3.106 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=246 \[ -\frac{63 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}-\frac{63 \tan (e+f x)}{128 a^3 c f (c-c \sec (e+f x))^{3/2}}-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3 \sec (e+f x)+a^3\right ) (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{5/2}} \]

[Out]

(-63*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(128*Sqrt[2]*a^3*c^(5/2)*f) - (21*Tan[
e + f*x])/(32*a^3*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(5*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x]
)^(5/2)) + (3*Tan[e + f*x])/(10*a*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2)) + (21*Tan[e + f*x])/(20
*f*(a^3 + a^3*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) - (63*Tan[e + f*x])/(128*a^3*c*f*(c - c*Sec[e + f*x])^
(3/2))

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Rubi [A]  time = 0.532018, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac{63 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}-\frac{63 \tan (e+f x)}{128 a^3 c f (c-c \sec (e+f x))^{3/2}}-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3 \sec (e+f x)+a^3\right ) (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(-63*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(128*Sqrt[2]*a^3*c^(5/2)*f) - (21*Tan[
e + f*x])/(32*a^3*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(5*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x]
)^(5/2)) + (3*Tan[e + f*x])/(10*a*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2)) + (21*Tan[e + f*x])/(20
*f*(a^3 + a^3*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) - (63*Tan[e + f*x])/(128*a^3*c*f*(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))^{5/2}} \, dx &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{9 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}} \, dx}{10 a}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}} \, dx}{20 a^2}\\ &=\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac{21 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{5/2}} \, dx}{8 a^3}\\ &=-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac{63 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{64 a^3 c}\\ &=-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{63 \tan (e+f x)}{128 a^3 c f (c-c \sec (e+f x))^{3/2}}+\frac{63 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{256 a^3 c^2}\\ &=-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{63 \tan (e+f x)}{128 a^3 c f (c-c \sec (e+f x))^{3/2}}-\frac{63 \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 )}{128 a^3 c^2 f}\\ &=-\frac{63 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}-\frac{21 \tan (e+f x)}{32 a^3 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{5/2}}+\frac{3 \tan (e+f x)}{10 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{21 \tan (e+f x)}{20 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{63 \tan (e+f x)}{128 a^3 c f (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 6.63869, size = 468, normalized size = 1.9 \[ \frac{\sin ^5\left (\frac{e}{2}+\frac{f x}{2}\right ) \cos ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^6(e+f x) \left (-\frac{257 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )}{10 f}+\frac{257 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right )}{10 f}-\frac{2 \sec ^5\left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f}+\frac{22 \sec ^3\left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f}-\frac{124 \sec \left (\frac{e}{2}+\frac{f x}{2}\right )}{5 f}-\frac{\cot \left (\frac{e}{2}\right ) \csc ^3\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f}+\frac{23 \cot \left (\frac{e}{2}\right ) \csc \left (\frac{e}{2}+\frac{f x}{2}\right )}{4 f}+\frac{\csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc ^4\left (\frac{e}{2}+\frac{f x}{2}\right )}{2 f}-\frac{23 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{4 f}\right )}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{5/2}}-\frac{63 e^{-\frac{1}{2} i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sin ^5\left (\frac{e}{2}+\frac{f x}{2}\right ) \cos ^6\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^{\frac{11}{2}}(e+f x) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{4 f (a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(-63*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]^(11/2)*Sin[e/2 + (f*x)/2]^5)/(4*
E^((I/2)*(e + f*x))*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(5/2)) + (Cos[e/2 + (f*x)/2]^6*Sec[e + f*x]^
6*((257*Cos[e/2]*Cos[(f*x)/2])/(10*f) + (23*Cot[e/2]*Csc[e/2 + (f*x)/2])/(4*f) - (Cot[e/2]*Csc[e/2 + (f*x)/2]^
3)/(2*f) - (124*Sec[e/2 + (f*x)/2])/(5*f) + (22*Sec[e/2 + (f*x)/2]^3)/(5*f) - (2*Sec[e/2 + (f*x)/2]^5)/(5*f) -
(23*Csc[e/2]*Csc[e/2 + (f*x)/2]^2*Sin[(f*x)/2])/(4*f) + (Csc[e/2]*Csc[e/2 + (f*x)/2]^4*Sin[(f*x)/2])/(2*f) -
(257*Sin[e/2]*Sin[(f*x)/2])/(10*f))*Sin[e/2 + (f*x)/2]^5)/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(5/2))

