3.200 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=213 \[ \frac{a^2 (3 c-2 d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{3/2} (c+d)^{7/2}}+\frac{a^2 (3 c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^3 (c+d \sec (e+f x))}+\frac{(3 c-2 d) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{6 f (c-d) (c+d)^2 (c+d \sec (e+f x))^2}-\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^3} \]

[Out]

(a^2*(3*c - 2*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(3/2)*(c + d)^(7/2)*f) - (d*(a
+ a*Sec[e + f*x])^2*Tan[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^3) + ((3*c - 2*d)*(a^2 + a^2*Sec[e + f
*x])*Tan[e + f*x])/(6*(c - d)*(c + d)^2*f*(c + d*Sec[e + f*x])^2) + (a^2*(3*c - 2*d)*Tan[e + f*x])/(2*(c - d)*
(c + d)^3*f*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.28264, antiderivative size = 268, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3987, 96, 94, 93, 205} \[ -\frac{a^3 (3 c-2 d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{f (c-d)^{3/2} (c+d)^{7/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{a^2 (3 c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^3 (c+d \sec (e+f x))}+\frac{(3 c-2 d) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{6 f (c-d) (c+d)^2 (c+d \sec (e+f x))^2}-\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^4,x]

[Out]

-((a^3*(3*c - 2*d)*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e
+ f*x])/((c - d)^(3/2)*(c + d)^(7/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])) - (d*(a + a*Sec[e
+ f*x])^2*Tan[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^3) + ((3*c - 2*d)*(a^2 + a^2*Sec[e + f*x])*Tan[e
+ f*x])/(6*(c - d)*(c + d)^2*f*(c + d*Sec[e + f*x])^2) + (a^2*(3*c - 2*d)*Tan[e + f*x])/(2*(c - d)*(c + d)^3*
f*(c + d*Sec[e + f*x]))

Rule 3987

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^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

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

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 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]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^4} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2}}{\sqrt{a-a x} (c+d x)^4} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}-\frac{\left (a^2 (3 c-2 d) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2}}{\sqrt{a-a x} (c+d x)^3} \, dx,x,\sec (e+f x)\right )}{3 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{(3 c-2 d) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}-\frac{\left (a^3 (3 c-2 d) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+a x}}{\sqrt{a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{2 (c+d) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{(3 c-2 d) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac{a^2 (3 c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^3 f (c+d \sec (e+f x))}-\frac{\left (a^4 (3 c-2 d) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{2 (c+d)^2 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{(3 c-2 d) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac{a^2 (3 c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^3 f (c+d \sec (e+f x))}-\frac{\left (a^4 (3 c-2 d) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a c-a d-(-a c-a d) x^2} \, dx,x,\frac{\sqrt{a+a \sec (e+f x)}}{\sqrt{a-a \sec (e+f x)}}\right )}{(c+d)^2 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{a^3 (3 c-2 d) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right ) \tan (e+f x)}{(c-d)^{3/2} (c+d)^{7/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{d (a+a \sec (e+f x))^2 \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{(3 c-2 d) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac{a^2 (3 c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^3 f (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 4.73488, size = 211, normalized size = 0.99 \[ \frac{a^2 (c-d)^2 \left (24 (3 c-2 d) (c \cos (e+f x)+d)^3 \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )-2 \sqrt{c^2-d^2} \sin (e+f x) \left (6 \left (6 c^2 d+c^3-7 c d^2-2 d^3\right ) \cos (e+f x)+\left (-7 c^2 d+12 c^3-6 c d^2-2 d^3\right ) \cos (2 (e+f x))-5 c^2 d+12 c^3+6 c d^2-22 d^3\right )\right )}{24 f (d-c)^3 (c+d)^3 \sqrt{c^2-d^2} (c \cos (e+f x)+d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^4,x]

[Out]

