3.182 \(\int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx\)

Optimal. Leaf size=179 \[ -\frac {2 g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{\sqrt {a} c f}+\frac {g^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {g^2 \cot (e+f x) \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}{a c f} \]

[Out]

-2*g^(5/2)*arctanh(a^(1/2)*g^(1/2)*tan(f*x+e)/(g*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2))/c/f/a^(1/2)+1/2*g^(
5/2)*arctanh(1/2*a^(1/2)*g^(1/2)*tan(f*x+e)*2^(1/2)/(g*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2))/c/f*2^(1/2)/a
^(1/2)+g^2*cot(f*x+e)*(g*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2)/a/c/f

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Rubi [A]  time = 0.36, antiderivative size = 242, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3964, 98, 157, 63, 217, 203, 93, 205} \[ \frac {2 g^{5/2} \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {g^{5/2} \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} \sqrt {c} f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {g^2 \tan (e+f x) \sqrt {g \sec (e+f x)}}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((g^2*Sqrt[g*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x]))) + (2*g^(5/2)*ArcT
an[(Sqrt[c]*Sqrt[g*Sec[e + f*x]])/(Sqrt[g]*Sqrt[c - c*Sec[e + f*x]])]*Tan[e + f*x])/(Sqrt[c]*f*Sqrt[a + a*Sec[
e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (g^(5/2)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[g*Sec[e + f*x]])/(Sqrt[g]*Sqrt[c -
c*Sec[e + f*x]])]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c]*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3964

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

Rubi steps

\begin {align*} \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx &=-\frac {(a c g \tan (e+f x)) \operatorname {Subst}\left (\int \frac {(g x)^{3/2}}{(a+a x) (c-c x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {g^2 \sqrt {g \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))}+\frac {(g \tan (e+f x)) \operatorname {Subst}\left (\int \frac {\frac {1}{2} a c g^2+a c g^2 x}{\sqrt {g x} (a+a x) \sqrt {c-c x}} \, dx,x,\sec (e+f x)\right )}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {g^2 \sqrt {g \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))}+\frac {\left (g^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {g x} \sqrt {c-c x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\left (a g^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {g x} (a+a x) \sqrt {c-c x}} \, dx,x,\sec (e+f x)\right )}{2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {g^2 \sqrt {g \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))}+\frac {\left (2 g^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {c x^2}{g}}} \, dx,x,\sqrt {g \sec (e+f x)}\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\left (a g^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{a g+2 a c x^2} \, dx,x,\frac {\sqrt {g \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}}\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {g^2 \sqrt {g \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))}-\frac {g^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right ) \tan (e+f x)}{\sqrt {2} \sqrt {c} f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\left (2 g^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{g}} \, dx,x,\frac {\sqrt {g \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}}\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {g^2 \sqrt {g \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))}+\frac {2 g^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right ) \tan (e+f x)}{\sqrt {c} f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {g^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right ) \tan (e+f x)}{\sqrt {2} \sqrt {c} f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.29, size = 328, normalized size = 1.83 \[ -\frac {\sin ^3(e+f x) \sqrt {\sec (e+f x)+1} (g \sec (e+f x))^{5/2} \left (8 \sqrt {\sec (e+f x)} \sqrt {\sec (e+f x)+1}+\sqrt {\tan ^2(e+f x)} \left (16 \log (\sec (e+f x)+1)-16 \log \left (\sec ^{\frac {3}{2}}(e+f x)+\sqrt {\sec (e+f x)}+\sqrt {\tan ^2(e+f x)} \sqrt {\sec (e+f x)+1}\right )+\sqrt {2} \left (\log \left (-3 \sec ^2(e+f x)-2 \sec (e+f x)-2 \sqrt {2} \sqrt {\tan ^2(e+f x)} \sqrt {\sec (e+f x)+1} \sqrt {\sec (e+f x)}+1\right )-\log \left (-3 \sec ^2(e+f x)-2 \sec (e+f x)+2 \sqrt {2} \sqrt {\tan ^2(e+f x)} \sqrt {\sec (e+f x)+1} \sqrt {\sec (e+f x)}+1\right )\right )\right )\right )}{8 c f (\cos (e+f x)-1) (\cos (e+f x)+1)^2 (\sec (e+f x)-1) \sec ^{\frac {5}{2}}(e+f x) \sqrt {a (\sec (e+f x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/8*((g*Sec[e + f*x])^(5/2)*Sqrt[1 + Sec[e + f*x]]*Sin[e + f*x]^3*(8*Sqrt[Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]
] + (16*Log[1 + Sec[e + f*x]] - 16*Log[Sqrt[Sec[e + f*x]] + Sec[e + f*x]^(3/2) + Sqrt[1 + Sec[e + f*x]]*Sqrt[T
an[e + f*x]^2]] + Sqrt[2]*(Log[1 - 2*Sec[e + f*x] - 3*Sec[e + f*x]^2 - 2*Sqrt[2]*Sqrt[Sec[e + f*x]]*Sqrt[1 + S
ec[e + f*x]]*Sqrt[Tan[e + f*x]^2]] - Log[1 - 2*Sec[e + f*x] - 3*Sec[e + f*x]^2 + 2*Sqrt[2]*Sqrt[Sec[e + f*x]]*
Sqrt[1 + Sec[e + f*x]]*Sqrt[Tan[e + f*x]^2]]))*Sqrt[Tan[e + f*x]^2]))/(c*f*(-1 + Cos[e + f*x])*(1 + Cos[e + f*
x])^2*(-1 + Sec[e + f*x])*Sec[e + f*x]^(5/2)*Sqrt[a*(1 + Sec[e + f*x])])

