3.232 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c+d \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=288 \[ \frac {\left (2 c^2-12 c d+45 d^2\right ) \tan (e+f x)}{15 f (c-d)^3 \left (a^3 \sec (e+f x)+a^3\right ) (c+d \sec (e+f x))}+\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sec (e+f x))}-\frac {2 d^3 (4 c+3 d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{a^3 f (c-d)^{9/2} (c+d)^{3/2}}+\frac {(2 c-9 d) \tan (e+f x)}{15 a f (c-d)^2 (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))}+\frac {\tan (e+f x)}{5 f (c-d) (a \sec (e+f x)+a)^3 (c+d \sec (e+f x))} \]

[Out]

-2*d^3*(4*c+3*d)*arctanh((c-d)^(1/2)*tan(1/2*e+1/2*f*x)/(c+d)^(1/2))/a^3/(c-d)^(9/2)/(c+d)^(3/2)/f+1/15*d*(2*c
^3-12*c^2*d+43*c*d^2+72*d^3)*tan(f*x+e)/a^3/(c-d)^4/(c+d)/f/(c+d*sec(f*x+e))+1/5*tan(f*x+e)/(c-d)/f/(a+a*sec(f
*x+e))^3/(c+d*sec(f*x+e))+1/15*(2*c-9*d)*tan(f*x+e)/a/(c-d)^2/f/(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))+1/15*(2*c^
2-12*c*d+45*d^2)*tan(f*x+e)/(c-d)^3/f/(a^3+a^3*sec(f*x+e))/(c+d*sec(f*x+e))

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Rubi [A]  time = 0.49, antiderivative size = 325, normalized size of antiderivative = 1.13, number of steps used = 8, 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} \[ \frac {\left (-12 c^2 d+2 c^3+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 f (c-d)^4 (c+d) \left (a^3 \sec (e+f x)+a^3\right )}+\frac {2 d^3 (4 c+3 d) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{a^2 f (c-d)^{9/2} (c+d)^{3/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {d \tan (e+f x)}{f \left (c^2-d^2\right ) (a \sec (e+f x)+a)^3 (c+d \sec (e+f x))}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a f (c-d)^3 (c+d) (a \sec (e+f x)+a)^2}+\frac {(c+6 d) \tan (e+f x)}{5 f (c-d)^2 (c+d) (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c + d*Sec[e + f*x])^2),x]

[Out]

((c + 6*d)*Tan[e + f*x])/(5*(c - d)^2*(c + d)*f*(a + a*Sec[e + f*x])^3) + ((2*c^2 - 10*c*d - 27*d^2)*Tan[e + f
*x])/(15*a*(c - d)^3*(c + d)*f*(a + a*Sec[e + f*x])^2) + (2*d^3*(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])/(a^2*(c - d)^(9/2)*(c + d)^(3/2)*f*Sqrt[a -
a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((2*c^3 - 12*c^2*d + 43*c*d^2 + 72*d^3)*Tan[e + f*x])/(15*(c - d)^
4*(c + d)*f*(a^3 + a^3*Sec[e + f*x])) - (d*Tan[e + f*x])/((c^2 - d^2)*f*(a + a*Sec[e + f*x])^3*(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 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 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 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]

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])

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (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)^{7/2} (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))^3 (c+d \sec (e+f x))}-\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {a^2 (c+3 d)-3 a^2 d x}{\sqrt {a-a x} (a+a x)^{7/2} (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+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^3 (c+d \sec (e+f x))}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {-a^4 \left (2 c^2-8 c d-15 d^2\right )-2 a^4 d (c+6 d) x}{\sqrt {a-a x} (a+a x)^{5/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{5 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+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a (c-d)^3 (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))^3 (c+d \sec (e+f x))}-\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {a^6 (c+d) \left (2 c^2-12 c d+45 d^2\right )+a^6 d \left (2 c^2-10 c d-27 d^2\right ) x}{\sqrt {a-a x} (a+a x)^{3/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{15 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+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a (c-d)^3 (c+d) f (a+a \sec (e+f x))^2}+\frac {\left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 (c-d)^4 (c+d) f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^3 (c+d \sec (e+f x))}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {15 a^8 d^3 (4 c+3 d)}{\sqrt {a-a x} \sqrt {a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{15 a^9 (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 {(c+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a (c-d)^3 (c+d) f (a+a \sec (e+f x))^2}+\frac {\left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 (c-d)^4 (c+d) f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^3 (c+d \sec (e+f x))}+\frac {\left (d^3 (4 c+3 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 )}{a (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 {(c+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a (c-d)^3 (c+d) f (a+a \sec (e+f x))^2}+\frac {\left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 (c-d)^4 (c+d) f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^3 (c+d \sec (e+f x))}+\frac {\left (2 d^3 (4 c+3 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 )}{a (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 {(c+6 d) \tan (e+f x)}{5 (c-d)^2 (c+d) f (a+a \sec (e+f x))^3}+\frac {\left (2 c^2-10 c d-27 d^2\right ) \tan (e+f x)}{15 a (c-d)^3 (c+d) f (a+a \sec (e+f x))^2}+\frac {2 d^3 (4 c+3 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^2 (c-d)^{9/2} (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 (c-d)^4 (c+d) f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x))^3 (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [C]  time = 7.09, size = 1772, normalized size = 6.15 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c + d*Sec[e + f*x])^2),x]

