\(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx\) [201]

Optimal result

Integrand size = 31, antiderivative size = 276 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\frac {a^2 \left (12 c^2-16 c d+7 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{4 (c-d)^{5/2} (c+d)^{9/2} f}-\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sec (e+f x))} \]

[Out]

1/4*a^2*(12*c^2-16*c*d+7*d^2)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/(c-d)^(5/2)/(c+d)^(9/2)/f-1/
4*a^2*(c-d)*tan(f*x+e)/d/(c+d)/f/(c+d*sec(f*x+e))^4+1/12*a^2*(c+8*d)*tan(f*x+e)/d/(c+d)^2/f/(c+d*sec(f*x+e))^3
+1/24*a^2*(2*c^2+16*c*d-21*d^2)*tan(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sec(f*x+e))^2+1/24*a^2*(2*c^3+16*c^2*d-59*c*
d^2+32*d^3)*tan(f*x+e)/(c-d)^2/d/(c+d)^4/f/(c+d*sec(f*x+e))

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4072, 100, 156, 12, 95, 211} \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=-\frac {a^3 \left (12 c^2-16 c d+7 d^2\right ) \tan (e+f x) \arctan \left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{4 f (c-d)^{5/2} (c+d)^{9/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 d f (c-d) (c+d)^3 (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 d f (c-d)^2 (c+d)^4 (c+d \sec (e+f x))}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d f (c+d)^2 (c+d \sec (e+f x))^3}-\frac {a^2 (c-d) \tan (e+f x)}{4 d f (c+d) (c+d \sec (e+f x))^4} \]

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^5,x]

[Out]

-1/4*(a^3*(12*c^2 - 16*c*d + 7*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)^(5/2)*(c + d)^(9/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) -
(a^2*(c - d)*Tan[e + f*x])/(4*d*(c + d)*f*(c + d*Sec[e + f*x])^4) + (a^2*(c + 8*d)*Tan[e + f*x])/(12*d*(c + d)
^2*f*(c + d*Sec[e + f*x])^3) + (a^2*(2*c^2 + 16*c*d - 21*d^2)*Tan[e + f*x])/(24*(c - d)*d*(c + d)^3*f*(c + d*S
ec[e + f*x])^2) + (a^2*(2*c^3 + 16*c^2*d - 59*c*d^2 + 32*d^3)*Tan[e + f*x])/(24*(c - d)^2*d*(c + d)^4*f*(c + d
*Sec[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 100

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

Rule 156

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] && ILtQ[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)^{3/2}}{\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 {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {(a \tan (e+f x)) \text {Subst}\left (\int \frac {-8 a^3 d-a^3 (c+7 d) x}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)^4} \, dx,x,\sec (e+f x)\right )}{4 d (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {-21 a^5 (c-d) d-2 a^5 (c-d) (c+8 d) x}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)^3} \, dx,x,\sec (e+f x)\right )}{12 a d (c+d) \left (c^2-d^2\right ) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {-2 a^7 (19 c-16 d) (c-d) d+a^7 (c-d) \left (21 d^2-2 c (c+8 d)\right ) x}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{24 a^3 d (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 {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sec (e+f x))}+\frac {\tan (e+f x) \text {Subst}\left (\int -\frac {3 a^9 (c-d) d \left (12 c^2-16 c d+7 d^2\right )}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{24 a^5 d (c+d) \left (c^2-d^2\right )^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sec (e+f x))}-\frac {\left (a^4 (c-d) \left (12 c^2-16 c d+7 d^2\right ) \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) \left (c^2-d^2\right )^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sec (e+f x))}-\frac {\left (a^4 (c-d) \left (12 c^2-16 c d+7 d^2\right ) \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) \left (c^2-d^2\right )^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = -\frac {a^3 \left (12 c^2-16 c d+7 d^2\right ) \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)^{5/2} (c+d)^{9/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^2 (c-d) \tan (e+f x)}{4 d (c+d) f (c+d \sec (e+f x))^4}+\frac {a^2 (c+8 d) \tan (e+f x)}{12 d (c+d)^2 f (c+d \sec (e+f x))^3}+\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \tan (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sec (e+f x))^2}+\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \tan (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sec (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.01 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\frac {a^2 \left (-\frac {24 \left (12 c^2-16 c d+7 d^2\right ) \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+\frac {\left (24 c^5+192 c^4 d-446 c^3 d^2+128 c^2 d^3-148 c d^4+160 d^5+\left (144 c^5-172 c^4 d+208 c^3 d^2-785 c^2 d^3+368 c d^4+102 d^5\right ) \cos (e+f x)+2 \left (12 c^5+96 c^4 d-227 c^3 d^2+32 c^2 d^3+44 c d^4+16 d^5\right ) \cos (2 (e+f x))+48 c^5 \cos (3 (e+f x))-68 c^4 d \cos (3 (e+f x))-16 c^3 d^2 \cos (3 (e+f x))+5 c^2 d^3 \cos (3 (e+f x))+16 c d^4 \cos (3 (e+f x))+6 d^5 \cos (3 (e+f x))\right ) \sin (e+f x)}{(d+c \cos (e+f x))^4}\right )}{96 (c-d)^2 (c+d)^4 f} \]

