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

Optimal. Leaf size=211 \[ \frac{d \left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sec (e+f x))}+\frac{2 d^2 (3 c+2 d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a^2 f (c-d)^{7/2} (c+d)^{3/2}}+\frac{(c-6 d) \tan (e+f x)}{3 a^2 f (c-d)^2 (\sec (e+f x)+1) (c+d \sec (e+f x))}+\frac{\tan (e+f x)}{3 f (c-d) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))} \]

[Out]

(2*d^2*(3*c + 2*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(a^2*(c - d)^(7/2)*(c + d)^(3/2)*f) +
(d*(c^2 - 6*c*d - 10*d^2)*Tan[e + f*x])/(3*a^2*(c - d)^3*(c + d)*f*(c + d*Sec[e + f*x])) + ((c - 6*d)*Tan[e +
f*x])/(3*a^2*(c - d)^2*f*(1 + Sec[e + f*x])*(c + d*Sec[e + f*x])) + Tan[e + f*x]/(3*(c - d)*f*(a + a*Sec[e + f
*x])^2*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.373458, antiderivative size = 260, normalized size of antiderivative = 1.23, 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, 103, 152, 12, 93, 205} \[ \frac{\left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 f (c-d)^3 (c+d) \left (a^2 \sec (e+f x)+a^2\right )}-\frac{d \tan (e+f x)}{f \left (c^2-d^2\right ) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))}-\frac{2 d^2 (3 c+2 d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{a f (c-d)^{7/2} (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c+4 d) \tan (e+f x)}{3 f (c-d)^2 (c+d) (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + 4*d)*Tan[e + f*x])/(3*(c - d)^2*(c + d)*f*(a + a*Sec[e + f*x])^2) - (2*d^2*(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])/(a*(c - d)^(7/2)*(c + d)^(3
/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c^2 - 6*c*d - 10*d^2)*Tan[e + f*x])/(3*(c - d)^3*
(c + d)*f*(a^2 + a^2*Sec[e + f*x])) - (d*Tan[e + f*x])/((c^2 - d^2)*f*(a + a*Sec[e + f*x])^2*(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 103

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] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[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))^2 (c+d \sec (e+f x))^2} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (a+a x)^{5/2} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^2 (c+2 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 )}{\left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+4 d) \tan (e+f x)}{3 (c-d)^2 (c+d) f (a+a \sec (e+f x))^2}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{-a^4 (c-6 d) (c+d)-a^4 d (c+4 d) x}{\sqrt{a-a x} (a+a x)^{3/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{3 a^3 (c-d) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+4 d) \tan (e+f x)}{3 (c-d)^2 (c+d) f (a+a \sec (e+f x))^2}+\frac{\left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 (c-d)^3 (c+d) f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{3 a^6 d^2 (3 c+2 d)}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{3 a^6 (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{(c+4 d) \tan (e+f x)}{3 (c-d)^2 (c+d) f (a+a \sec (e+f x))^2}+\frac{\left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 (c-d)^3 (c+d) f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}-\frac{\left (d^2 (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 )}{(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{(c+4 d) \tan (e+f x)}{3 (c-d)^2 (c+d) f (a+a \sec (e+f x))^2}+\frac{\left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 (c-d)^3 (c+d) f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}-\frac{\left (2 d^2 (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{(c+4 d) \tan (e+f x)}{3 (c-d)^2 (c+d) f (a+a \sec (e+f x))^2}-\frac{2 d^2 (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)}{a (c-d)^{7/2} (c+d)^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{\left (c^2-6 c d-10 d^2\right ) \tan (e+f x)}{3 (c-d)^3 (c+d) f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^2 (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [C]  time = 3.91091, size = 376, normalized size = 1.78 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) (c \cos (e+f x)+d) \left (\frac{12 d^2 (3 c+2 d) (\sin (e)+i \cos (e)) \cos ^3\left (\frac{1}{2} (e+f x)\right ) (c \cos (e+f x)+d) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{(c+d) \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{6 d^3 \cos ^3\left (\frac{1}{2} (e+f x)\right ) (c \sin (f x)-d \sin (e))}{c (c+d) \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right )}+(c-d) \tan \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) (c \cos (e+f x)+d)-4 (c-4 d) \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \cos ^2\left (\frac{1}{2} (e+f x)\right ) (c \cos (e+f x)+d)+(c-d) \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (c \cos (e+f x)+d)\right )}{3 a^2 f (d-c)^3 (\sec (e+f x)+1)^2 (c+d \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(e + f*x)/2]*(d + c*Cos[e + f*x])*Sec[e + f*x]^4*((12*d^2*(3*c + 2*d)*ArcTan[((I*Cos[e] + Sin[e])*(c*Si
n[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]^3*(d + c
*Cos[e + f*x])*(I*Cos[e] + Sin[e]))/((c + d)*Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + (c - d)*(d + c*Cos
[e + f*x])*Sec[e/2]*Sin[(f*x)/2] - 4*(c - 4*d)*Cos[(e + f*x)/2]^2*(d + c*Cos[e + f*x])*Sec[e/2]*Sin[(f*x)/2] +
(6*d^3*Cos[(e + f*x)/2]^3*(-(d*Sin[e]) + c*Sin[f*x]))/(c*(c + d)*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2]))
+ (c - d)*Cos[(e + f*x)/2]*(d + c*Cos[e + f*x])*Tan[e/2]))/(3*a^2*(-c + d)^3*f*(1 + Sec[e + f*x])^2*(c + d*Se
c[e + f*x])^2)

