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

Optimal. Leaf size=207 \[ -\frac{3 d \left (2 c^2+2 c d+d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a f (c-d)^{7/2} (c+d)^{5/2}}+\frac{d (2 c+d) (c+4 d) \tan (e+f x)}{2 a f (c-d)^3 (c+d)^2 (c+d \sec (e+f x))}+\frac{d (2 c+3 d) \tan (e+f x)}{2 a f (c-d)^2 (c+d) (c+d \sec (e+f x))^2}+\frac{\tan (e+f x)}{f (c-d) (a \sec (e+f x)+a) (c+d \sec (e+f x))^2} \]

[Out]

(-3*d*(2*c^2 + 2*c*d + d^2)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(a*(c - d)^(7/2)*(c + d)^(5/2
)*f) + (d*(2*c + 3*d)*Tan[e + f*x])/(2*a*(c - d)^2*(c + d)*f*(c + d*Sec[e + f*x])^2) + Tan[e + f*x]/((c - d)*f
*(a + a*Sec[e + f*x])*(c + d*Sec[e + f*x])^2) + (d*(2*c + d)*(c + 4*d)*Tan[e + f*x])/(2*a*(c - d)^3*(c + d)^2*
f*(c + d*Sec[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.4, antiderivative size = 268, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3987, 103, 151, 152, 12, 93, 205} \[ \frac{3 d \left (2 c^2+2 c d+d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{f (c-d)^{7/2} (c+d)^{5/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{d (4 c+d) \tan (e+f x)}{2 f \left (c^2-d^2\right )^2 (a \sec (e+f x)+a) (c+d \sec (e+f x))}-\frac{d \tan (e+f x)}{2 f \left (c^2-d^2\right ) (a \sec (e+f x)+a) (c+d \sec (e+f x))^2}+\frac{(2 c+d) (c+4 d) \tan (e+f x)}{2 f (c-d)^3 (c+d)^2 (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c + d)*(c + 4*d)*Tan[e + f*x])/(2*(c - d)^3*(c + d)^2*f*(a + a*Sec[e + f*x])) + (3*d*(2*c^2 + 2*c*d + d^2)
*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)^
(7/2)*(c + d)^(5/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d*Tan[e + f*x])/(2*(c^2 - d^2)*f*(
a + a*Sec[e + f*x])*(c + d*Sec[e + f*x])^2) - (d*(4*c + d)*Tan[e + f*x])/(2*(c^2 - d^2)^2*f*(a + a*Sec[e + f*x
])*(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 151

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] && IntegerQ[m]