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c + d*Sec[e + f*x])^5,x]

[Out]

(a^2*((-24*(12*c^2 - 16*c*d + 7*d^2)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] + (
(24*c^5 + 192*c^4*d - 446*c^3*d^2 + 128*c^2*d^3 - 148*c*d^4 + 160*d^5 + (144*c^5 - 172*c^4*d + 208*c^3*d^2 - 7
85*c^2*d^3 + 368*c*d^4 + 102*d^5)*Cos[e + f*x] + 2*(12*c^5 + 96*c^4*d - 227*c^3*d^2 + 32*c^2*d^3 + 44*c*d^4 +
16*d^5)*Cos[2*(e + f*x)] + 48*c^5*Cos[3*(e + f*x)] - 68*c^4*d*Cos[3*(e + f*x)] - 16*c^3*d^2*Cos[3*(e + f*x)] +
5*c^2*d^3*Cos[3*(e + f*x)] + 16*c*d^4*Cos[3*(e + f*x)] + 6*d^5*Cos[3*(e + f*x)])*Sin[e + f*x])/(d + c*Cos[e +
f*x])^4))/(96*(c - d)^2*(c + d)^4*f)

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(\frac {8 a^{2} \left (-\frac {\frac {\left (12 c^{2}-16 c d +7 d^{2}\right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{32 c^{4}+128 c^{3} d +192 c^{2} d^{2}+128 c \,d^{3}+32 d^{4}}-\frac {11 \left (12 c^{2}-16 c d +7 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{96 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}+\frac {\left (156 c^{2}-272 c d +83 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{96 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}-\frac {\left (20 c^{2}-48 c d +25 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{32 \left (c +d \right ) \left (c^{2}-2 c d +d^{2}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{4}}+\frac {\left (12 c^{2}-16 c d +7 d^{2}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{32 \left (c^{6}+2 c^{5} d -c^{4} d^{2}-4 c^{3} d^{3}-c^{2} d^{4}+2 c \,d^{5}+d^{6}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\)	\(352\)
default	\(\frac {8 a^{2} \left (-\frac {\frac {\left (12 c^{2}-16 c d +7 d^{2}\right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{32 c^{4}+128 c^{3} d +192 c^{2} d^{2}+128 c \,d^{3}+32 d^{4}}-\frac {11 \left (12 c^{2}-16 c d +7 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{96 \left (c^{3}+3 c^{2} d +3 c \,d^{2}+d^{3}\right )}+\frac {\left (156 c^{2}-272 c d +83 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{96 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}-\frac {\left (20 c^{2}-48 c d +25 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{32 \left (c +d \right ) \left (c^{2}-2 c d +d^{2}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{4}}+\frac {\left (12 c^{2}-16 c d +7 d^{2}\right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{32 \left (c^{6}+2 c^{5} d -c^{4} d^{2}-4 c^{3} d^{3}-c^{2} d^{4}+2 c \,d^{5}+d^{6}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\)	\(352\)
risch	\(\text {Expression too large to display}\)	\(1557\)

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^5,x,method=_RETURNVERBOSE)

