\(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx\) [209]

Optimal result

Integrand size = 31, antiderivative size = 266 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\frac {5 a^3 (4 c-3 d) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{4 (c-d)^{3/2} (c+d)^{9/2} f}-\frac {d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}-\frac {5 a^3 (4 c-3 d) \tan (e+f x)}{24 d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {5 a^3 (4 c-3 d) (c+4 d) \tan (e+f x)}{24 (c-d) d (c+d)^4 f (c+d \sec (e+f x))} \]

[Out]

5/4*a^3*(4*c-3*d)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/(c-d)^(3/2)/(c+d)^(9/2)/f-1/4*d*(a+a*sec
(f*x+e))^3*tan(f*x+e)/(c^2-d^2)/f/(c+d*sec(f*x+e))^4+1/12*a*(4*c-3*d)*(a+a*sec(f*x+e))^2*tan(f*x+e)/(c-d)/(c+d
)^2/f/(c+d*sec(f*x+e))^3-5/24*a^3*(4*c-3*d)*tan(f*x+e)/d/(c+d)^3/f/(c+d*sec(f*x+e))^2+5/24*a^3*(4*c-3*d)*(c+4*
d)*tan(f*x+e)/(c-d)/d/(c+d)^4/f/(c+d*sec(f*x+e))

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4072, 98, 96, 95, 211} \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=-\frac {5 a^4 (4 c-3 d) \tan (e+f x) \arctan \left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{4 f (c-d)^{3/2} (c+d)^{9/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {5 a^3 (4 c-3 d) \tan (e+f x)}{8 f (c-d) (c+d)^4 (c+d \sec (e+f x))}+\frac {5 (4 c-3 d) \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{24 f (c-d) (c+d)^3 (c+d \sec (e+f x))^2}-\frac {d \tan (e+f x) (a \sec (e+f x)+a)^3}{4 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) \tan (e+f x) (a \sec (e+f x)+a)^2}{12 f (c-d) (c+d)^2 (c+d \sec (e+f x))^3} \]

[In]

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

[Out]

(-5*a^4*(4*c - 3*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])/(4*(c - d)^(3/2)*(c + d)^(9/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d*(a + a*Sec[
e + f*x])^3*Tan[e + f*x])/(4*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^4) + (a*(4*c - 3*d)*(a + a*Sec[e + f*x])^2*Tan
[e + f*x])/(12*(c - d)*(c + d)^2*f*(c + d*Sec[e + f*x])^3) + (5*(4*c - 3*d)*(a^3 + a^3*Sec[e + f*x])*Tan[e + f
*x])/(24*(c - d)*(c + d)^3*f*(c + d*Sec[e + f*x])^2) + (5*a^3*(4*c - 3*d)*Tan[e + f*x])/(8*(c - d)*(c + d)^4*f
*(c + d*Sec[e + f*x]))

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 4072

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{5/2}}{\sqrt {a-a x} (c+d x)^5} \, 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))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}-\frac {\left (a^2 (4 c-3 d) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{5/2}}{\sqrt {a-a x} (c+d x)^4} \, dx,x,\sec (e+f x)\right )}{4 \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))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}-\frac {\left (5 a^3 (4 c-3 d) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{3/2}}{\sqrt {a-a x} (c+d x)^3} \, dx,x,\sec (e+f x)\right )}{12 (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))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}-\frac {\left (5 a^4 (4 c-3 d) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+a x}}{\sqrt {a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{8 (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))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))}-\frac {\left (5 a^5 (4 c-3 d) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{8 (c+d)^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))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))}-\frac {\left (5 a^5 (4 c-3 d) \tan (e+f x)\right ) \text {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 )}{4 (c+d)^3 \left (c^2-d^2\right ) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {5 a^4 (4 c-3 d) \arctan \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)}{4 (c-d)^{3/2} (c+d)^{9/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac {a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.45 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.03 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\frac {a^3 \left (-\frac {120 (4 c-3 d) \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}-\frac {\left (-72 c^4-478 c^3 d+336 c^2 d^2+28 c d^3+336 d^4+\left (-296 c^4-84 c^3 d-577 c^2 d^2+984 c d^3+198 d^4\right ) \cos (e+f x)+\left (-72 c^4-470 c^3 d+384 c^2 d^2+200 c d^3+48 d^4\right ) \cos (2 (e+f x))-88 c^4 \cos (3 (e+f x))+36 c^3 d \cos (3 (e+f x))+37 c^2 d^2 \cos (3 (e+f x))+24 c d^3 \cos (3 (e+f x))+6 d^4 \cos (3 (e+f x))\right ) \sin (e+f x)}{(d+c \cos (e+f x))^4}\right )}{96 (c-d) (c+d)^4 f} \]