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Maple [A]  time = 0.09, size = 203, normalized size = 1. \begin{align*}{\frac{1}{2\,f{a}^{2}} \left ( -{\frac{1}{ \left ({c}^{2}-2\,cd+{d}^{2} \right ) \left ( c-d \right ) } \left ({\frac{c}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{d}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-c\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +5\,\tan \left ( 1/2\,fx+e/2 \right ) d \right ) }-4\,{\frac{{d}^{2}}{ \left ( c-d \right ) ^{3}} \left ( -{\frac{\tan \left ( 1/2\,fx+e/2 \right ) d}{ \left ( c+d \right ) \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 ) }}-{\frac{3\,c+2\,d}{ \left ( c+d \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 ) } \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))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.645012, size = 2631, normalized size = 12.47 \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))^2,x, algorithm="fricas")

[Out]

[-1/6*(3*(3*c*d^3 + 2*d^4 + (3*c^2*d^2 + 2*c*d^3)*cos(f*x + e)^3 + (6*c^2*d^2 + 7*c*d^3 + 2*d^4)*cos(f*x + e)^
2 + (3*c^2*d^2 + 8*c*d^3 + 4*d^4)*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*(c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*c*d^4 + 10*d^5 + (2*c^5 - 6*c^4*d - 10*c^3*d^2 + 3*c^
2*d^3 + 8*c*d^4 + 3*d^5)*cos(f*x + e)^2 + (c^5 - 4*c^4*d - 14*c^3*d^2 - 10*c^2*d^3 + 13*c*d^4 + 14*d^5)*cos(f*
x + e))*sin(f*x + e))/((a^2*c^7 - 2*a^2*c^6*d - a^2*c^5*d^2 + 4*a^2*c^4*d^3 - a^2*c^3*d^4 - 2*a^2*c^2*d^5 + a^
2*c*d^6)*f*cos(f*x + e)^3 + (2*a^2*c^7 - 3*a^2*c^6*d - 4*a^2*c^5*d^2 + 7*a^2*c^4*d^3 + 2*a^2*c^3*d^4 - 5*a^2*c
^2*d^5 + a^2*d^7)*f*cos(f*x + e)^2 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^2*c^4*d^3 + 7*a^2*c^3*d^4 - 4*a^2*c^2*d^5
- 3*a^2*c*d^6 + 2*a^2*d^7)*f*cos(f*x + e) + (a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2
*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f), 1/3*(3*(3*c*d^3 + 2*d^4 + (3*c^2*d^2 + 2*c*d^3)*cos(f*x + e)^3 + (6*c^2*d^2
+ 7*c*d^3 + 2*d^4)*cos(f*x + e)^2 + (3*c^2*d^2 + 8*c*d^3 + 4*d^4)*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))) + (c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*c*d^4 + 10*
d^5 + (2*c^5 - 6*c^4*d - 10*c^3*d^2 + 3*c^2*d^3 + 8*c*d^4 + 3*d^5)*cos(f*x + e)^2 + (c^5 - 4*c^4*d - 14*c^3*d^
2 - 10*c^2*d^3 + 13*c*d^4 + 14*d^5)*cos(f*x + e))*sin(f*x + e))/((a^2*c^7 - 2*a^2*c^6*d - a^2*c^5*d^2 + 4*a^2*
c^4*d^3 - a^2*c^3*d^4 - 2*a^2*c^2*d^5 + a^2*c*d^6)*f*cos(f*x + e)^3 + (2*a^2*c^7 - 3*a^2*c^6*d - 4*a^2*c^5*d^2
+ 7*a^2*c^4*d^3 + 2*a^2*c^3*d^4 - 5*a^2*c^2*d^5 + a^2*d^7)*f*cos(f*x + e)^2 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^