(a^2*(c - d)^2*(24*(3*c - 2*d)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d + c*Cos[e + f*x])^3 - 2
*Sqrt[c^2 - d^2]*(12*c^3 - 5*c^2*d + 6*c*d^2 - 22*d^3 + 6*(c^3 + 6*c^2*d - 7*c*d^2 - 2*d^3)*Cos[e + f*x] + (12
*c^3 - 7*c^2*d - 6*c*d^2 - 2*d^3)*Cos[2*(e + f*x)])*Sin[e + f*x]))/(24*(-c + d)^3*(c + d)^3*Sqrt[c^2 - d^2]*f*
(d + c*Cos[e + f*x])^3)

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Maple [A]  time = 0.128, size = 228, normalized size = 1.1 \begin{align*} 8\,{\frac{{a}^{2}}{f} \left ( -{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{3}} \left ( 1/8\,{\frac{ \left ( 3\,c-2\,d \right ) \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{5}}{{c}^{3}+3\,{c}^{2}d+3\,{d}^{2}c+{d}^{3}}}-1/3\,{\frac{ \left ( 3\,c-2\,d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{{c}^{2}+2\,cd+{d}^{2}}}+1/8\,{\frac{ \left ( 5\,c-6\,d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( c-d \right ) }} \right ) }+1/8\,{\frac{3\,c-2\,d}{ \left ({c}^{4}+2\,{c}^{3}d-2\,c{d}^{3}-{d}^{4} \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^4,x)

[Out]

8/f*a^2*(-(1/8*(3*c-2*d)*(c-d)/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5-1/3*(3*c-2*d)/(c^2+2*c*d+d^2)*ta
n(1/2*f*x+1/2*e)^3+1/8*(5*c-6*d)/(c+d)/(c-d)*tan(1/2*f*x+1/2*e))/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*
d-c-d)^3+1/8*(3*c-2*d)/(c^4+2*c^3*d-2*c*d^3-d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(
c-d))^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.67826, size = 2585, normalized size = 12.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(3*a^2*c*d^3 - 2*a^2*d^4 + (3*a^2*c^4 - 2*a^2*c^3*d)*cos(f*x + e)^3 + 3*(3*a^2*c^3*d - 2*a^2*c^2*d^2)
*cos(f*x + e)^2 + 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e))*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2
- 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e
)^2 + 2*c*d*cos(f*x + e) + d^2)) + 2*(a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a^2*d^5 +
(12*a^2*c^5 - 7*a^2*c^4*d - 18*a^2*c^3*d^2 + 5*a^2*c^2*d^3 + 6*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e)^2 + 3*(a^2*
c^5 + 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e))*sin(f*x + e))/((c^9
+ 2*c^8*d - c^7*d^2 - 4*c^6*d^3 - c^5*d^4 + 2*c^4*d^5 + c^3*d^6)*f*cos(f*x + e)^3 + 3*(c^8*d + 2*c^7*d^2 - c^
6*d^3 - 4*c^5*d^4 - c^4*d^5 + 2*c^3*d^6 + c^2*d^7)*f*cos(f*x + e)^2 + 3*(c^7*d^2 + 2*c^6*d^3 - c^5*d^4 - 4*c^4
*d^5 - c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e) + (c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^3*d^6 - c^2*d^7 + 2*
c*d^8 + d^9)*f), 1/6*(3*(3*a^2*c*d^3 - 2*a^2*d^4 + (3*a^2*c^4 - 2*a^2*c^3*d)*cos(f*x + e)^3 + 3*(3*a^2*c^3*d -
2*a^2*c^2*d^2)*cos(f*x + e)^2 + 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-sqrt(-
c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*
a^2*c*d^4 + 10*a^2*d^5 + (12*a^2*c^5 - 7*a^2*c^4*d - 18*a^2*c^3*d^2 + 5*a^2*c^2*d^3 + 6*a^2*c*d^4 + 2*a^2*d^5)
*cos(f*x + e)^2 + 3*(a^2*c^5 + 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x
+ e))*sin(f*x + e))/((c^9 + 2*c^8*d - c^7*d^2 - 4*c^6*d^3 - c^5*d^4 + 2*c^4*d^5 + c^3*d^6)*f*cos(f*x + e)^3 +
3*(c^8*d + 2*c^7*d^2 - c^6*d^3 - 4*c^5*d^4 - c^4*d^5 + 2*c^3*d^6 + c^2*d^7)*f*cos(f*x + e)^2 + 3*(c^7*d^2 + 2*
c^6*d^3 - c^5*d^4 - 4*c^4*d^5 - c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e) + (c^6*d^3 + 2*c^5*d^4 - c^4*d^5 -
4*c^3*d^6 - c^2*d^7 + 2*c*d^8 + d^9)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{\sec{\left (e + f x \right )}}{c^{4} + 4 c^{3} d \sec{\left (e + f x \right )} + 6 c^{2} d^{2} \sec ^{2}{\left (e + f x \right )} + 4 c d^{3} \sec ^{3}{\left (e + f x \right )} + d^{4} \sec ^{4}{\left (e + f x \right )}}\, dx + \int \frac{2 \sec ^{2}{\left (e + f x \right )}}{c^{4} + 4 c^{3} d \sec{\left (e + f x \right )} + 6 c^{2} d^{2} \sec ^{2}{\left (e + f x \right )} + 4 c d^{3} \sec ^{3}{\left (e + f x \right )} + d^{4} \sec ^{4}{\left (e + f x \right )}}\, dx + \int \frac{\sec ^{3}{\left (e + f x \right )}}{c^{4} + 4 c^{3} d \sec{\left (e + f x \right )} + 6 c^{2} d^{2} \sec ^{2}{\left (e + f x \right )} + 4 c d^{3} \sec ^{3}{\left (e + f x \right )} + d^{4} \sec ^{4}{\left (e + f x \right )}}\, dx\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))**2/(c+d*sec(f*x+e))**4,x)