[In]

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

[Out]

(a^3*((-120*(4*c - 3*d)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] - ((-72*c^4 - 47
8*c^3*d + 336*c^2*d^2 + 28*c*d^3 + 336*d^4 + (-296*c^4 - 84*c^3*d - 577*c^2*d^2 + 984*c*d^3 + 198*d^4)*Cos[e +
f*x] + (-72*c^4 - 470*c^3*d + 384*c^2*d^2 + 200*c*d^3 + 48*d^4)*Cos[2*(e + f*x)] - 88*c^4*Cos[3*(e + f*x)] +
36*c^3*d*Cos[3*(e + f*x)] + 37*c^2*d^2*Cos[3*(e + f*x)] + 24*c*d^3*Cos[3*(e + f*x)] + 6*d^4*Cos[3*(e + f*x)])*
Sin[e + f*x])/(d + c*Cos[e + f*x])^4))/(96*(c - d)*(c + d)^4*f)

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {16 a^{3} \left (\frac {-\frac {5 \left (4 c -3 d \right ) \left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{64 \left (c^{4}+4 c^{3} d +6 c^{2} d^{2}+4 c \,d^{3}+d^{4}\right )}+\frac {55 \left (c -d \right ) \left (4 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{192 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {73 \left (4 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{192 \left (c^{2}+2 c d +d^{2}\right )}+\frac {\left (44 c -49 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{64 \left (c +d \right ) \left (c -d \right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{4}}+\frac {5 \left (4 c -3 d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{64 \left (c^{5}+3 c^{4} d +2 c^{3} d^{2}-2 c^{2} d^{3}-3 c \,d^{4}-d^{5}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\)	\(303\)
default	\(\frac {16 a^{3} \left (\frac {-\frac {5 \left (4 c -3 d \right ) \left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{64 \left (c^{4}+4 c^{3} d +6 c^{2} d^{2}+4 c \,d^{3}+d^{4}\right )}+\frac {55 \left (c -d \right ) \left (4 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{192 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}-\frac {73 \left (4 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{192 \left (c^{2}+2 c d +d^{2}\right )}+\frac {\left (44 c -49 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{64 \left (c +d \right ) \left (c -d \right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{4}}+\frac {5 \left (4 c -3 d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{64 \left (c^{5}+3 c^{4} d +2 c^{3} d^{2}-2 c^{2} d^{3}-3 c \,d^{4}-d^{5}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\)	\(303\)
risch	\(\text {Expression too large to display}\)	\(1235\)

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^5,x,method=_RETURNVERBOSE)

[Out]

16/f*a^3*((-5/64*(4*c-3*d)*(c^2-2*c*d+d^2)/(c^4+4*c^3*d+6*c^2*d^2+4*c*d^3+d^4)*tan(1/2*f*x+1/2*e)^7+55/192*(c-
d)*(4*c-3*d)/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5-73/192*(4*c-3*d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e
)^3+1/64*(44*c-49*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)^4+5/6
4*(4*c-3*d)/(c^5+3*c^4*d+2*c^3*d^2-2*c^2*d^3-3*c*d^4-d^5)/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tan(1/2*f*x+1/2*e)
/((c+d)*(c-d))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (247) = 494\).