[Out]

((4*c + 3*d)*Cos[e/2 + (f*x)/2]^6*(d + c*Cos[e + f*x])^2*Sec[e + f*x]^5*(((16*I)*d^3*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]]) + (16*d^3
*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[Cos[2*e
] - I*Sin[2*e]])))/((-c + d)^4*(c + d)*(a + a*Sec[e + f*x])^3*(c + d*Sec[e + f*x])^2) + (Cos[e/2 + (f*x)/2]*(d
+ c*Cos[e + f*x])*Sec[e/2]*Sec[e]*Sec[e + f*x]^5*(-55*c^5*Sin[(f*x)/2] + 135*c^4*d*Sin[(f*x)/2] - 20*c^3*d^2*
Sin[(f*x)/2] - 810*c^2*d^3*Sin[(f*x)/2] - 450*c*d^4*Sin[(f*x)/2] + 150*d^5*Sin[(f*x)/2] + 47*c^5*Sin[(3*f*x)/2
] - 137*c^4*d*Sin[(3*f*x)/2] + 88*c^3*d^2*Sin[(3*f*x)/2] + 812*c^2*d^3*Sin[(3*f*x)/2] + 690*c*d^4*Sin[(3*f*x)/
2] + 75*d^5*Sin[(3*f*x)/2] - 50*c^5*Sin[e - (f*x)/2] + 130*c^4*d*Sin[e - (f*x)/2] - 10*c^3*d^2*Sin[e - (f*x)/2
] - 1030*c^2*d^3*Sin[e - (f*x)/2] - 990*c*d^4*Sin[e - (f*x)/2] - 150*d^5*Sin[e - (f*x)/2] + 50*c^5*Sin[e + (f*
x)/2] - 130*c^4*d*Sin[e + (f*x)/2] + 10*c^3*d^2*Sin[e + (f*x)/2] + 1030*c^2*d^3*Sin[e + (f*x)/2] + 765*c*d^4*S
in[e + (f*x)/2] - 150*d^5*Sin[e + (f*x)/2] - 55*c^5*Sin[2*e + (f*x)/2] + 135*c^4*d*Sin[2*e + (f*x)/2] - 20*c^3
*d^2*Sin[2*e + (f*x)/2] - 810*c^2*d^3*Sin[2*e + (f*x)/2] - 675*c*d^4*Sin[2*e + (f*x)/2] - 150*d^5*Sin[2*e + (f
*x)/2] - 30*c^5*Sin[e + (3*f*x)/2] + 90*c^4*d*Sin[e + (3*f*x)/2] - 60*c^3*d^2*Sin[e + (3*f*x)/2] - 360*c^2*d^3
*Sin[e + (3*f*x)/2] - 30*c*d^4*Sin[e + (3*f*x)/2] + 75*d^5*Sin[e + (3*f*x)/2] + 47*c^5*Sin[2*e + (3*f*x)/2] -
