\(\int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx\) [163]

Optimal result

Integrand size = 27, antiderivative size = 203 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 a^{7/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c d^{3/2} \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]

[Out]

2*a^3*tan(f*x+e)/d/f/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c/f/(
a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-2*a^(7/2)*(c-d)^2*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/
(c+d)^(1/2))*tan(f*x+e)/c/d^(3/2)/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4025, 186, 65, 212, 214} \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=-\frac {2 a^{7/2} (c-d)^2 \tan (e+f x) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{c d^{3/2} f \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{7/2} \tan (e+f x) \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a \sec (e+f x)+a}} \]

[In]

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

[Out]

(2*a^3*Tan[e + f*x])/(d*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan
[e + f*x])/(c*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*a^(7/2)*(c - d)^2*ArcTanh[(Sqrt[d]*Sqr
t[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(c*d^(3/2)*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*
Sqrt[a + a*Sec[e + f*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 186

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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 214

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

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(a + b*x)^(m - 1/2)*((c
+ d*x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m - 1/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^2}{x \sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \left (\frac {a^2}{d \sqrt {a-a x}}+\frac {a^2}{c x \sqrt {a-a x}}-\frac {a^2 (c-d)^2}{c d \sqrt {a-a x} (c+d x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}-\frac {\left (a^4 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (a^4 (c-d)^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{c d f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}+\frac {\left (2 a^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (2 a^3 (c-d)^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+d-\frac {d x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{c d f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 a^{7/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c d^{3/2} \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 5.71 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.69 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\frac {\cos ^{\frac {3}{2}}(e+f x) (d+c \cos (e+f x)) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (1+\sec (e+f x)))^{5/2} \left (\frac {10 (c-d)^2 (c+3 d+2 c \cos (e+f x)) \csc \left (\frac {1}{2} (e+f x)\right ) \left (-\text {arctanh}\left (\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )+\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )}{d (c+d) \sqrt {\cos (e+f x)} \sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}}+\frac {20 (3 c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}-\frac {16 (c-d)^2 d (d+c \cos (e+f x)) \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\frac {2 d \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )}{c+d}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right )}{(c+d)^3 \cos ^{\frac {5}{2}}(e+f x)}+10 c \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right )-\frac {2 \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )\right )}{40 c^2 f (c+d \sec (e+f x))} \]

[In]

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

[Out]

(Cos[e + f*x]^(3/2)*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^5*(a*(1 + Sec[e + f*x]))^(5/2)*((10*(c - d)^2*(c + 3
*d + 2*c*Cos[e + f*x])*Csc[(e + f*x)/2]*(-ArcTanh[Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]] + Sqrt[-((d*(-1 +
Sec[e + f*x]))/(c + d))]))/(d*(c + d)*Sqrt[Cos[e + f*x]]*Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]) + (20*(3*c
- d)*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x]] - (16*(c - d)^2*d*(d + c*Cos[e + f*x])*Hypergeometric2F1[2, 5/2, 7/2
, (-2*d*Sec[e + f*x]*Sin[(e + f*x)/2]^2)/(c + d)]*Sin[(e + f*x)/2]^3)/((c + d)^3*Cos[e + f*x]^(5/2)) + 10*c*(S
qrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]] - (2*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x]])))/(40*c^2*f*(c + d*Sec[e +
f*x]))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1401\) vs. \(2(173)=346\).

Time = 23.45 (sec) , antiderivative size = 1402, normalized size of antiderivative = 6.91

method	result	size
default	\(\text {Expression too large to display}\)	\(1402\)

[In]

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

[Out]

1/2/f*a^2/c/((c+d)*(c-d))^(1/2)/d/(d/(c-d))^(1/2)*(2*((c+d)*(c-d))^(1/2)*2^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^
2-1)^(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(-cot(f*x+e)+csc(f*x+e)))*(d/(c-d))^(1/2)*d
-2^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-
d))^(1/2)*c-2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)
+csc(f*x+e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*c^2+2*2^(1/2)*
((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)
*c-2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+
e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*c*d-2^(1/2)*((1-cos(f*x
+e))^2*csc(f*x+e)^2-1)^(1/2)*ln(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c-2^(1/2)*
(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))-c+d)/(-
c*(-cot(f*x+e)+csc(f*x+e))+(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*d^2+2^(1/2)*((1-cos(f*x+e))^2*csc(
f*x+e)^2-1)^(1/2)*ln(-2*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c+2^(1/2)*(d/(c-d))^
(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))+c-d)/(c*(-cot(f*x
+e)+csc(f*x+e))-(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*c^2-2*2^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-
1)^(1/2)*ln(-2*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c+2^(1/2)*(d/(c-d))^(1/2)*((1
-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))+c-d)/(c*(-cot(f*x+e)+csc(f
*x+e))-(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*c*d+2^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*ln
(-2*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c+2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e)
)^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))+c-d)/(c*(-cot(f*x+e)+csc(f*x+e))-(-co
t(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))*d^2+4*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c*(-cot(f*x+e)+csc(f*x+
e)))*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 2.93 (sec) , antiderivative size = 1140, normalized size of antiderivative = 5.62 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[(2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2
*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x
+ e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + (a^2*d*cos(f*x + e) + a^2*d)*sqrt(-a)*log((2*a*cos(f*x
+ e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(c
os(f*x + e) + 1)))/(c*d*f*cos(f*x + e) + c*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)
- 2*(a^2*d*cos(f*x + e) + a^2*d)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)
*sin(f*x + e))) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(-a/(c*d
+ d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x +
e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e)
+ d)))/(c*d*f*cos(f*x + e) + c*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + 2*(a^2*c^
2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqr
t(a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*sin(f*x + e))) + (a^2*d*cos(f*x + e)
+ a^2*d)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*si
n(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/(c*d*f*cos(f*x + e) + c*d*f), 2*(a^2*c*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sin(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x
+ e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqrt(a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x
+ e)/(a*sin(f*x + e))) - (a^2*d*cos(f*x + e) + a^2*d)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c
os(f*x + e)/(sqrt(a)*sin(f*x + e))))/(c*d*f*cos(f*x + e) + c*d*f)]

Sympy [F]

\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}{c + d \sec {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

Integral((a*(sec(e + f*x) + 1))**(5/2)/(c + d*sec(e + f*x)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((a + a/cos(e + f*x))^(5/2)/(c + d/cos(e + f*x)),x)

[Out]

int((a + a/cos(e + f*x))^(5/2)/(c + d/cos(e + f*x)), x)