Time = 0.36 (sec) , antiderivative size = 1714, normalized size of antiderivative = 6.44 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\text {Too large to display} \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

[1/48*(15*(4*a^3*c*d^4 - 3*a^3*d^5 + (4*a^3*c^5 - 3*a^3*c^4*d)*cos(f*x + e)^4 + 4*(4*a^3*c^4*d - 3*a^3*c^3*d^2
)*cos(f*x + e)^3 + 6*(4*a^3*c^3*d^2 - 3*a^3*c^2*d^3)*cos(f*x + e)^2 + 4*(4*a^3*c^2*d^3 - 3*a^3*c*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*(2*a^3*c^5*d + 12*a^3
*c^4*d^2 + 41*a^3*c^3*d^3 - 84*a^3*c^2*d^4 - 43*a^3*c*d^5 + 72*a^3*d^6 + (88*a^3*c^6 - 36*a^3*c^5*d - 125*a^3*
c^4*d^2 + 12*a^3*c^3*d^3 + 31*a^3*c^2*d^4 + 24*a^3*c*d^5 + 6*a^3*d^6)*cos(f*x + e)^3 + (36*a^3*c^6 + 235*a^3*c
^5*d - 228*a^3*c^4*d^2 - 335*a^3*c^3*d^3 + 168*a^3*c^2*d^4 + 100*a^3*c*d^5 + 24*a^3*d^6)*cos(f*x + e)^2 + (8*a
^3*c^6 + 48*a^3*c^5*d + 164*a^3*c^4*d^2 - 276*a^3*c^3*d^3 - 217*a^3*c^2*d^4 + 228*a^3*c*d^5 + 45*a^3*d^6)*cos(
f*x + e))*sin(f*x + e))/((c^11 + 3*c^10*d + c^9*d^2 - 5*c^8*d^3 - 5*c^7*d^4 + c^6*d^5 + 3*c^5*d^6 + c^4*d^7)*f
*cos(f*x + e)^4 + 4*(c^10*d + 3*c^9*d^2 + c^8*d^3 - 5*c^7*d^4 - 5*c^6*d^5 + c^5*d^6 + 3*c^4*d^7 + c^3*d^8)*f*c
os(f*x + e)^3 + 6*(c^9*d^2 + 3*c^8*d^3 + c^7*d^4 - 5*c^6*d^5 - 5*c^5*d^6 + c^4*d^7 + 3*c^3*d^8 + c^2*d^9)*f*co
s(f*x + e)^2 + 4*(c^8*d^3 + 3*c^7*d^4 + c^6*d^5 - 5*c^5*d^6 - 5*c^4*d^7 + c^3*d^8 + 3*c^2*d^9 + c*d^10)*f*cos(
f*x + e) + (c^7*d^4 + 3*c^6*d^5 + c^5*d^6 - 5*c^4*d^7 - 5*c^3*d^8 + c^2*d^9 + 3*c*d^10 + d^11)*f), 1/24*(15*(4
*a^3*c*d^4 - 3*a^3*d^5 + (4*a^3*c^5 - 3*a^3*c^4*d)*cos(f*x + e)^4 + 4*(4*a^3*c^4*d - 3*a^3*c^3*d^2)*cos(f*x +
e)^3 + 6*(4*a^3*c^3*d^2 - 3*a^3*c^2*d^3)*cos(f*x + e)^2 + 4*(4*a^3*c^2*d^3 - 3*a^3*c*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))) + (2*a^3*c^5*d + 12*a^3*c
^4*d^2 + 41*a^3*c^3*d^3 - 84*a^3*c^2*d^4 - 43*a^3*c*d^5 + 72*a^3*d^6 + (88*a^3*c^6 - 36*a^3*c^5*d - 125*a^3*c^
4*d^2 + 12*a^3*c^3*d^3 + 31*a^3*c^2*d^4 + 24*a^3*c*d^5 + 6*a^3*d^6)*cos(f*x + e)^3 + (36*a^3*c^6 + 235*a^3*c^5
*d - 228*a^3*c^4*d^2 - 335*a^3*c^3*d^3 + 168*a^3*c^2*d^4 + 100*a^3*c*d^5 + 24*a^3*d^6)*cos(f*x + e)^2 + (8*a^3
*c^6 + 48*a^3*c^5*d + 164*a^3*c^4*d^2 - 276*a^3*c^3*d^3 - 217*a^3*c^2*d^4 + 228*a^3*c*d^5 + 45*a^3*d^6)*cos(f*
x + e))*sin(f*x + e))/((c^11 + 3*c^10*d + c^9*d^2 - 5*c^8*d^3 - 5*c^7*d^4 + c^6*d^5 + 3*c^5*d^6 + c^4*d^7)*f*c
os(f*x + e)^4 + 4*(c^10*d + 3*c^9*d^2 + c^8*d^3 - 5*c^7*d^4 - 5*c^6*d^5 + c^5*d^6 + 3*c^4*d^7 + c^3*d^8)*f*cos
(f*x + e)^3 + 6*(c^9*d^2 + 3*c^8*d^3 + c^7*d^4 - 5*c^6*d^5 - 5*c^5*d^6 + c^4*d^7 + 3*c^3*d^8 + c^2*d^9)*f*cos(
f*x + e)^2 + 4*(c^8*d^3 + 3*c^7*d^4 + c^6*d^5 - 5*c^5*d^6 - 5*c^4*d^7 + c^3*d^8 + 3*c^2*d^9 + c*d^10)*f*cos(f*
x + e) + (c^7*d^4 + 3*c^6*d^5 + c^5*d^6 - 5*c^4*d^7 - 5*c^3*d^8 + c^2*d^9 + 3*c*d^10 + d^11)*f)]

