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 \[ \frac{d \left (-12 c^2 d+2 c^3+43 c d^2+72 d^3\right ) \tan (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sec (e+f x))}+\frac{\left (2 c^2-12 c d+45 d^2\right ) \tan (e+f x)}{15 f (c-d)^3 \left (a^3 \sec (e+f x)+a^3\right ) (c+d \sec (e+f x))}-\frac{2 d^3 (4 c+3 d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a^3 f (c-d)^{9/2} (c+d)^{3/2}}+\frac{(2 c-9 d) \tan (e+f x)}{15 a f (c-d)^2 (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))}+\frac{\tan (e+f x)}{5 f (c-d) (a \sec (e+f x)+a)^3 (c+d \sec (e+f x))} \]
[Out]
(-2*d^3*(4*c + 3*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(a^3*(c - d)^(9/2)*(c + d)^(3/2)*f) +
(d*(2*c^3 - 12*c^2*d + 43*c*d^2 + 72*d^3)*Tan[e + f*x])/(15*a^3*(c - d)^4*(c + d)*f*(c + d*Sec[e + f*x])) + T
an[e + f*x]/(5*(c - d)*f*(a + a*Sec[e + f*x])^3*(c + d*Sec[e + f*x])) + ((2*c - 9*d)*Tan[e + f*x])/(15*a*(c -
d)^2*f*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])) + ((2*c^2 - 12*c*d + 45*d^2)*Tan[e + f*x])/(15*(c - d)^3*f
*(a^3 + a^3*Sec[e + f*x])*(c + d*Sec[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.486091, 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 3987
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])
Rule 103
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])
Rule 152
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 93
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^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*}
Mathematica [C] time = 7.01324, 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)
________________________________________________________________________________________
Maple [A] time = 0.104, size = 284, normalized size = 1. \begin{align*}{\frac{1}{4\,f{a}^{3}} \left ({\frac{1}{ \left ({c}^{2}-2\,cd+{d}^{2} \right ) \left ( c-d \right ) ^{2}} \left ({\frac{{c}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,cd}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{{d}^{2}}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,{c}^{2}}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{8\,cd}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}{d}^{2}+\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ){c}^{2}-6\,cd\tan \left ( 1/2\,fx+e/2 \right ) +17\,\tan \left ( 1/2\,fx+e/2 \right ){d}^{2} \right ) }+16\,{\frac{{d}^{3}}{ \left ( c-d \right ) ^{4}} \left ( -1/2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) d}{ \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}-1/2\,{\frac{4\,c+3\,d}{ \left ( c+d \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)/(a+a*sec(f*x+e))^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*f*x+1/2*e)^5*c^2-2/5*tan(1/2*f*x+1/2*e)^5*c*d+1/5*tan(1/2*f*
x+1/2*e)^5*d^2-2/3*tan(1/2*f*x+1/2*e)^3*c^2+8/3*tan(1/2*f*x+1/2*e)^3*c*d-2*tan(1/2*f*x+1/2*e)^3*d^2+tan(1/2*f*
x+1/2*e)*c^2-6*c*d*tan(1/2*f*x+1/2*e)+17*tan(1/2*f*x+1/2*e)*d^2)+16*d^3/(c-d)^4*(-1/2*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)-1/2*(4*c+3*d)/(c+d)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2
*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 0.706203, size = 3638, normalized size = 12.63 \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))^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)]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.21708, size = 1284, normalized size = 4.46 \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))^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