3.231 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^3 (c+d \sec (e+f x))} \, dx\)

Optimal. Leaf size=181 \[ \frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 f (c-d)^3 \left (a^3 \sec (e+f x)+a^3\right )}-\frac{2 d^3 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a^3 f (c-d)^{7/2} \sqrt{c+d}}+\frac{(2 c-7 d) \tan (e+f x)}{15 a f (c-d)^2 (a \sec (e+f x)+a)^2}+\frac{\tan (e+f x)}{5 f (c-d) (a \sec (e+f x)+a)^3} \]

[Out]

(-2*d^3*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(a^3*(c - d)^(7/2)*Sqrt[c + d]*f) + Tan[e + f*x]/
(5*(c - d)*f*(a + a*Sec[e + f*x])^3) + ((2*c - 7*d)*Tan[e + f*x])/(15*a*(c - d)^2*f*(a + a*Sec[e + f*x])^2) +
((2*c^2 - 9*c*d + 22*d^2)*Tan[e + f*x])/(15*(c - d)^3*f*(a^3 + a^3*Sec[e + f*x]))

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Rubi [A]  time = 0.322949, antiderivative size = 235, normalized size of antiderivative = 1.3, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3987, 104, 152, 12, 93, 205} \[ \frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 f (c-d)^3 \left (a^3 \sec (e+f x)+a^3\right )}+\frac{2 d^3 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{a^2 f (c-d)^{7/2} \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(2 c-7 d) \tan (e+f x)}{15 a f (c-d)^2 (a \sec (e+f x)+a)^2}+\frac{\tan (e+f x)}{5 f (c-d) (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[e + f*x]/(5*(c - d)*f*(a + a*Sec[e + f*x])^3) + ((2*c - 7*d)*Tan[e + f*x])/(15*a*(c - d)^2*f*(a + a*Sec[e
+ f*x])^2) + (2*d^3*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])/(a^2*(c - d)^(7/2)*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((2*c^2 - 9*c*
d + 22*d^2)*Tan[e + f*x])/(15*(c - d)^3*f*(a^3 + a^3*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 104

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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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))^3 (c+d \sec (e+f x))} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (a+a x)^{7/2} (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{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{-a^2 (2 c-5 d)-2 a^2 d x}{\sqrt{a-a x} (a+a x)^{5/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{5 a (c-d) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{(2 c-7 d) \tan (e+f x)}{15 a (c-d)^2 f (a+a \sec (e+f x))^2}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^4 \left (2 c^2-7 c d+15 d^2\right )+a^4 (2 c-7 d) d x}{\sqrt{a-a x} (a+a x)^{3/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{15 a^4 (c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{(2 c-7 d) \tan (e+f x)}{15 a (c-d)^2 f (a+a \sec (e+f x))^2}+\frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{15 a^6 d^3}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{15 a^7 (c-d)^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{(2 c-7 d) \tan (e+f x)}{15 a (c-d)^2 f (a+a \sec (e+f x))^2}+\frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{\left (d^3 \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 )}{a (c-d)^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{(2 c-7 d) \tan (e+f x)}{15 a (c-d)^2 f (a+a \sec (e+f x))^2}+\frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{\left (2 d^3 \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 )}{a (c-d)^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{5 (c-d) f (a+a \sec (e+f x))^3}+\frac{(2 c-7 d) \tan (e+f x)}{15 a (c-d)^2 f (a+a \sec (e+f x))^2}+\frac{2 d^3 \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)}{a^2 (c-d)^{7/2} \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 c^2-9 c d+22 d^2\right ) \tan (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.146, size = 345, normalized size = 1.91 \[ \frac{\cos \left (\frac{1}{2} (e+f x)\right ) \left (\sec \left (\frac{e}{2}\right ) \left (-15 \left (2 c^2-7 c d+9 d^2\right ) \sin \left (e+\frac{f x}{2}\right )+5 \left (8 c^2-27 c d+37 d^2\right ) \sin \left (\frac{f x}{2}\right )+20 c^2 \sin \left (e+\frac{3 f x}{2}\right )-15 c^2 \sin \left (2 e+\frac{3 f x}{2}\right )+7 c^2 \sin \left (2 e+\frac{5 f x}{2}\right )-75 c d \sin \left (e+\frac{3 f x}{2}\right )+45 c d \sin \left (2 e+\frac{3 f x}{2}\right )-24 c d \sin \left (2 e+\frac{5 f x}{2}\right )+115 d^2 \sin \left (e+\frac{3 f x}{2}\right )-45 d^2 \sin \left (2 e+\frac{3 f x}{2}\right )+32 d^2 \sin \left (2 e+\frac{5 f x}{2}\right )\right )+\frac{480 d^3 (\sin (e)+i \cos (e)) \cos ^5\left (\frac{1}{2} (e+f x)\right ) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{30 a^3 f (c-d)^3 (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*((480*d^3*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 -
d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*Cos[(e + f*x)/2]^5*(I*Cos[e] + Sin[e]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*
Sin[e])^2]) + Sec[e/2]*(5*(8*c^2 - 27*c*d + 37*d^2)*Sin[(f*x)/2] - 15*(2*c^2 - 7*c*d + 9*d^2)*Sin[e + (f*x)/2]
+ 20*c^2*Sin[e + (3*f*x)/2] - 75*c*d*Sin[e + (3*f*x)/2] + 115*d^2*Sin[e + (3*f*x)/2] - 15*c^2*Sin[2*e + (3*f*
x)/2] + 45*c*d*Sin[2*e + (3*f*x)/2] - 45*d^2*Sin[2*e + (3*f*x)/2] + 7*c^2*Sin[2*e + (5*f*x)/2] - 24*c*d*Sin[2*
e + (5*f*x)/2] + 32*d^2*Sin[2*e + (5*f*x)/2])))/(30*a^3*(c - d)^3*f*(1 + Cos[e + f*x])^3)