Sympy [F]

\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec {\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec {\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec {\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec {\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx\right ) \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c+d*sec(f*x+e))**5,x)

[Out]

a**3*(Integral(sec(e + f*x)/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e
+ f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*sec(e + f*x)**5), x) + Integral(3*sec(e + f*x)**2/(c**5 + 5*c**4*d
*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*
sec(e + f*x)**5), x) + Integral(3*sec(e + f*x)**3/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2
+ 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*sec(e + f*x)**5), x) + Integral(sec(e + f*x)
**4/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec
(e + f*x)**4 + d**5*sec(e + f*x)**5), x))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (247) = 494\).

Time = 0.48 (sec) , antiderivative size = 601, normalized size of antiderivative = 2.26 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\frac {\frac {15 \, {\left (4 \, a^{3} c - 3 \, a^{3} 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^{5} + 3 \, c^{4} d + 2 \, c^{3} d^{2} - 2 \, c^{2} d^{3} - 3 \, c d^{4} - d^{5}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {60 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 225 \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 315 \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 195 \, a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 45 \, a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 220 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 385 \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 55 \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 385 \, a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 165 \, a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 292 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 73 \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 511 \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 73 \, a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 219 \, a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 132 \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 249 \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 45 \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 309 \, a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 \, a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{5} + 3 \, c^{4} d + 2 \, c^{3} d^{2} - 2 \, c^{2} d^{3} - 3 \, c d^{4} - 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 )}^{4}}}{12 \, f} \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^5,x, algorithm="giac")

[Out]