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Maple [B]  time = 0.278, size = 631, normalized size = 2.6 \begin{align*} \text{result too large to display} \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))^(5/2),x)

[Out]

-1/160/a^3/f*(-1+cos(f*x+e))^3*(45*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(11/2)*cos(f*x+e)^2+20*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(11/2)*cos(f*x+e)+35*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)-25*(-2*cos(f*x+e)/(1+cos(f*x+
e)))^(11/2)-70*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)-45*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x
+e)^2+35*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+90*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)+63*(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(5/2)*cos(f*x+e)^2-45*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)-126*cos(f*x+e)*(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(5/2)-105*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+63*(-2*cos(f*x+e)/(1+cos(f*x+e)))^
(5/2)+210*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+315*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2
)+315*cos(f*x+e)^2*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)))^(3/2)-630
*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)-630*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))
+315*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+315*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e
))/cos(f*x+e))^(5/2)/sin(f*x+e)^5/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)

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Maxima [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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.835112, size = 1207, normalized size = 4.91 \begin{align*} \left [-\frac{315 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 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 (257 \, \cos \left (f x + e\right )^{5} - 354 \, \cos \left (f x + e\right )^{4} - 588 \, \cos \left (f x + e\right )^{3} + 210 \, \cos \left (f x + e\right )^{2} + 315 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2560 \,{\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} f\right )} \sin \left (f x + e\right )}, \frac{315 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 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 (257 \, \cos \left (f x + e\right )^{5} - 354 \, \cos \left (f x + e\right )^{4} - 588 \, \cos \left (f x + e\right )^{3} + 210 \, \cos \left (f x + e\right )^{2} + 315 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{1280 \,{\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

[-1/2560*(315*sqrt(2)*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 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*(257*cos(f*x + e)^5 - 354*cos(f*x + e)^4 - 588*cos(f*x + e)^3 + 210*cos(
f*x + e)^2 + 315*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^3*f*cos(f*x + e)^4 - 2*a^3*c^3
*f*cos(f*x + e)^2 + a^3*c^3*f)*sin(f*x + e)), 1/1280*(315*sqrt(2)*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 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*(257*cos(f*x + e)^5 - 354*cos(f*x + e)^4 - 588*cos(f*x + e)^3 + 210*cos(f*x + e)^2 + 315*cos(f*x + e))*sqrt
((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^3*f*cos(f*x + e)^4 - 2*a^3*c^3*f*cos(f*x + e)^2 + a^3*c^3*f)*sin(
f*x + e))]

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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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.89017, size = 298, normalized size = 1.21 \begin{align*} -\frac{\sqrt{2}{\left (315 \, \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{5 \,{\left (17 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c + 15 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2}\right )}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}} - \frac{8 \,{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} c^{8} - 5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{9} + 30 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{10}\right )}}{c^{10}}\right )}}{1280 \, a^{3} c^{3} 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))^(5/2),x, algorithm="giac")

[Out]

-1/1280*sqrt(2)*(315*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) - 5*(17*(c*tan(1/2*f*x + 1/2*e
)^2 - c)^(3/2)*c + 15*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(c^2*tan(1/2*f*x + 1/2*e)^4) - 8*((c*tan(1/2*f*x
+ 1/2*e)^2 - c)^(5/2)*c^8 - 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^9 + 30*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c
)*c^10)/c^10)/(a^3*c^3*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))