137*c^4*d*Sin[2*e + (3*f*x)/2] + 88*c^3*d^2*Sin[2*e + (3*f*x)/2] + 812*c^2*d^3*Sin[2*e + (3*f*x)/2] + 525*c*d^
4*Sin[2*e + (3*f*x)/2] - 75*d^5*Sin[2*e + (3*f*x)/2] - 30*c^5*Sin[3*e + (3*f*x)/2] + 90*c^4*d*Sin[3*e + (3*f*x
)/2] - 60*c^3*d^2*Sin[3*e + (3*f*x)/2] - 360*c^2*d^3*Sin[3*e + (3*f*x)/2] - 195*c*d^4*Sin[3*e + (3*f*x)/2] - 7
5*d^5*Sin[3*e + (3*f*x)/2] + 20*c^5*Sin[e + (5*f*x)/2] - 76*c^4*d*Sin[e + (5*f*x)/2] + 106*c^3*d^2*Sin[e + (5*
f*x)/2] + 346*c^2*d^3*Sin[e + (5*f*x)/2] + 219*c*d^4*Sin[e + (5*f*x)/2] + 15*d^5*Sin[e + (5*f*x)/2] - 15*c^5*S
in[2*e + (5*f*x)/2] + 45*c^4*d*Sin[2*e + (5*f*x)/2] - 30*c^3*d^2*Sin[2*e + (5*f*x)/2] - 90*c^2*d^3*Sin[2*e + (
5*f*x)/2] + 75*c*d^4*Sin[2*e + (5*f*x)/2] + 15*d^5*Sin[2*e + (5*f*x)/2] + 20*c^5*Sin[3*e + (5*f*x)/2] - 76*c^4
*d*Sin[3*e + (5*f*x)/2] + 106*c^3*d^2*Sin[3*e + (5*f*x)/2] + 346*c^2*d^3*Sin[3*e + (5*f*x)/2] + 144*c*d^4*Sin[
3*e + (5*f*x)/2] - 15*d^5*Sin[3*e + (5*f*x)/2] - 15*c^5*Sin[4*e + (5*f*x)/2] + 45*c^4*d*Sin[4*e + (5*f*x)/2] -
30*c^3*d^2*Sin[4*e + (5*f*x)/2] - 90*c^2*d^3*Sin[4*e + (5*f*x)/2] - 15*d^5*Sin[4*e + (5*f*x)/2] + 7*c^5*Sin[2
*e + (7*f*x)/2] - 27*c^4*d*Sin[2*e + (7*f*x)/2] + 38*c^3*d^2*Sin[2*e + (7*f*x)/2] + 72*c^2*d^3*Sin[2*e + (7*f*
x)/2] + 15*c*d^4*Sin[2*e + (7*f*x)/2] + 15*c*d^4*Sin[3*e + (7*f*x)/2] + 7*c^5*Sin[4*e + (7*f*x)/2] - 27*c^4*d*
Sin[4*e + (7*f*x)/2] + 38*c^3*d^2*Sin[4*e + (7*f*x)/2] + 72*c^2*d^3*Sin[4*e + (7*f*x)/2]))/(120*c*(-c + d)^4*(
c + d)*f*(a + a*Sec[e + f*x])^3*(c + d*Sec[e + f*x])^2)