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Maple [A]  time = 0.084, size = 203, normalized size = 1.1 \begin{align*}{\frac{1}{4\,f{a}^{3}} \left ({\frac{1}{ \left ( c-d \right ) ^{3}} \left ({\frac{{c}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,cd}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{{d}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,{c}^{2}}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}cd-{\frac{4\,{d}^{2}}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ){c}^{2}-4\,cd\tan \left ( 1/2\,fx+e/2 \right ) +7\,\tan \left ( 1/2\,fx+e/2 \right ){d}^{2} \right ) }-8\,{\frac{{d}^{3}}{ \left ( c-d \right ) ^{3}\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))^3/(c+d*sec(f*x+e)),x)

[Out]

1/4/f/a^3*(1/(c-d)^3*(1/5*tan(1/2*f*x+1/2*e)^5*c^2-2/5*tan(1/2*f*x+1/2*e)^5*c*d+1/5*tan(1/2*f*x+1/2*e)^5*d^2-2
/3*tan(1/2*f*x+1/2*e)^3*c^2+2*tan(1/2*f*x+1/2*e)^3*c*d-4/3*tan(1/2*f*x+1/2*e)^3*d^2+tan(1/2*f*x+1/2*e)*c^2-4*c
*d*tan(1/2*f*x+1/2*e)+7*tan(1/2*f*x+1/2*e)*d^2)-8*d^3/(c-d)^3/((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))^3/(c+d*sec(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.571845, size = 2187, normalized size = 12.08 \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))^3/(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[-1/30*(15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^2 + 3*d^3*cos(f*x + e) + d^3)*sqrt(c^2 - d^2)*log((2*c*d*c
os(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*(2*c^4 - 9*c^3*d + 20*c^2*d^2 + 9*c*d^3 - 22*d^4 + (7*c
^4 - 24*c^3*d + 25*c^2*d^2 + 24*c*d^3 - 32*d^4)*cos(f*x + e)^2 + 3*(2*c^4 - 9*c^3*d + 15*c^2*d^2 + 9*c*d^3 - 1
7*d^4)*cos(f*x + e))*sin(f*x + e))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3
*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*c
os(f*x + e)^2 + 3*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x +
e) + (a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f), -1/15*(15*(d^3*cos(f*
x + e)^3 + 3*d^3*cos(f*x + e)^2 + 3*d^3*cos(f*x + e) + d^3)*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))) - (2*c^4 - 9*c^3*d + 20*c^2*d^2 + 9*c*d^3 - 22*d^4 + (7*c^4 - 24*c^3*
d + 25*c^2*d^2 + 24*c*d^3 - 32*d^4)*cos(f*x + e)^2 + 3*(2*c^4 - 9*c^3*d + 15*c^2*d^2 + 9*c*d^3 - 17*d^4)*cos(f
*x + e))*sin(f*x + e))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(
f*x + e)^3 + 3*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^
2 + 3*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) + (a^3*c^
5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f)]

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

[Out]

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

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Giac [B]  time = 1.1888, size = 660, normalized size = 3.65 \begin{align*} -\frac{\frac{120 \,{\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 )} d^{3}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} \sqrt{-c^{2} + d^{2}}} - \frac{3 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 12 \, a^{12} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 18 \, a^{12} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 12 \, a^{12} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3 \, a^{12} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 10 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 50 \, a^{12} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 90 \, a^{12} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 70 \, a^{12} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 20 \, a^{12} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 90 \, a^{12} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 240 \, a^{12} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 270 \, a^{12} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 105 \, a^{12} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15} c^{5} - 5 \, a^{15} c^{4} d + 10 \, a^{15} c^{3} d^{2} - 10 \, a^{15} c^{2} d^{3} + 5 \, a^{15} c d^{4} - a^{15} d^{5}}}{60 \, 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))^3/(c+d*sec(f*x+e)),x, algorithm="giac")

[Out]

-1/60*(120*(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)))*d^3/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*sqrt(-c^2 + d^2)) - (3*a^12
*c^4*tan(1/2*f*x + 1/2*e)^5 - 12*a^12*c^3*d*tan(1/2*f*x + 1/2*e)^5 + 18*a^12*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 -
12*a^12*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 3*a^12*d^4*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^4*tan(1/2*f*x + 1/2*e)^3
+ 50*a^12*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 90*a^12*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 70*a^12*c*d^3*tan(1/2*f*x +
1/2*e)^3 - 20*a^12*d^4*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^4*tan(1/2*f*x + 1/2*e) - 90*a^12*c^3*d*tan(1/2*f*x +
1/2*e) + 240*a^12*c^2*d^2*tan(1/2*f*x + 1/2*e) - 270*a^12*c*d^3*tan(1/2*f*x + 1/2*e) + 105*a^12*d^4*tan(1/2*f
*x + 1/2*e))/(a^15*c^5 - 5*a^15*c^4*d + 10*a^15*c^3*d^2 - 10*a^15*c^2*d^3 + 5*a^15*c*d^4 - a^15*d^5))/f