3.105 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}-\frac{7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3 \sec (e+f x)+a^3\right ) (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{3/2}} \]

[Out]

(-7*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(16*Sqrt[2]*a^3*c^(3/2)*f) - (7*Tan[e +
f*x])/(16*a^3*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(5*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(
3/2)) + (7*Tan[e + f*x])/(30*a*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)) + (7*Tan[e + f*x])/(12*f*(
a^3 + a^3*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.477317, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac{7 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}-\frac{7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3 \sec (e+f x)+a^3\right ) (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(-7*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(16*Sqrt[2]*a^3*c^(3/2)*f) - (7*Tan[e +
f*x])/(16*a^3*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(5*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(
3/2)) + (7*Tan[e + f*x])/(30*a*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)) + (7*Tan[e + f*x])/(12*f*(
a^3 + a^3*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2))

Rule 3960

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

Rule 3796

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

Rule 3795

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

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])

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}} \, dx &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx}{10 a}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{7 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2}} \, dx}{12 a^2}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac{7 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{8 a^3}\\ &=-\frac{7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac{7 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{32 a^3 c}\\ &=-\frac{7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{2 c+x^2} \, dx,x,\frac{c \tan (e+f x)}{\sqrt{c-c \sec (e+f x)}}\right )}{16 a^3 c f}\\ &=-\frac{7 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}-\frac{7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac{7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 6.44551, size = 398, normalized size = 1.88 \[ \frac{\sin ^3\left (\frac{e}{2}+\frac{f x}{2}\right ) \cos ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^5(e+f x) \left (\frac{278 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )}{15 f}-\frac{278 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right )}{15 f}+\frac{2 \sec ^5\left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f}-\frac{56 \sec ^3\left (\frac{e}{2}+\frac{f x}{2}\right )}{15 f}+\frac{242 \sec \left (\frac{e}{2}+\frac{f x}{2}\right )}{15 f}-\frac{\cot \left (\frac{e}{2}\right ) \csc \left (\frac{e}{2}+\frac{f x}{2}\right )}{f}+\frac{\csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{f}\right )}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{3/2}}+\frac{7 e^{-\frac{1}{2} i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sin ^3\left (\frac{e}{2}+\frac{f x}{2}\right ) \cos ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^{\frac{9}{2}}(e+f x) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(7*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))
/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]*Cos[e/2 + (f*x)/2]^6*Sec[e + f*x]^(9/2)*Sin[e/2 + (f*x)/2]^3)/(E^((I
/2)*(e + f*x))*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2)) + (Cos[e/2 + (f*x)/2]^6*Sec[e + f*x]^5*((-
278*Cos[e/2]*Cos[(f*x)/2])/(15*f) - (Cot[e/2]*Csc[e/2 + (f*x)/2])/f + (242*Sec[e/2 + (f*x)/2])/(15*f) - (56*Se
c[e/2 + (f*x)/2]^3)/(15*f) + (2*Sec[e/2 + (f*x)/2]^5)/(5*f) + (Csc[e/2]*Csc[e/2 + (f*x)/2]^2*Sin[(f*x)/2])/f +
(278*Sin[e/2]*Sin[(f*x)/2])/(15*f))*Sin[e/2 + (f*x)/2]^3)/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.27, size = 370, normalized size = 1.8 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{120\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 15\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{9/2}+15\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{9/2}+15\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{7/2}\cos \left ( fx+e \right ) -15\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{7/2}-21\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}+21\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}+35\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}-35\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}-105\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) -105\,\cos \left ( fx+e \right ) \arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) +105\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}+105\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2),x)

[Out]

1/120/a^3/f*(-1+cos(f*x+e))^2*(15*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+15*(-2*cos(f*x+e)/(1+cos(f*x
+e)))^(9/2)+15*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)-15*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)-21*cos(
f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+21*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+35*cos(f*x+e)*(-2*cos(f*x+
e)/(1+cos(f*x+e)))^(3/2)-35*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-105*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(
f*x+e)-105*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))+105*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+
105*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)^3/(-2*cos(
f*x+e)/(1+cos(f*x+e)))^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )}^{3}{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^3*(-c*sec(f*x + e) + c)^(3/2)), x)

________________________________________________________________________________________

Fricas [A]  time = 0.6973, size = 1242, normalized size = 5.86 \begin{align*} \left [-\frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt{-c} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{-c} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} +{\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \,{\left (139 \, \cos \left (f x + e\right )^{4} + 21 \, \cos \left (f x + e\right )^{3} - 175 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{960 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )}, \frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt{c} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \,{\left (139 \, \cos \left (f x + e\right )^{4} + 21 \, \cos \left (f x + e\right )^{3} - 175 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{480 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/960*(105*sqrt(2)*(cos(f*x + e)^3 + cos(f*x + e)^2 - cos(f*x + e) - 1)*sqrt(-c)*log((2*sqrt(2)*(cos(f*x + e
)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((
cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) + 4*(139*cos(f*x + e)^4 + 21*cos(f*x + e)^3 - 175*cos(f*x + e)^2
- 105*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^2*f*cos(f*x + e)^3 + a^3*c^2*f*cos(f*x +
e)^2 - a^3*c^2*f*cos(f*x + e) - a^3*c^2*f)*sin(f*x + e)), 1/480*(105*sqrt(2)*(cos(f*x + e)^3 + cos(f*x + e)^2
- cos(f*x + e) - 1)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(
f*x + e)))*sin(f*x + e) - 2*(139*cos(f*x + e)^4 + 21*cos(f*x + e)^3 - 175*cos(f*x + e)^2 - 105*cos(f*x + e))*s
qrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^2*f*cos(f*x + e)^3 + a^3*c^2*f*cos(f*x + e)^2 - a^3*c^2*f*cos(
f*x + e) - a^3*c^2*f)*sin(f*x + e))]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.75236, size = 257, normalized size = 1.21 \begin{align*} -\frac{\sqrt{2}{\left (105 \, \sqrt{c} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right ) - \frac{15 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}} - \frac{2 \,{\left (3 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} c^{8} - 10 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{9} + 45 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{10}\right )}}{c^{10}}\right )}}{480 \, a^{3} c^{2} f \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/480*sqrt(2)*(105*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) - 15*sqrt(c*tan(1/2*f*x + 1/2*e
)^2 - c)/tan(1/2*f*x + 1/2*e)^2 - 2*(3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2)*c^8 - 10*(c*tan(1/2*f*x + 1/2*e)^2
- c)^(3/2)*c^9 + 45*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^10)/c^10)/(a^3*c^2*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)
*sgn(tan(1/2*f*x + 1/2*e)))