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fricas [B]  time = 0.54, size = 1693, normalized size = 5.88 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

[1/30*(15*(4*c*d^4 + 3*d^5 + (4*c^2*d^3 + 3*c*d^4)*cos(f*x + e)^4 + (12*c^2*d^3 + 13*c*d^4 + 3*d^5)*cos(f*x +
e)^3 + 3*(4*c^2*d^3 + 7*c*d^4 + 3*d^5)*cos(f*x + e)^2 + (4*c^2*d^3 + 15*c*d^4 + 9*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*c^5*d - 12*c^4*d^2 + 41*c^3*d^3 +
84*c^2*d^4 - 43*c*d^5 - 72*d^6 + (7*c^6 - 27*c^5*d + 31*c^4*d^2 + 99*c^3*d^3 - 23*c^2*d^4 - 72*c*d^5 - 15*d^6
)*cos(f*x + e)^3 + (6*c^6 - 29*c^5*d + 51*c^4*d^2 + 193*c^3*d^3 + 60*c^2*d^4 - 164*c*d^5 - 117*d^6)*cos(f*x +
e)^2 + (2*c^6 - 6*c^5*d + 5*c^4*d^2 + 147*c^3*d^3 + 164*c^2*d^4 - 141*c*d^5 - 171*d^6)*cos(f*x + e))*sin(f*x +
e))/((a^3*c^8 - 3*a^3*c^7*d + a^3*c^6*d^2 + 5*a^3*c^5*d^3 - 5*a^3*c^4*d^4 - a^3*c^3*d^5 + 3*a^3*c^2*d^6 - a^3
*c*d^7)*f*cos(f*x + e)^4 + (3*a^3*c^8 - 8*a^3*c^7*d + 16*a^3*c^5*d^3 - 10*a^3*c^4*d^4 - 8*a^3*c^3*d^5 + 8*a^3*
c^2*d^6 - a^3*d^8)*f*cos(f*x + e)^3 + 3*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5
+ 2*a^3*c^2*d^6 + 2*a^3*c*d^7 - a^3*d^8)*f*cos(f*x + e)^2 + (a^3*c^8 - 8*a^3*c^6*d^2 + 8*a^3*c^5*d^3 + 10*a^3
*c^4*d^4 - 16*a^3*c^3*d^5 + 8*a^3*c*d^7 - 3*a^3*d^8)*f*cos(f*x + e) + (a^3*c^7*d - 3*a^3*c^6*d^2 + a^3*c^5*d^3
+ 5*a^3*c^4*d^4 - 5*a^3*c^3*d^5 - a^3*c^2*d^6 + 3*a^3*c*d^7 - a^3*d^8)*f), -1/15*(15*(4*c*d^4 + 3*d^5 + (4*c^
2*d^3 + 3*c*d^4)*cos(f*x + e)^4 + (12*c^2*d^3 + 13*c*d^4 + 3*d^5)*cos(f*x + e)^3 + 3*(4*c^2*d^3 + 7*c*d^4 + 3*
d^5)*cos(f*x + e)^2 + (4*c^2*d^3 + 15*c*d^4 + 9*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^5*d - 12*c^4*d^2 + 41*c^3*d^3 + 84*c^2*d^4 - 43*c*d^5 -
72*d^6 + (7*c^6 - 27*c^5*d + 31*c^4*d^2 + 99*c^3*d^3 - 23*c^2*d^4 - 72*c*d^5 - 15*d^6)*cos(f*x + e)^3 + (6*c^
6 - 29*c^5*d + 51*c^4*d^2 + 193*c^3*d^3 + 60*c^2*d^4 - 164*c*d^5 - 117*d^6)*cos(f*x + e)^2 + (2*c^6 - 6*c^5*d
+ 5*c^4*d^2 + 147*c^3*d^3 + 164*c^2*d^4 - 141*c*d^5 - 171*d^6)*cos(f*x + e))*sin(f*x + e))/((a^3*c^8 - 3*a^3*c
^7*d + a^3*c^6*d^2 + 5*a^3*c^5*d^3 - 5*a^3*c^4*d^4 - a^3*c^3*d^5 + 3*a^3*c^2*d^6 - a^3*c*d^7)*f*cos(f*x + e)^4
+ (3*a^3*c^8 - 8*a^3*c^7*d + 16*a^3*c^5*d^3 - 10*a^3*c^4*d^4 - 8*a^3*c^3*d^5 + 8*a^3*c^2*d^6 - a^3*d^8)*f*cos
(f*x + e)^3 + 3*(a^3*c^8 - 2*a^3*c^7*d - 2*a^3*c^6*d^2 + 6*a^3*c^5*d^3 - 6*a^3*c^3*d^5 + 2*a^3*c^2*d^6 + 2*a^3
*c*d^7 - a^3*d^8)*f*cos(f*x + e)^2 + (a^3*c^8 - 8*a^3*c^6*d^2 + 8*a^3*c^5*d^3 + 10*a^3*c^4*d^4 - 16*a^3*c^3*d^
5 + 8*a^3*c*d^7 - 3*a^3*d^8)*f*cos(f*x + e) + (a^3*c^7*d - 3*a^3*c^6*d^2 + a^3*c^5*d^3 + 5*a^3*c^4*d^4 - 5*a^3
*c^3*d^5 - a^3*c^2*d^6 + 3*a^3*c*d^7 - a^3*d^8)*f)]

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giac [B]  time = 0.97, size = 951, normalized size = 3.30 \[ \text {result too large to display} \]

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))^2,x, algorithm="giac")

[Out]