[Out]

8/f*a^2*(-(1/32*(12*c^2-16*c*d+7*d^2)*(c-d)/(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)^7-11/96*(12
*c^2-16*c*d+7*d^2)/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5+1/96*(156*c^2-272*c*d+83*d^2)/(c-d)/(c^2+2*c
*d+d^2)*tan(1/2*f*x+1/2*e)^3-1/32*(20*c^2-48*c*d+25*d^2)/(c+d)/(c^2-2*c*d+d^2)*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+1/32*(12*c^2-16*c*d+7*d^2)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2
*c*d^5+d^6)/((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. 925 vs. \(2 (257) = 514\).

Time = 0.40 (sec) , antiderivative size = 1908, normalized size of antiderivative = 6.91 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\text {Too large to display} \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

[1/48*(3*(12*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6 + (12*a^2*c^6 - 16*a^2*c^5*d + 7*a^2*c^4*d^2)*cos(f*x + e)
^4 + 4*(12*a^2*c^5*d - 16*a^2*c^4*d^2 + 7*a^2*c^3*d^3)*cos(f*x + e)^3 + 6*(12*a^2*c^4*d^2 - 16*a^2*c^3*d^3 + 7
*a^2*c^2*d^4)*cos(f*x + e)^2 + 4*(12*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*d^5)*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^2*c^5*d^2 + 16*a^2*c^4*d^3 - 61*a^2*c
^3*d^4 + 16*a^2*c^2*d^5 + 59*a^2*c*d^6 - 32*a^2*d^7 + (48*a^2*c^7 - 68*a^2*c^6*d - 64*a^2*c^5*d^2 + 73*a^2*c^4
*d^3 + 32*a^2*c^3*d^4 + a^2*c^2*d^5 - 16*a^2*c*d^6 - 6*a^2*d^7)*cos(f*x + e)^3 + (12*a^2*c^7 + 96*a^2*c^6*d -
239*a^2*c^5*d^2 - 64*a^2*c^4*d^3 + 271*a^2*c^3*d^4 - 16*a^2*c^2*d^5 - 44*a^2*c*d^6 - 16*a^2*d^7)*cos(f*x + e)^
2 + (8*a^2*c^6*d + 64*a^2*c^5*d^2 - 208*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 221*a^2*c^2*d^5 - 80*a^2*c*d^6 - 21*a^2
*d^7)*cos(f*x + e))*sin(f*x + e))/((c^12 + 2*c^11*d - 2*c^10*d^2 - 6*c^9*d^3 + 6*c^7*d^5 + 2*c^6*d^6 - 2*c^5*d
^7 - c^4*d^8)*f*cos(f*x + e)^4 + 4*(c^11*d + 2*c^10*d^2 - 2*c^9*d^3 - 6*c^8*d^4 + 6*c^6*d^6 + 2*c^5*d^7 - 2*c^
