\(\int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [570]

Optimal result

Integrand size = 23, antiderivative size = 427 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}+\frac {2 \left (16 a^4+12 a^3 b-16 a^2 b^2-9 a b^3-b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]

[Out]

8/3*a*(4*a^4-7*a^2*b^2+2*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*
(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^5/(a+b)^(3/2)/d+2/3*(16*a^4+12*a^3*b-16*a^
2*b^2-9*a*b^3-b^4)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+
c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^4/(a+b)^(3/2)/d-2/3*a^2*sec(d*x+c)^2*tan(d*x+c)/b/(a^
2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+4/3*a^3*(3*a^2-5*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/3*
(2*a^2-b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3930, 4175, 4167, 4090, 3917, 4089} \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^5 d (a-b) (a+b)^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (16 a^4+12 a^3 b-16 a^2 b^2-9 a b^3-b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b^4 d (a-b) (a+b)^{3/2}} \]

[In]

Int[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(8*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/
(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*(a - b)*b^5*(a + b)^
(3/2)*d) + (2*(16*a^4 + 12*a^3*b - 16*a^2*b^2 - 9*a*b^3 - b^4)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c
+ d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a -
b))])/(3*(a - b)*b^4*(a + b)^(3/2)*d) - (2*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^(3/2)) + (4*a^3*(3*a^2 - 5*b^2)*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(2*
a^2 - b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(a^2 - b^2)*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3930

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)
*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist
[d^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b
*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4175

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e
+ f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 a^2-\frac {3}{2} a b \sec (c+d x)-\frac {3}{2} \left (2 a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\sec (c+d x) \left (\frac {1}{2} a^2 b \left (3 a^2-5 b^2\right )+\frac {1}{2} a \left (6 a^4-11 a^2 b^2+3 b^4\right ) \sec (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2} \\ & = -\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (\frac {3}{8} b^2 \left (4 a^4-7 a^2 b^2-b^4\right )+\frac {3}{2} a b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2} \\ & = -\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (16 a^4+12 a^3 b-16 a^2 b^2-9 a b^3-b^4\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b^3 (a+b)^2}-\frac {\left (4 a \left (4 a^4-7 a^2 b^2+2 b^4\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2} \\ & = \frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}+\frac {2 \left (16 a^4+12 a^3 b-16 a^2 b^2-9 a b^3-b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 13.95 (sec) , antiderivative size = 578, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {4 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (4 a \left (4 a^5+4 a^4 b-7 a^3 b^2-7 a^2 b^3+2 a b^4+2 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b \left (-16 a^5-4 a^4 b+28 a^3 b^2+7 a^2 b^3-8 a b^4+b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+2 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}}+\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sin (c+d x)}{3 b^4 \left (-a^2+b^2\right )^2}-\frac {2 a^3 \sin (c+d x)}{3 b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (-7 a^5 \sin (c+d x)+11 a^3 b^2 \sin (c+d x)\right )}{3 b^3 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}+\frac {2 \tan (c+d x)}{3 b^3}\right )}{d (a+b \sec (c+d x))^{5/2}} \]

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(4*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(4*a*(4*a^5 + 4*a^4*b - 7*a
^3*b^2 - 7*a^2*b^3 + 2*a*b^4 + 2*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)
*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-16*a^5 - 4*a^4*b + 28*a^3*b^2
+ 7*a^2*b^3 - 8*a*b^4 + b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co
s[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Cos[c + d
*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(a^2 - b^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]
*(a + b*Sec[c + d*x])^(5/2)) + ((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-8*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Sin[c
+ d*x])/(3*b^4*(-a^2 + b^2)^2) - (2*a^3*Sin[c + d*x])/(3*b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) - (2*(-7*a^
5*Sin[c + d*x] + 11*a^3*b^2*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (2*Tan[c + d*x])/(3*b
^3)))/(d*(a + b*Sec[c + d*x])^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(5025\) vs. \(2(393)=786\).

Time = 14.62 (sec) , antiderivative size = 5026, normalized size of antiderivative = 11.77

method	result	size
default	\(\text {Expression too large to display}\)	\(5026\)

[In]

int(sec(d*x+c)^5/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

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

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^5/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*
x + c) + a^3), x)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**5/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**5/(a + b*sec(c + d*x))**(5/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

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

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(b*sec(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))^(5/2)), x)