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

Mathematica [C]  time = 6.78144, size = 1422, normalized size = 6.87 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*c^2 + 2*c*d + d^2)*Cos[e/2 + (f*x)/2]^2*(d + c*Cos[e + f*x])^3*Sec[e + f*x]^4*(((-6*I)*d*ArcTan[Sec[(f*x)/
2]*(Cos[e]/(Sqrt[c^2 - d^2]*Sqrt[Cos[2*e] - I*Sin[2*e]]) - (I*Sin[e])/(Sqrt[c^2 - d^2]*Sqrt[Cos[2*e] - I*Sin[2
*e]]))*((-I)*d*Sin[(f*x)/2] + I*c*Sin[e + (f*x)/2])]*Cos[e])/(Sqrt[c^2 - d^2]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) -
(6*d*ArcTan[Sec[(f*x)/2]*(Cos[e]/(Sqrt[c^2 - d^2]*Sqrt[Cos[2*e] - I*Sin[2*e]]) - (I*Sin[e])/(Sqrt[c^2 - d^2]*
Sqrt[Cos[2*e] - I*Sin[2*e]]))*((-I)*d*Sin[(f*x)/2] + I*c*Sin[e + (f*x)/2])]*Sin[e])/(Sqrt[c^2 - d^2]*f*Sqrt[Co
s[2*e] - I*Sin[2*e]])))/((-c + d)^3*(c + d)^2*(a + a*Sec[e + f*x])*(c + d*Sec[e + f*x])^3) + (Cos[e/2 + (f*x)/
2]*(d + c*Cos[e + f*x])*Sec[e/2]*Sec[e]*Sec[e + f*x]^4*(8*c^5*d*Sin[(f*x)/2] + 10*c^4*d^2*Sin[(f*x)/2] - 11*c^
3*d^3*Sin[(f*x)/2] - 17*c^2*d^4*Sin[(f*x)/2] - 2*c*d^5*Sin[(f*x)/2] + 2*d^6*Sin[(f*x)/2] - 8*c^5*d*Sin[(3*f*x)
/2] - 22*c^4*d^2*Sin[(3*f*x)/2] - 27*c^3*d^3*Sin[(3*f*x)/2] - 5*c^2*d^4*Sin[(3*f*x)/2] + 2*c*d^5*Sin[(3*f*x)/2
] + 4*c^6*Sin[e - (f*x)/2] + 8*c^5*d*Sin[e - (f*x)/2] + 18*c^4*d^2*Sin[e - (f*x)/2] + 35*c^3*d^3*Sin[e - (f*x)
/2] + 25*c^2*d^4*Sin[e - (f*x)/2] + 2*c*d^5*Sin[e - (f*x)/2] - 2*d^6*Sin[e - (f*x)/2] - 4*c^6*Sin[e + (f*x)/2]
- 8*c^5*d*Sin[e + (f*x)/2] - 6*c^4*d^2*Sin[e + (f*x)/2] - 7*c^3*d^3*Sin[e + (f*x)/2] + 5*c^2*d^4*Sin[e + (f*x
)/2] + 2*c*d^5*Sin[e + (f*x)/2] - 2*d^6*Sin[e + (f*x)/2] + 8*c^5*d*Sin[2*e + (f*x)/2] + 22*c^4*d^2*Sin[2*e + (
f*x)/2] + 17*c^3*d^3*Sin[2*e + (f*x)/2] + 13*c^2*d^4*Sin[2*e + (f*x)/2] + 2*c*d^5*Sin[2*e + (f*x)/2] - 2*d^6*S
in[2*e + (f*x)/2] + 2*c^6*Sin[e + (3*f*x)/2] + 4*c^5*d*Sin[e + (3*f*x)/2] - 4*c^4*d^2*Sin[e + (3*f*x)/2] - 19*
c^3*d^3*Sin[e + (3*f*x)/2] - 5*c^2*d^4*Sin[e + (3*f*x)/2] + 2*c*d^5*Sin[e + (3*f*x)/2] - 8*c^5*d*Sin[2*e + (3*
f*x)/2] - 16*c^4*d^2*Sin[2*e + (3*f*x)/2] - c^3*d^3*Sin[2*e + (3*f*x)/2] + 2*c^2*d^4*Sin[2*e + (3*f*x)/2] - 2*
c*d^5*Sin[2*e + (3*f*x)/2] + 2*c^6*Sin[3*e + (3*f*x)/2] + 4*c^5*d*Sin[3*e + (3*f*x)/2] + 2*c^4*d^2*Sin[3*e + (
3*f*x)/2] + 7*c^3*d^3*Sin[3*e + (3*f*x)/2] + 2*c^2*d^4*Sin[3*e + (3*f*x)/2] - 2*c*d^5*Sin[3*e + (3*f*x)/2] - 2
*c^6*Sin[e + (5*f*x)/2] - 4*c^5*d*Sin[e + (5*f*x)/2] - 8*c^4*d^2*Sin[e + (5*f*x)/2] - 2*c^3*d^3*Sin[e + (5*f*x
)/2] + c^2*d^4*Sin[e + (5*f*x)/2] - 6*c^4*d^2*Sin[2*e + (5*f*x)/2] - 2*c^3*d^3*Sin[2*e + (5*f*x)/2] + c^2*d^4*
Sin[2*e + (5*f*x)/2] - 2*c^6*Sin[3*e + (5*f*x)/2] - 4*c^5*d*Sin[3*e + (5*f*x)/2] - 2*c^4*d^2*Sin[3*e + (5*f*x)
/2]))/(8*c^2*(-c + d)^3*(c + d)^2*f*(a + a*Sec[e + f*x])*(c + d*Sec[e + f*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.1, size = 221, normalized size = 1.1 \begin{align*}{\frac{1}{fa} \left ({\frac{1}{{c}^{3}-3\,{c}^{2}d+3\,{d}^{2}c-{d}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+2\,{\frac{d}{ \left ( c-d \right ) ^{3}} \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 ) ^{2}} \left ( -3/2\,{\frac{d \left ( 2\,{c}^{2}-cd-{d}^{2} \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{{c}^{2}+2\,cd+{d}^{2}}}+1/2\,{\frac{d \left ( 6\,c+d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{c+d}} \right ) }-3/2\,{\frac{2\,{c}^{2}+2\,cd+{d}^{2}}{ \left ({c}^{2}+2\,cd+{d}^{2} \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))/(c+d*sec(f*x+e))^3,x)

[Out]