4*d^8 - c^3*d^9)*f*cos(f*x + e)^3 + 6*(c^10*d^2 + 2*c^9*d^3 - 2*c^8*d^4 - 6*c^7*d^5 + 6*c^5*d^7 + 2*c^4*d^8 -
2*c^3*d^9 - c^2*d^10)*f*cos(f*x + e)^2 + 4*(c^9*d^3 + 2*c^8*d^4 - 2*c^7*d^5 - 6*c^6*d^6 + 6*c^4*d^8 + 2*c^3*d^
9 - 2*c^2*d^10 - c*d^11)*f*cos(f*x + e) + (c^8*d^4 + 2*c^7*d^5 - 2*c^6*d^6 - 6*c^5*d^7 + 6*c^3*d^9 + 2*c^2*d^1
0 - 2*c*d^11 - d^12)*f), 1/24*(3*(12*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6 + (12*a^2*c^6 - 16*a^2*c^5*d + 7*a
^2*c^4*d^2)*cos(f*x + e)^4 + 4*(12*a^2*c^5*d - 16*a^2*c^4*d^2 + 7*a^2*c^3*d^3)*cos(f*x + e)^3 + 6*(12*a^2*c^4*
d^2 - 16*a^2*c^3*d^3 + 7*a^2*c^2*d^4)*cos(f*x + e)^2 + 4*(12*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*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*a^2*c
^5*d^2 + 16*a^2*c^4*d^3 - 61*a^2*c^3*d^4 + 16*a^2*c^2*d^5 + 59*a^2*c*d^6 - 32*a^2*d^7 + (48*a^2*c^7 - 68*a^2*c
^6*d - 64*a^2*c^5*d^2 + 73*a^2*c^4*d^3 + 32*a^2*c^3*d^4 + a^2*c^2*d^5 - 16*a^2*c*d^6 - 6*a^2*d^7)*cos(f*x + e)
^3 + (12*a^2*c^7 + 96*a^2*c^6*d - 239*a^2*c^5*d^2 - 64*a^2*c^4*d^3 + 271*a^2*c^3*d^4 - 16*a^2*c^2*d^5 - 44*a^2
*c*d^6 - 16*a^2*d^7)*cos(f*x + e)^2 + (8*a^2*c^6*d + 64*a^2*c^5*d^2 - 208*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 221*a
^2*c^2*d^5 - 80*a^2*c*d^6 - 21*a^2*d^7)*cos(f*x + e))*sin(f*x + e))/((c^12 + 2*c^11*d - 2*c^10*d^2 - 6*c^9*d^3
+ 6*c^7*d^5 + 2*c^6*d^6 - 2*c^5*d^7 - c^4*d^8)*f*cos(f*x + e)^4 + 4*(c^11*d + 2*c^10*d^2 - 2*c^9*d^3 - 6*c^8*
d^4 + 6*c^6*d^6 + 2*c^5*d^7 - 2*c^4*d^8 - c^3*d^9)*f*cos(f*x + e)^3 + 6*(c^10*d^2 + 2*c^9*d^3 - 2*c^8*d^4 - 6*
c^7*d^5 + 6*c^5*d^7 + 2*c^4*d^8 - 2*c^3*d^9 - c^2*d^10)*f*cos(f*x + e)^2 + 4*(c^9*d^3 + 2*c^8*d^4 - 2*c^7*d^5
- 6*c^6*d^6 + 6*c^4*d^8 + 2*c^3*d^9 - 2*c^2*d^10 - c*d^11)*f*cos(f*x + e) + (c^8*d^4 + 2*c^7*d^5 - 2*c^6*d^6 -
6*c^5*d^7 + 6*c^3*d^9 + 2*c^2*d^10 - 2*c*d^11 - d^12)*f)]

