3.1.98 \(\int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [98]

Optimal. Leaf size=138 \[ -\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{5/2} f}-\frac {(5 a+2 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f} \]

[Out]

-3/8*a^2*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(5/2)/f-1/8*(5*a+2*b)*cot(f*x+e)*csc(f
*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)^2/f-1/4*cot(f*x+e)^3*csc(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/(a+b)/f

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4219, 481, 541, 12, 385, 213} \begin {gather*} -\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 f (a+b)^{5/2}}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 f (a+b)}-\frac {(5 a+2 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(8*(a + b)^(5/2)*f) - ((5*a + 2*b)*Cot
[e + f*x]*Csc[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(8*(a + b)^2*f) - (Cot[e + f*x]^3*Csc[e + f*x]*Sqrt[a + b*S
ec[e + f*x]^2])/(4*(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 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*} \int \frac {\csc ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^3 \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f}-\frac {\text {Subst}\left (\int \frac {-a-2 (2 a+b) x^2}{\left (-1+x^2\right )^2 \sqrt {a+b x^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) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f}-\frac {\text {Subst}\left (\int -\frac {3 a^2}{\left (-1+x^2\right ) \sqrt {a+b x^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) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f}+\frac {\left (3 a^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)^2 f}\\ &=-\frac {(5 a+2 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f}+\frac {\left (3 a^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)^2 f}\\ &=-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{5/2} f}-\frac {(5 a+2 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b)^2 f}-\frac {\cot ^3(e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 (a+b) f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.17, size = 78, normalized size = 0.57 \begin {gather*} -\frac {a^2 (a+2 b+a \cos (2 (e+f x))) \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \sec (e+f x)}{2 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(4982\) vs. \(2(122)=244\).
time = 0.16, size = 4983, normalized size = 36.11

method result size
default \(\text {Expression too large to display}\) \(4983\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/f*(-3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(
f*x+e)^2/(a+b)^(1/2))*a^4-3*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a
+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^2*b^2-6*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^3*b-3*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f
*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^2*b^2-6*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)
/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1
/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^3*b-3*
cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/si
n(f*x+e)^2/(a+b)^(1/2))*a^2*b^2-6*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^3*b-3*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^2*b^2+12*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln
(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+
cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^3*b+6*cos(f*x+e)^3*((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+
b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^2
*b^2+12*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+
e)-1))*a^3*b+6*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(c
os(f*x+e)-1))*a^2*b^2-10*(a+b)^(5/2)*a*b-3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a
+b)^(1/2)+b)/(cos(f*x+e)-1))*a^4+6*(a+b)^(5/2)*cos(f*x+e)^4*a^2-10*(a+b)^(5/2)*cos(f*x+e)^2*a^2+2*(a+b)^(5/2)*
cos(f*x+e)^2*a*b-6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)
-1))*a^3*b-3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*co
s(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a
^2*b^2-3*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e
)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^4-3*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^4-3*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(co
s(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^4-3*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+
cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(cos(f*x+e)-1))*a^4+6*cos(f*x+e)^3*((b+a*cos(f*x+e)^2
)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(cos(f*x+e)-1)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(
1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/sin(f*x+e)^2/(a+b)^(1/2))*a^4+6*c
os(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos
(f*x+e)*(a+b)^(1/2)+cos(f*x+e)*a+((b+a*cos(f*x+...

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 3.15, size = 515, normalized size = 3.73 \begin {gather*} \left [\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 7 \, 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}}}}{16 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, \frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) + {\left (3 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + 7 \, 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}}}}{8 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^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*(a^2 + a*b)*
cos(f*x + e)^3 - (5*a^2 + 7*a*b + 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^3 + 3*
a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*f), 1/8*(3*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^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*(a^2 + a*b)*cos(f*x + e)^3 - (5*a^2 +
7*a*b + 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*c
os(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**5/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (122) = 244\).
time = 1.01, size = 785, normalized size = 5.69 \begin {gather*} -\frac {\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} {\left (\frac {{\left (a + b\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{2} + 2 \, a b + b^{2}} + \frac {3 \, {\left (3 \, a + b\right )}}{a^{2} + 2 \, a b + b^{2}}\right )} - \frac {24 \, a^{2} \arctan \left (-\frac {\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}}{\sqrt {-a - b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a - b}} - \frac {12 \, a^{2} \log \left ({\left | {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} \sqrt {a + b} - a + b \right |}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} + \frac {4 \, {\left (2 \, {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )}^{3} {\left (2 \, a^{2} - 2 \, a b - b^{2}\right )} - 3 \, {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b} - 2 \, {\left (3 \, a^{3} + a^{2} b - 2 \, a b^{2}\right )} {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} + {\left (5 \, a^{3} + 11 \, a^{2} b + 7 \, a b^{2} + b^{3}\right )} \sqrt {a + b}\right )}}{{\left ({\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )}^{2} - a - b\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{64 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

-1/64*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x
 + 1/2*e)^2 + a + b)*((a + b)*tan(1/2*f*x + 1/2*e)^2/(a^2 + 2*a*b + b^2) + 3*(3*a + b)/(a^2 + 2*a*b + b^2)) -
24*a^2*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4
- 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/((a^2 + 2*a*b + b^2)*sqrt(-a
 - b)) - 12*a^2*log(abs((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x +
1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) - a + b))/((a^2 + 2*a
*b + b^2)*sqrt(a + b)) + 4*(2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*
f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^3*(2*a^2 - 2*a*b - b^2) - 3
*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*
f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2*(a^2 + 2*a*b + b^2)*sqrt(a + b) - 2*(3*a^3 + a^2*b - 2
*a*b^2)*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*t
an(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)) + (5*a^3 + 11*a^2*b + 7*a*b^2 + b^3)*sqrt(a + b))
/(((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/
2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - a - b)^2*(a^2 + 2*a*b + b^2)))/(f*sgn(cos(f*x + e)
))

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\sin \left (e+f\,x\right )}^5\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________