-1/60*(120*d^4*tan(1/2*f*x + 1/2*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)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)) + 120*(4*c*d^3 + 3*d^4)*(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^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)*sqrt(-c^2 + d^2)) - (3*a^1
2*c^8*tan(1/2*f*x + 1/2*e)^5 - 24*a^12*c^7*d*tan(1/2*f*x + 1/2*e)^5 + 84*a^12*c^6*d^2*tan(1/2*f*x + 1/2*e)^5 -
168*a^12*c^5*d^3*tan(1/2*f*x + 1/2*e)^5 + 210*a^12*c^4*d^4*tan(1/2*f*x + 1/2*e)^5 - 168*a^12*c^3*d^5*tan(1/2*
f*x + 1/2*e)^5 + 84*a^12*c^2*d^6*tan(1/2*f*x + 1/2*e)^5 - 24*a^12*c*d^7*tan(1/2*f*x + 1/2*e)^5 + 3*a^12*d^8*ta
n(1/2*f*x + 1/2*e)^5 - 10*a^12*c^8*tan(1/2*f*x + 1/2*e)^3 + 100*a^12*c^7*d*tan(1/2*f*x + 1/2*e)^3 - 420*a^12*c
^6*d^2*tan(1/2*f*x + 1/2*e)^3 + 980*a^12*c^5*d^3*tan(1/2*f*x + 1/2*e)^3 - 1400*a^12*c^4*d^4*tan(1/2*f*x + 1/2*
e)^3 + 1260*a^12*c^3*d^5*tan(1/2*f*x + 1/2*e)^3 - 700*a^12*c^2*d^6*tan(1/2*f*x + 1/2*e)^3 + 220*a^12*c*d^7*tan
(1/2*f*x + 1/2*e)^3 - 30*a^12*d^8*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^8*tan(1/2*f*x + 1/2*e) - 180*a^12*c^7*d*t
an(1/2*f*x + 1/2*e) + 1020*a^12*c^6*d^2*tan(1/2*f*x + 1/2*e) - 3180*a^12*c^5*d^3*tan(1/2*f*x + 1/2*e) + 5850*a
^12*c^4*d^4*tan(1/2*f*x + 1/2*e) - 6540*a^12*c^3*d^5*tan(1/2*f*x + 1/2*e) + 4380*a^12*c^2*d^6*tan(1/2*f*x + 1/
2*e) - 1620*a^12*c*d^7*tan(1/2*f*x + 1/2*e) + 255*a^12*d^8*tan(1/2*f*x + 1/2*e))/(a^15*c^10 - 10*a^15*c^9*d +
45*a^15*c^8*d^2 - 120*a^15*c^7*d^3 + 210*a^15*c^6*d^4 - 252*a^15*c^5*d^5 + 210*a^15*c^4*d^6 - 120*a^15*c^3*d^7
+ 45*a^15*c^2*d^8 - 10*a^15*c*d^9 + a^15*d^10))/f

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maple [A]  time = 0.94, size = 284, normalized size = 0.99 \[ \frac {\frac {\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{2}}{5}-\frac {2 \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c d}{5}+\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d^{2}}{5}-\frac {2 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{2}}{3}+\frac {8 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c d}{3}-2 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d^{2}+\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{2}-6 c d \tan \left (\frac {e}{2}+\frac {f x}{2}\right )+17 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d^{2}}{\left (c^{2}-2 c d +d^{2}\right ) \left (c -d \right )^{2}}+\frac {16 d^{3} \left (-\frac {d \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{2 \left (c +d \right ) \left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )}-\frac {\left (4 c +3 d \right ) \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c -d \right )^{4}}}{4 f \,a^{3}} \]

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))^2,x)

[Out]

