3.210 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx\)
Optimal. Leaf size=183 \[ \frac{d \left (-12 c^2 d+8 c^3+12 c d^2-3 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{2 a f}-\frac{d \tan (e+f x) \left (d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)+4 \left (-16 c^2 d+3 c^3+12 c d^2-4 d^3\right )\right )}{6 a f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^3}{f (a \sec (e+f x)+a)}-\frac{d (3 c-4 d) \tan (e+f x) (c+d \sec (e+f x))^2}{3 a f} \]
[Out]
(d*(8*c^3 - 12*c^2*d + 12*c*d^2 - 3*d^3)*ArcTanh[Sin[e + f*x]])/(2*a*f) - ((3*c - 4*d)*d*(c + d*Sec[e + f*x])^
2*Tan[e + f*x])/(3*a*f) + ((c - d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) - (d*(4*(3*c^
3 - 16*c^2*d + 12*c*d^2 - 4*d^3) + d*(6*c^2 - 20*c*d + 9*d^2)*Sec[e + f*x])*Tan[e + f*x])/(6*a*f)
________________________________________________________________________________________
Rubi [A] time = 0.335273, antiderivative size = 236, normalized size of antiderivative =
1.29, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules
used = {3987, 98, 153, 147, 63, 217, 203} \[ \frac{d \left (-12 c^2 d+8 c^3+12 c d^2-3 d^3\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{d \tan (e+f x) \left (d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)+4 \left (-16 c^2 d+3 c^3+12 c d^2-4 d^3\right )\right )}{6 a f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^3}{f (a \sec (e+f x)+a)}-\frac{d (3 c-4 d) \tan (e+f x) (c+d \sec (e+f x))^2}{3 a f} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + a*Sec[e + f*x]),x]
[Out]
(d*(8*c^3 - 12*c^2*d + 12*c*d^2 - 3*d^3)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f
*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - ((3*c - 4*d)*d*(c + d*Sec[e + f*x])^2*Tan[e + f*x
])/(3*a*f) + ((c - d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) - (d*(4*(3*c^3 - 16*c^2*d
+ 12*c*d^2 - 4*d^3) + d*(6*c^2 - 20*c*d + 9*d^2)*Sec[e + f*x])*Tan[e + f*x])/(6*a*f)
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 98
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 153
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
Rule 147
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^4}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^2 \left (-a^2 (4 c-3 d) d+a^2 (3 c-4 d) d x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(3 c-4 d) d (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a f}+\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x) \left (a^4 d \left (12 c^2-15 c d+8 d^2\right )-a^4 d \left (6 c^2-20 c d+9 d^2\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{3 a^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(3 c-4 d) d (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a f}+\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{d \left (4 \left (3 c^3-16 c^2 d+12 c d^2-4 d^3\right )+d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a f}-\frac{\left (a d \left (8 c^3-12 c^2 d+12 c d^2-3 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(3 c-4 d) d (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a f}+\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{d \left (4 \left (3 c^3-16 c^2 d+12 c d^2-4 d^3\right )+d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a f}+\frac{\left (d \left (8 c^3-12 c^2 d+12 c d^2-3 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{(3 c-4 d) d (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a f}+\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{d \left (4 \left (3 c^3-16 c^2 d+12 c d^2-4 d^3\right )+d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a f}+\frac{\left (d \left (8 c^3-12 c^2 d+12 c d^2-3 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d \left (8 c^3-12 c^2 d+12 c d^2-3 d^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{(3 c-4 d) d (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a f}+\frac{(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{d \left (4 \left (3 c^3-16 c^2 d+12 c d^2-4 d^3\right )+d \left (6 c^2-20 c d+9 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a f}\\ \end{align*}