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fricas [A]  time = 0.60, size = 569, normalized size = 3.18 \[ \left [\frac {\sqrt {2} a g^{2} \sqrt {\frac {g}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - g \cos \left (f x + e\right )^{2} + 2 \, g \cos \left (f x + e\right ) + 3 \, g}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 2 \, a g^{2} \sqrt {\frac {g}{a}} \log \left (\frac {g \cos \left (f x + e\right )^{3} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, g \cos \left (f x + e\right )^{2} + 8 \, g}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2}}\right ) \sin \left (f x + e\right ) + 4 \, g^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, a c f \sin \left (f x + e\right )}, -\frac {\sqrt {2} a g^{2} \sqrt {-\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{g \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, a g^{2} \sqrt {-\frac {g}{a}} \arctan \left (\frac {2 \, \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{g \cos \left (f x + e\right )^{2} - g \cos \left (f x + e\right ) - 2 \, g}\right ) \sin \left (f x + e\right ) - 2 \, g^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, a c f \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*a*g^2*sqrt(g/a)*log((2*sqrt(2)*sqrt(g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x
+ e))*cos(f*x + e)*sin(f*x + e) - g*cos(f*x + e)^2 + 2*g*cos(f*x + e) + 3*g)/(cos(f*x + e)^2 + 2*cos(f*x + e)
+ 1))*sin(f*x + e) + 2*a*g^2*sqrt(g/a)*log((g*cos(f*x + e)^3 + 4*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt(g/a)*s
qrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) - 7*g*cos(f*x + e)^2 + 8*g)/(cos(f*x
+ e)^3 + cos(f*x + e)^2))*sin(f*x + e) + 4*g^2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*co
s(f*x + e))/(a*c*f*sin(f*x + e)), -1/2*(sqrt(2)*a*g^2*sqrt(-g/a)*arctan(sqrt(2)*sqrt(-g/a)*sqrt((a*cos(f*x + e
) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)/(g*sin(f*x + e)))*sin(f*x + e) + 2*a*g^2*sqrt(-g/a)*arc
tan(2*sqrt(-g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(g*cos
(f*x + e)^2 - g*cos(f*x + e) - 2*g))*sin(f*x + e) - 2*g^2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f
*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\left (g \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (c \sec \left (f x + e\right ) - c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 2.95, size = 294, normalized size = 1.64 \[ \frac {\left (\cos \left (f x +e \right ) \sqrt {2}\, \arcsinh \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\sqrt {2}\, \arcsinh \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \sin \left (f x +e \right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}+2 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right )}{2}\right ) \cos \left (f x +e \right )+2 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (-\cos \left (f x +e \right )-1+\sin \left (f x +e \right )\right )}{2}\right ) \cos \left (f x +e \right )-2 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right )}{2}\right )-2 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (-\cos \left (f x +e \right )-1+\sin \left (f x +e \right )\right )}{2}\right )\right ) \left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 c f \left (\frac {1}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{6} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x)

[Out]

1/2/c/f*(cos(f*x+e)*2^(1/2)*arcsinh((-1+cos(f*x+e))/sin(f*x+e))-2^(1/2)*arcsinh((-1+cos(f*x+e))/sin(f*x+e))+2*
sin(f*x+e)*(1/(1+cos(f*x+e)))^(1/2)+2*arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))*cos(f*x+
e)+2*arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)-1+sin(f*x+e)))*cos(f*x+e)-2*arctanh(1/2*(1/(1+cos(f*x+e
)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))-2*arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)-1+sin(f*x+e))))*(g/co
s(f*x+e))^(5/2)*(-1+cos(f*x+e))^2*cos(f*x+e)^3*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/(1/(1+cos(f*x+e)))^(5/2)/si
n(f*x+e)^6/a

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maxima [B]  time = 1.22, size = 1400, normalized size = 7.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*g^2*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) - 4*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 4*g^2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (sqrt(2)*g^2*cos(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + sqrt(2)*g^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(
2)*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + sqrt(2)*g^2)*log(2*cos(1/4*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sqrt(2)*cos(1/4*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2)
+ (sqrt(2)*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sqrt(2)*g^2*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(2)*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + sqrt(2)*g^
2)*log(2*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*sqrt(2)*sin(1/4*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 2) - (sqrt(2)*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 +
sqrt(2)*g^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(2)*g^2*cos(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + sqrt(2)*g^2)*log(2*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(
1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2) + (sqrt(2)*g^2*cos(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sqrt(2)*g^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*s
qrt(2)*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + sqrt(2)*g^2)*log(2*cos(1/4*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(2)*cos(1/4*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
2) + (g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + g^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e)))^2 - 2*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + g^2)*log(cos(1/4*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/4*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + g^2
*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*g^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) + g^2)*log(cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/4*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e)))^2 - 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(g)/((sqrt(2)*c*cos(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sqrt(2)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
))^2 - 2*sqrt(2)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + sqrt(2)*c)*sqrt(a)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g/cos(e + f*x))^(5/2)/((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))),x)

[Out]

int((g/cos(e + f*x))^(5/2)/((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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