2*c^4*d^3 + 7*a^2*c^3*d^4 - 4*a^2*c^2*d^5 - 3*a^2*c*d^6 + 2*a^2*d^7)*f*cos(f*x + e) + (a^2*c^6*d - 2*a^2*c^5*d
^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*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^{2} \sec ^{2}{\left (e + f x \right )} + 2 c^{2} \sec{\left (e + f x \right )} + c^{2} + 2 c d \sec ^{3}{\left (e + f x \right )} + 4 c d \sec ^{2}{\left (e + f x \right )} + 2 c d \sec{\left (e + f x \right )} + d^{2} \sec ^{4}{\left (e + f x \right )} + 2 d^{2} \sec ^{3}{\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx}{a^{2}} \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))**2,x)

[Out]

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

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Giac [B]  time = 1.25995, size = 662, normalized size = 3.14 \begin{align*} \frac{\frac{12 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} 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 )}} + \frac{12 \,{\left (3 \, c d^{2} + 2 \, d^{3}\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 (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} \sqrt{-c^{2} + d^{2}}} - \frac{a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4 \, a^{4} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, a^{4} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4 \, a^{4} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a^{4} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 24 \, a^{4} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 54 \, a^{4} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 48 \, a^{4} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 15 \, a^{4} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6} c^{6} - 6 \, a^{6} c^{5} d + 15 \, a^{6} c^{4} d^{2} - 20 \, a^{6} c^{3} d^{3} + 15 \, a^{6} c^{2} d^{4} - 6 \, a^{6} c d^{5} + a^{6} d^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(12*d^3*tan(1/2*f*x + 1/2*e)/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 -
d*tan(1/2*f*x + 1/2*e)^2 - c - d)) + 12*(3*c*d^2 + 2*d^3)*(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)))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c
*d^3 - a^2*d^4)*sqrt(-c^2 + d^2)) - (a^4*c^4*tan(1/2*f*x + 1/2*e)^3 - 4*a^4*c^3*d*tan(1/2*f*x + 1/2*e)^3 + 6*a
^4*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a^4*c*d^3*tan(1/2*f*x + 1/2*e)^3 + a^4*d^4*tan(1/2*f*x + 1/2*e)^3 - 3*a^
4*c^4*tan(1/2*f*x + 1/2*e) + 24*a^4*c^3*d*tan(1/2*f*x + 1/2*e) - 54*a^4*c^2*d^2*tan(1/2*f*x + 1/2*e) + 48*a^4*
c*d^3*tan(1/2*f*x + 1/2*e) - 15*a^4*d^4*tan(1/2*f*x + 1/2*e))/(a^6*c^6 - 6*a^6*c^5*d + 15*a^6*c^4*d^2 - 20*a^6
*c^3*d^3 + 15*a^6*c^2*d^4 - 6*a^6*c*d^5 + a^6*d^6))/f