e)^5 - (78*a^3*b - 71*a^2*b^2 - 61*a*b^3 + 88*b^4)*cos(f*x + e)^3 - 5*(11*a^2*b^2 + a*b^3 - 10*b^4)*cos(f*x +
e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*
b^5)*f*cos(f*x + e)^8 - 2*(a^7 + 4*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7
+ a^6*b - 9*a^5*b^2 - 25*a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b
^2 + 5*a^4*b^3 - 5*a^2*b^5 - 4*a*b^6 - b^7)*f*cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5
+ 5*a*b^6 + b^7)*f), 1/24*(3*((3*a^4 - 24*a^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(3*a^4 - 27*a^3*b + 32*a^2*b^2
- 8*a*b^3)*cos(f*x + e)^6 + (3*a^4 - 36*a^3*b + 107*a^2*b^2 - 56*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 -
24*a*b^3 + 8*b^4 + 2*(3*a^3*b - 27*a^2*b^2 + 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)*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)) + (3*(3*a^4 - 21*a^3*b - 16*a^2*b^2 + 8*a*
b^3)*cos(f*x + e)^7 - (15*a^4 - 117*a^3*b + 4*a^2*b^2 + 104*a*b^3 - 32*b^4)*cos(f*x + e)^5 - (78*a^3*b - 71*a^
2*b^2 - 61*a*b^3 + 88*b^4)*cos(f*x + e)^3 - 5*(11*a^2*b^2 + a*b^3 - 10*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)
^2 + b)/cos(f*x + e)^2))/((a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^8 - 2
*(a^7 + 4*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6*b - 9*a^5*b^2 - 25*
a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5*a^4*b^3 - 5*a^2*b^
5 - 4*a*b^6 - b^7)*f*cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(1/(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^(5/2)),x)

[Out]

\text{Hanged}