1/12*(15*(4*a^3*c - 3*a^3*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^5 + 3*c^4*d + 2*c^3*d^2 - 2*c^2*d^3 - 3*c*d^4 - d^5)*sqrt(
-c^2 + d^2)) - (60*a^3*c^4*tan(1/2*f*x + 1/2*e)^7 - 225*a^3*c^3*d*tan(1/2*f*x + 1/2*e)^7 + 315*a^3*c^2*d^2*tan
(1/2*f*x + 1/2*e)^7 - 195*a^3*c*d^3*tan(1/2*f*x + 1/2*e)^7 + 45*a^3*d^4*tan(1/2*f*x + 1/2*e)^7 - 220*a^3*c^4*t
an(1/2*f*x + 1/2*e)^5 + 385*a^3*c^3*d*tan(1/2*f*x + 1/2*e)^5 + 55*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 - 385*a^3
*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 165*a^3*d^4*tan(1/2*f*x + 1/2*e)^5 + 292*a^3*c^4*tan(1/2*f*x + 1/2*e)^3 + 73*a
^3*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 511*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 73*a^3*c*d^3*tan(1/2*f*x + 1/2*e)^3
+ 219*a^3*d^4*tan(1/2*f*x + 1/2*e)^3 - 132*a^3*c^4*tan(1/2*f*x + 1/2*e) - 249*a^3*c^3*d*tan(1/2*f*x + 1/2*e)
+ 45*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e) + 309*a^3*c*d^3*tan(1/2*f*x + 1/2*e) + 147*a^3*d^4*tan(1/2*f*x + 1/2*e))
/((c^5 + 3*c^4*d + 2*c^3*d^2 - 2*c^2*d^3 - 3*c*d^4 - d^5)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2
- c - d)^4))/f

Mupad [B] (verification not implemented)

Time = 17.09 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.45 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx=\frac {\frac {55\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (4\,a^3\,c^2-7\,a^3\,c\,d+3\,a^3\,d^2\right )}{12\,{\left (c+d\right )}^3}-\frac {73\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,a^3\,c-3\,a^3\,d\right )}{12\,{\left (c+d\right )}^2}-\frac {5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (4\,a^3\,c^3-11\,a^3\,c^2\,d+10\,a^3\,c\,d^2-3\,a^3\,d^3\right )}{4\,{\left (c+d\right )}^4}+\frac {a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (44\,c-49\,d\right )}{4\,\left (c+d\right )\,\left (c-d\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (6\,c^4-12\,c^2\,d^2+6\,d^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-4\,c^4-8\,c^3\,d+8\,c\,d^3+4\,d^4\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (4\,c^4-8\,c^3\,d+8\,c\,d^3-4\,d^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (c^4-4\,c^3\,d+6\,c^2\,d^2-4\,c\,d^3+d^4\right )+4\,c\,d^3+4\,c^3\,d+c^4+d^4+6\,c^2\,d^2\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )\,\left (4\,c-3\,d\right )}{4\,f\,{\left (c+d\right )}^{9/2}\,{\left (c-d\right )}^{3/2}} \]

[In]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c + d/cos(e + f*x))^5),x)

[Out]

((55*tan(e/2 + (f*x)/2)^5*(4*a^3*c^2 + 3*a^3*d^2 - 7*a^3*c*d))/(12*(c + d)^3) - (73*tan(e/2 + (f*x)/2)^3*(4*a^
3*c - 3*a^3*d))/(12*(c + d)^2) - (5*tan(e/2 + (f*x)/2)^7*(4*a^3*c^3 - 3*a^3*d^3 + 10*a^3*c*d^2 - 11*a^3*c^2*d)
)/(4*(c + d)^4) + (a^3*tan(e/2 + (f*x)/2)*(44*c - 49*d))/(4*(c + d)*(c - d)))/(f*(tan(e/2 + (f*x)/2)^4*(6*c^4
+ 6*d^4 - 12*c^2*d^2) + tan(e/2 + (f*x)/2)^2*(8*c*d^3 - 8*c^3*d - 4*c^4 + 4*d^4) - tan(e/2 + (f*x)/2)^6*(8*c*d
^3 - 8*c^3*d + 4*c^4 - 4*d^4) + tan(e/2 + (f*x)/2)^8*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2) + 4*c*d^3 + 4
*c^3*d + c^4 + d^4 + 6*c^2*d^2)) + (5*a^3*atanh((tan(e/2 + (f*x)/2)*(c - d)^(1/2))/(c + d)^(1/2))*(4*c - 3*d))
/(4*f*(c + d)^(9/2)*(c - d)^(3/2))