1/f/a*(1/(c^3-3*c^2*d+3*c*d^2-d^3)*tan(1/2*f*x+1/2*e)+2*d/(c-d)^3*((-3/2*d*(2*c^2-c*d-d^2)/(c^2+2*c*d+d^2)*tan
(1/2*f*x+1/2*e)^3+1/2*d*(6*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)^
2-3/2*(2*c^2+2*c*d+d^2)/(c^2+2*c*d+d^2)/((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))/(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.692185, size = 2865, normalized size = 13.84 \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))/(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/4*(3*(2*c^2*d^3 + 2*c*d^4 + d^5 + (2*c^4*d + 2*c^3*d^2 + c^2*d^3)*cos(f*x + e)^3 + (2*c^4*d + 6*c^3*d^2 +
5*c^2*d^3 + 2*c*d^4)*cos(f*x + e)^2 + (4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*cos(f*x + e))*sqrt(c^2 - d^2)*lo
g((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*(2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5
- 4*d^6 + (2*c^6 + 4*c^5*d + 6*c^4*d^2 - 2*c^3*d^3 - 9*c^2*d^4 - 2*c*d^5 + d^6)*cos(f*x + e)^2 + (4*c^5*d + 14
*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^9 - a*c^8*d - 3*a*c^7*d^
2 + 3*a*c^6*d^3 + 3*a*c^5*d^4 - 3*a*c^4*d^5 - a*c^3*d^6 + a*c^2*d^7)*f*cos(f*x + e)^3 + (a*c^9 + a*c^8*d - 5*a
*c^7*d^2 - 3*a*c^6*d^3 + 9*a*c^5*d^4 + 3*a*c^4*d^5 - 7*a*c^3*d^6 - a*c^2*d^7 + 2*a*c*d^8)*f*cos(f*x + e)^2 + (
2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5 - 3*a*c^3*d^6 - 5*a*c^2*d^7 + a*c*d^8 + a*d^9)
*f*cos(f*x + e) + (a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d^7 - a*c*d^8 + a
*d^9)*f), -1/2*(3*(2*c^2*d^3 + 2*c*d^4 + d^5 + (2*c^4*d + 2*c^3*d^2 + c^2*d^3)*cos(f*x + e)^3 + (2*c^4*d + 6*c
^3*d^2 + 5*c^2*d^3 + 2*c*d^4)*cos(f*x + e)^2 + (4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*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))) - (2*c^4*d^2 + 9*c^3*d^3 + 2
*c^2*d^4 - 9*c*d^5 - 4*d^6 + (2*c^6 + 4*c^5*d + 6*c^4*d^2 - 2*c^3*d^3 - 9*c^2*d^4 - 2*c*d^5 + d^6)*cos(f*x + e
)^2 + (4*c^5*d + 14*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^9 - a
*c^8*d - 3*a*c^7*d^2 + 3*a*c^6*d^3 + 3*a*c^5*d^4 - 3*a*c^4*d^5 - a*c^3*d^6 + a*c^2*d^7)*f*cos(f*x + e)^3 + (a*
c^9 + a*c^8*d - 5*a*c^7*d^2 - 3*a*c^6*d^3 + 9*a*c^5*d^4 + 3*a*c^4*d^5 - 7*a*c^3*d^6 - a*c^2*d^7 + 2*a*c*d^8)*f
*cos(f*x + e)^2 + (2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5 - 3*a*c^3*d^6 - 5*a*c^2*d^7
+ a*c*d^8 + a*d^9)*f*cos(f*x + e) + (a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^
2*d^7 - a*c*d^8 + a*d^9)*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^{3} \sec{\left (e + f x \right )} + c^{3} + 3 c^{2} d \sec ^{2}{\left (e + f x \right )} + 3 c^{2} d \sec{\left (e + f x \right )} + 3 c d^{2} \sec ^{3}{\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{4}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.30518, size = 505, normalized size = 2.44 \begin{align*} -\frac{\frac{3 \,{\left (2 \, c^{2} d + 2 \, c d^{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 c^{5} - a c^{4} d - 2 \, a c^{3} d^{2} + 2 \, a c^{2} d^{3} + a c d^{4} - a d^{5}\right )} \sqrt{-c^{2} + d^{2}}} - \frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}} + \frac{6 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 7 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (a c^{5} - a c^{4} d - 2 \, a c^{3} d^{2} + 2 \, a c^{2} d^{3} + a c d^{4} - a d^{5}\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 )}^{2}}}{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))/(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-(3*(2*c^2*d + 2*c*d^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*c^5 - a*c^4*d - 2*a*c^3*d^2 + 2*a*c^2*d^3 + a*c*d^4 - a*d
^5)*sqrt(-c^2 + d^2)) - tan(1/2*f*x + 1/2*e)/(a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3) + (6*c^2*d^2*tan(1/2*f*x
+ 1/2*e)^3 - 3*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 3*d^4*tan(1/2*f*x + 1/2*e)^3 - 6*c^2*d^2*tan(1/2*f*x + 1/2*e) -
7*c*d^3*tan(1/2*f*x + 1/2*e) - d^4*tan(1/2*f*x + 1/2*e))/((a*c^5 - a*c^4*d - 2*a*c^3*d^2 + 2*a*c^2*d^3 + a*c*d
^4 - a*d^5)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^2))/f