1/4/f/a^3*(1/(c^2-2*c*d+d^2)/(c-d)^2*(1/5*tan(1/2*e+1/2*f*x)^5*c^2-2/5*tan(1/2*e+1/2*f*x)^5*c*d+1/5*tan(1/2*e+
1/2*f*x)^5*d^2-2/3*tan(1/2*e+1/2*f*x)^3*c^2+8/3*tan(1/2*e+1/2*f*x)^3*c*d-2*tan(1/2*e+1/2*f*x)^3*d^2+tan(1/2*e+
1/2*f*x)*c^2-6*c*d*tan(1/2*e+1/2*f*x)+17*tan(1/2*e+1/2*f*x)*d^2)+16*d^3/(c-d)^4*(-1/2*d/(c+d)*tan(1/2*e+1/2*f*
x)/(tan(1/2*e+1/2*f*x)^2*c-tan(1/2*e+1/2*f*x)^2*d-c-d)-1/2*(4*c+3*d)/(c+d)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2
*e+1/2*f*x)*(c-d)/((c+d)*(c-d))^(1/2))))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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))^2,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 details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 2.12, size = 464, normalized size = 1.61 \[ \frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{20\,a^3\,f\,{\left (c-d\right )}^2}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {2\,\left (c^2-d^2\right )\,\left (\frac {1}{a^3\,{\left (c-d\right )}^2}-\frac {c^2-d^2}{2\,a^3\,{\left (c-d\right )}^4}\right )}{{\left (c-d\right )}^2}-\frac {3}{2\,a^3\,{\left (c-d\right )}^2}+\frac {{\left (c+d\right )}^2}{4\,a^3\,{\left (c-d\right )}^4}\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {1}{3\,a^3\,{\left (c-d\right )}^2}-\frac {c^2-d^2}{6\,a^3\,{\left (c-d\right )}^4}\right )}{f}+\frac {2\,d^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (c+d\right )\,\left (a^3\,c^5-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a^3\,c^5-5\,a^3\,c^4\,d+10\,a^3\,c^3\,d^2-10\,a^3\,c^2\,d^3+5\,a^3\,c\,d^4-a^3\,d^5\right )+a^3\,d^5-3\,a^3\,c\,d^4-3\,a^3\,c^4\,d+2\,a^3\,c^2\,d^3+2\,a^3\,c^3\,d^2\right )}+\frac {d^3\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^5-5{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^4\,d+10{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^3\,d^2-10{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,d^3+5{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c\,d^4-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,d^5}{\sqrt {c+d}\,{\left (c-d\right )}^{9/2}}\right )\,\left (4\,c+3\,d\right )\,2{}\mathrm {i}}{a^3\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c + d/cos(e + f*x))^2),x)

[Out]

tan(e/2 + (f*x)/2)^5/(20*a^3*f*(c - d)^2) - (tan(e/2 + (f*x)/2)*((2*(c^2 - d^2)*(1/(a^3*(c - d)^2) - (c^2 - d^
2)/(2*a^3*(c - d)^4)))/(c - d)^2 - 3/(2*a^3*(c - d)^2) + (c + d)^2/(4*a^3*(c - d)^4)))/f - (tan(e/2 + (f*x)/2)
^3*(1/(3*a^3*(c - d)^2) - (c^2 - d^2)/(6*a^3*(c - d)^4)))/f + (2*d^4*tan(e/2 + (f*x)/2))/(f*(c + d)*(a^3*c^5 -
tan(e/2 + (f*x)/2)^2*(a^3*c^5 - a^3*d^5 + 5*a^3*c*d^4 - 5*a^3*c^4*d - 10*a^3*c^2*d^3 + 10*a^3*c^3*d^2) + a^3*
d^5 - 3*a^3*c*d^4 - 3*a^3*c^4*d + 2*a^3*c^2*d^3 + 2*a^3*c^3*d^2)) + (d^3*atan((c^5*tan(e/2 + (f*x)/2)*1i - d^5
*tan(e/2 + (f*x)/2)*1i + c*d^4*tan(e/2 + (f*x)/2)*5i - c^4*d*tan(e/2 + (f*x)/2)*5i - c^2*d^3*tan(e/2 + (f*x)/2
)*10i + c^3*d^2*tan(e/2 + (f*x)/2)*10i)/((c + d)^(1/2)*(c - d)^(9/2)))*(4*c + 3*d)*2i)/(a^3*f*(c + d)^(3/2)*(c
- d)^(9/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sec ^{3}{\left (e + f x \right )} + 3 c^{2} \sec ^{2}{\left (e + f x \right )} + 3 c^{2} \sec {\left (e + f x \right )} + c^{2} + 2 c d \sec ^{4}{\left (e + f x \right )} + 6 c d \sec ^{3}{\left (e + f x \right )} + 6 c d \sec ^{2}{\left (e + f x \right )} + 2 c d \sec {\left (e + f x \right )} + d^{2} \sec ^{5}{\left (e + f x \right )} + 3 d^{2} \sec ^{4}{\left (e + f x \right )} + 3 d^{2} \sec ^{3}{\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx}{a^{3}} \]

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))**2,x)

[Out]

Integral(sec(e + f*x)/(c**2*sec(e + f*x)**3 + 3*c**2*sec(e + f*x)**2 + 3*c**2*sec(e + f*x) + c**2 + 2*c*d*sec(
e + f*x)**4 + 6*c*d*sec(e + f*x)**3 + 6*c*d*sec(e + f*x)**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**5 + 3*d*
*2*sec(e + f*x)**4 + 3*d**2*sec(e + f*x)**3 + d**2*sec(e + f*x)**2), x)/a**3

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