3.163 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx\)
Optimal. Leaf size=203 \[ -\frac {2 a^{7/2} (c-d)^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{c d^{3/2} f \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{7/2} \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a \sec (e+f x)+a}} \]
[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)
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Rubi [A] time = 0.23, antiderivative size = 203, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used =
{3940, 180, 63, 206, 208} \[ -\frac {2 a^{7/2} (c-d)^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{c d^{3/2} f \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{7/2} \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
[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 63
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 180
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 3940
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*} \int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tanh ^{-1}\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 \tanh ^{-1}\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*}
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Mathematica [C] time = 6.55, size = 343, normalized size = 1.69 \[ \frac {\cos ^{\frac {3}{2}}(e+f x) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (\sec (e+f x)+1))^{5/2} (c \cos (e+f x)+d) \left (-\frac {16 d (c-d)^2 \sin ^3\left (\frac {1}{2} (e+f x)\right ) (c \cos (e+f x)+d) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {2 d \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )}{c+d}\right )}{(c+d)^3 \cos ^{\frac {5}{2}}(e+f x)}+\frac {20 (3 c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}+\frac {10 (c-d)^2 \csc \left (\frac {1}{2} (e+f x)\right ) (2 c \cos (e+f x)+c+3 d) \left (\sqrt {-\frac {d (\sec (e+f x)-1)}{c+d}}-\tanh ^{-1}\left (\sqrt {-\frac {d (\sec (e+f x)-1)}{c+d}}\right )\right )}{d (c+d) \sqrt {\cos (e+f x)} \sqrt {-\frac {d (\sec (e+f x)-1)}{c+d}}}+10 c \left (\sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right )-\frac {2 \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )\right )}{40 c^2 f (c+d \sec (e+f x))} \]
Warning: Unable to verify antiderivative.
[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]))
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fricas [A] time = 5.11, size = 1140, normalized size = 5.62 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e)),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2*(sqrt(2)*a^3*sqr
t(-a*tan(1/2*(f*x+exp(1)))^2+a)*sign(cos(f*x+exp(1)))*tan(1/2*(f*x+exp(1)))/d/(-a*tan(1/2*(f*x+exp(1)))^2+a)+2
*(1/512*(-256*sqrt(2)*a^3*sqrt(-a)*c^2*sign(cos(f*x+exp(1)))-256*sqrt(2)*a^3*sqrt(-a)*d^2*sign(cos(f*x+exp(1))
)+512*sqrt(2)*a^3*sqrt(-a)*c*d*sign(cos(f*x+exp(1))))*atan(1/2*(c*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)
*tan(1/2*(f*x+exp(1))))^2-d*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+a*c+3*a*d)/s
qrt(2)/sqrt(-d^2-c*d)/a)/sqrt(2)/sqrt(-d^2-c*d)/a/c/d+1/4*a^2*sqrt(-a)*sign(cos(f*x+exp(1)))*ln(abs((sqrt(-a*t
an(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+a*(2*sqrt(2)-3)))/c-1/4*a^2*sqrt(-a)*sign(cos(f*x+
exp(1)))*ln(abs((sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+a*(-2*sqrt(2)-3)))/c))/f
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maple [B] time = 1.50, size = 1487, normalized size = 7.33 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e)),x)
[Out]
-1/2/f*(2*(d/(c-d))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*((c+
d)*(c-d))^(1/2)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)-ln(2*((d/(c-d))^(1/2)*(-2*cos(f*x+e)
/(1+cos(f*x+e)))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin
(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin
(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*c^2*sin(f*x+e)+2*ln(2*((d
/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)
/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)-((c+d)*(c-d))^(1/
2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/
2)*c*d*sin(f*x+e)-ln(2*((d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(
c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-c*sin(f*x+e)+d*si
n(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(
f*x+e)/(1+cos(f*x+e)))^(1/2)*d^2*sin(f*x+e)+ln(-2*((d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2
)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)-c*sin(f*x+e)+d*sin(f*
x+e)-((c+d)*(c-d))^(1/2)*cos(f*x+e)+((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*
x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*c^2*sin(f*x+e)-2*ln(-2*((d/(c-d))^(1/2)*(-2*cos(f*x+e)
/(1+cos(f*x+e)))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*d*sin
(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)-((c+d)*(c-d))^(1/2)*cos(f*x+e)+((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin
(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*c*d*sin(f*x+e)+ln(-2*((d/
(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)-((c+d)*(c-d))^(1/2)*cos(f*x+e)+((c+d)*(c-d))^(1/2
))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2
)*d^2*sin(f*x+e)+4*(d/(c-d))^(1/2)*((c+d)*(c-d))^(1/2)*c*cos(f*x+e)-4*(d/(c-d))^(1/2)*((c+d)*(c-d))^(1/2)*c)*(
a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)*a^2/c/((c+d)*(c-d))^(1/2)/d/(d/(c-d))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}{c + d \sec {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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