[Out]

a**2*(Integral(sec(e + f*x)/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x
)**3 + d**4*sec(e + f*x)**4), x) + Integral(2*sec(e + f*x)**2/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(
e + f*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**4*sec(e + f*x)**4), x) + Integral(sec(e + f*x)**3/(c**4 + 4*c**3*d
*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**4*sec(e + f*x)**4), x))

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Giac [B]  time = 1.44826, size = 567, normalized size = 2.66 \begin{align*} -\frac{\frac{3 \,{\left (3 \, a^{2} c - 2 \, a^{2} d\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )}}{{\left (c^{4} + 2 \, c^{3} d - 2 \, c d^{3} - d^{4}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{9 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 24 \, a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 21 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 6 \, a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 24 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 16 \, a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 24 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 16 \, a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 21 \, a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 18 \, a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (c^{4} + 2 \, c^{3} d - 2 \, c d^{3} - d^{4}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/3*(3*(3*a^2*c - 2*a^2*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e)
- d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/((c^4 + 2*c^3*d - 2*c*d^3 - d^4)*sqrt(-c^2 + d^2)) + (9*a^2*c^3*t
an(1/2*f*x + 1/2*e)^5 - 24*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^5 + 21*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^5 - 6*a^2*d^3*
tan(1/2*f*x + 1/2*e)^5 - 24*a^2*c^3*tan(1/2*f*x + 1/2*e)^3 + 16*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*c*d^
2*tan(1/2*f*x + 1/2*e)^3 - 16*a^2*d^3*tan(1/2*f*x + 1/2*e)^3 + 15*a^2*c^3*tan(1/2*f*x + 1/2*e) + 12*a^2*c^2*d*
tan(1/2*f*x + 1/2*e) - 21*a^2*c*d^2*tan(1/2*f*x + 1/2*e) - 18*a^2*d^3*tan(1/2*f*x + 1/2*e))/((c^4 + 2*c^3*d -
2*c*d^3 - d^4)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^3))/f