Sympy [F]

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

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2/(c+d*sec(f*x+e))**5,x)

[Out]

a**2*(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(2*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(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))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(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. 710 vs. \(2 (257) = 514\).

Time = 0.51 (sec) , antiderivative size = 710, normalized size of antiderivative = 2.57 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\frac {\frac {3 \, {\left (12 \, a^{2} c^{2} - 16 \, a^{2} c d + 7 \, a^{2} d^{2}\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^{6} + 2 \, c^{5} d - c^{4} d^{2} - 4 \, c^{3} d^{3} - c^{2} d^{4} + 2 \, c d^{5} + d^{6}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {36 \, a^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 156 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 273 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 243 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 111 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 21 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 132 \, a^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 308 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 121 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 231 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 253 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 77 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 156 \, a^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 116 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 345 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 199 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 189 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 83 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 60 \, a^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 36 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 177 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 81 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 75 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{6} + 2 \, c^{5} d - c^{4} d^{2} - 4 \, c^{3} d^{3} - c^{2} d^{4} + 2 \, c d^{5} + d^{6}\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))^2/(c+d*sec(f*x+e))^5,x, algorithm="giac")

[Out]

1/12*(3*(12*a^2*c^2 - 16*a^2*c*d + 7*a^2*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(-2*c + 2*d) + arctan(-(c*t
an(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/((c^6 + 2*c^5*d - c^4*d^2 - 4*c^3*d^3 - c^2*d
^4 + 2*c*d^5 + d^6)*sqrt(-c^2 + d^2)) - (36*a^2*c^5*tan(1/2*f*x + 1/2*e)^7 - 156*a^2*c^4*d*tan(1/2*f*x + 1/2*e
)^7 + 273*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^7 - 243*a^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^7 + 111*a^2*c*d^4*tan(1/2*
f*x + 1/2*e)^7 - 21*a^2*d^5*tan(1/2*f*x + 1/2*e)^7 - 132*a^2*c^5*tan(1/2*f*x + 1/2*e)^5 + 308*a^2*c^4*d*tan(1/
2*f*x + 1/2*e)^5 - 121*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 231*a^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 + 253*a^2*c
*d^4*tan(1/2*f*x + 1/2*e)^5 - 77*a^2*d^5*tan(1/2*f*x + 1/2*e)^5 + 156*a^2*c^5*tan(1/2*f*x + 1/2*e)^3 - 116*a^2
*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 345*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 199*a^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^
3 + 189*a^2*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 83*a^2*d^5*tan(1/2*f*x + 1/2*e)^3 - 60*a^2*c^5*tan(1/2*f*x + 1/2*e)
- 36*a^2*c^4*d*tan(1/2*f*x + 1/2*e) + 177*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e) + 147*a^2*c^2*d^3*tan(1/2*f*x + 1/
2*e) - 81*a^2*c*d^4*tan(1/2*f*x + 1/2*e) - 75*a^2*d^5*tan(1/2*f*x + 1/2*e))/((c^6 + 2*c^5*d - c^4*d^2 - 4*c^3*
d^3 - c^2*d^4 + 2*c*d^5 + d^6)*(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 = 17.02 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.59 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^5} \, dx=\frac {\frac {11\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (12\,a^2\,c^2-16\,a^2\,c\,d+7\,a^2\,d^2\right )}{12\,{\left (c+d\right )}^3}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (12\,a^2\,c^3-28\,a^2\,c^2\,d+23\,a^2\,c\,d^2-7\,a^2\,d^3\right )}{4\,{\left (c+d\right )}^4}-\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (156\,c^2-272\,c\,d+83\,d^2\right )}{12\,{\left (c+d\right )}^2\,\left (c-d\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (20\,c^2-48\,c\,d+25\,d^2\right )}{4\,\left (c+d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (6\,c^4-12\,c^2\,d^2+6\,d^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-4\,c^4-8\,c^3\,d+8\,c\,d^3+4\,d^4\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (4\,c^4-8\,c^3\,d+8\,c\,d^3-4\,d^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (c^4-4\,c^3\,d+6\,c^2\,d^2-4\,c\,d^3+d^4\right )+4\,c\,d^3+4\,c^3\,d+c^4+d^4+6\,c^2\,d^2\right )}+\frac {a^2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-2\,d\right )\,\left (c^2-2\,c\,d+d^2\right )}{2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,\left (12\,c^2-16\,c\,d+7\,d^2\right )}{4\,f\,{\left (c+d\right )}^{9/2}\,{\left (c-d\right )}^{5/2}} \]

[In]

int((a + a/cos(e + f*x))^2/(cos(e + f*x)*(c + d/cos(e + f*x))^5),x)

[Out]

((11*tan(e/2 + (f*x)/2)^5*(12*a^2*c^2 + 7*a^2*d^2 - 16*a^2*c*d))/(12*(c + d)^3) - (tan(e/2 + (f*x)/2)^7*(12*a^
2*c^3 - 7*a^2*d^3 + 23*a^2*c*d^2 - 28*a^2*c^2*d))/(4*(c + d)^4) - (a^2*tan(e/2 + (f*x)/2)^3*(156*c^2 - 272*c*d
+ 83*d^2))/(12*(c + d)^2*(c - d)) + (a^2*tan(e/2 + (f*x)/2)*(20*c^2 - 48*c*d + 25*d^2))/(4*(c + d)*(c^2 - 2*c
*d + d^2)))/(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)) + (a^2*atanh((tan(e/2 + (f*x)/2)*
(2*c - 2*d)*(c^2 - 2*c*d + d^2))/(2*(c + d)^(1/2)*(c - d)^(5/2)))*(12*c^2 - 16*c*d + 7*d^2))/(4*f*(c + d)^(9/2
)*(c - d)^(5/2))