Mathematica [B] time = 6.45251, size = 1243, normalized size = 6.79 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + a*Sec[e + f*x]),x]
[Out]
((-8*c^3*d + 12*c^2*d^2 - 12*c*d^3 + 3*d^4)*Cos[e/2 + (f*x)/2]^2*Cos[e + f*x]^3*Log[Cos[e/2 + (f*x)/2] - Sin[e
/2 + (f*x)/2]]*(c + d*Sec[e + f*x])^4)/(f*(d + c*Cos[e + f*x])^4*(a + a*Sec[e + f*x])) + ((8*c^3*d - 12*c^2*d^
2 + 12*c*d^3 - 3*d^4)*Cos[e/2 + (f*x)/2]^2*Cos[e + f*x]^3*Log[Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]]*(c + d*
Sec[e + f*x])^4)/(f*(d + c*Cos[e + f*x])^4*(a + a*Sec[e + f*x])) + (Cos[e/2 + (f*x)/2]*Sec[e/2]*Sec[e]*(c + d*
Sec[e + f*x])^4*(-18*c^4*Sin[(f*x)/2] + 72*c^3*d*Sin[(f*x)/2] - 36*c^2*d^2*Sin[(f*x)/2] + 24*c*d^3*Sin[(f*x)/2
] + 6*d^4*Sin[(f*x)/2] + 18*c^4*Sin[(3*f*x)/2] - 72*c^3*d*Sin[(3*f*x)/2] + 180*c^2*d^2*Sin[(3*f*x)/2] - 108*c*
d^3*Sin[(3*f*x)/2] + 39*d^4*Sin[(3*f*x)/2] - 72*c^2*d^2*Sin[e - (f*x)/2] + 48*c*d^3*Sin[e - (f*x)/2] - 24*d^4*
Sin[e - (f*x)/2] - 36*c^2*d^2*Sin[e + (f*x)/2] + 24*c*d^3*Sin[e + (f*x)/2] - 6*d^4*Sin[e + (f*x)/2] - 18*c^4*S
in[2*e + (f*x)/2] + 72*c^3*d*Sin[2*e + (f*x)/2] - 144*c^2*d^2*Sin[2*e + (f*x)/2] + 96*c*d^3*Sin[2*e + (f*x)/2]
- 24*d^4*Sin[2*e + (f*x)/2] + 72*c^2*d^2*Sin[e + (3*f*x)/2] - 36*c*d^3*Sin[e + (3*f*x)/2] + 21*d^4*Sin[e + (3
*f*x)/2] + 18*c^4*Sin[2*e + (3*f*x)/2] - 72*c^3*d*Sin[2*e + (3*f*x)/2] + 72*c^2*d^2*Sin[2*e + (3*f*x)/2] - 36*
c*d^3*Sin[2*e + (3*f*x)/2] + 9*d^4*Sin[2*e + (3*f*x)/2] - 36*c^2*d^2*Sin[3*e + (3*f*x)/2] + 36*c*d^3*Sin[3*e +
(3*f*x)/2] - 9*d^4*Sin[3*e + (3*f*x)/2] + 36*c^2*d^2*Sin[e + (5*f*x)/2] - 12*c*d^3*Sin[e + (5*f*x)/2] + 7*d^4
*Sin[e + (5*f*x)/2] - 6*c^4*Sin[2*e + (5*f*x)/2] + 24*c^3*d*Sin[2*e + (5*f*x)/2] + 12*c*d^3*Sin[2*e + (5*f*x)/
2] + d^4*Sin[2*e + (5*f*x)/2] + 12*c*d^3*Sin[3*e + (5*f*x)/2] - 3*d^4*Sin[3*e + (5*f*x)/2] - 6*c^4*Sin[4*e + (
5*f*x)/2] + 24*c^3*d*Sin[4*e + (5*f*x)/2] - 36*c^2*d^2*Sin[4*e + (5*f*x)/2] + 36*c*d^3*Sin[4*e + (5*f*x)/2] -
9*d^4*Sin[4*e + (5*f*x)/2] + 6*c^4*Sin[2*e + (7*f*x)/2] - 24*c^3*d*Sin[2*e + (7*f*x)/2] + 72*c^2*d^2*Sin[2*e +
(7*f*x)/2] - 48*c*d^3*Sin[2*e + (7*f*x)/2] + 16*d^4*Sin[2*e + (7*f*x)/2] + 36*c^2*d^2*Sin[3*e + (7*f*x)/2] -
24*c*d^3*Sin[3*e + (7*f*x)/2] + 10*d^4*Sin[3*e + (7*f*x)/2] + 6*c^4*Sin[4*e + (7*f*x)/2] - 24*c^3*d*Sin[4*e +
(7*f*x)/2] + 36*c^2*d^2*Sin[4*e + (7*f*x)/2] - 24*c*d^3*Sin[4*e + (7*f*x)/2] + 6*d^4*Sin[4*e + (7*f*x)/2]))/(4
8*f*(d + c*Cos[e + f*x])^4*(a + a*Sec[e + f*x]))
________________________________________________________________________________________
Maple [B] time = 0.079, size = 596, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e)),x)
[Out]
6/a/f*d^3/(tan(1/2*f*x+1/2*e)+1)*c-2/a/f*d^3/(tan(1/2*f*x+1/2*e)+1)^2*c-4/a/f*tan(1/2*f*x+1/2*e)*c*d^3+4/a/f*l
n(tan(1/2*f*x+1/2*e)+1)*c^3*d-6/a/f*ln(tan(1/2*f*x+1/2*e)+1)*c^2*d^2+6/a/f*ln(tan(1/2*f*x+1/2*e)+1)*c*d^3+6/a/
f*tan(1/2*f*x+1/2*e)*c^2*d^2-4/a/f*ln(tan(1/2*f*x+1/2*e)-1)*c^3*d+6/a/f*ln(tan(1/2*f*x+1/2*e)-1)*c^2*d^2-6/a/f
*ln(tan(1/2*f*x+1/2*e)-1)*c*d^3-6/a/f*d^2/(tan(1/2*f*x+1/2*e)-1)*c^2+6/a/f*d^3/(tan(1/2*f*x+1/2*e)-1)*c+2/a/f*
d^3/(tan(1/2*f*x+1/2*e)-1)^2*c-5/2/a/f*d^4/(tan(1/2*f*x+1/2*e)-1)-1/a/f*d^4/(tan(1/2*f*x+1/2*e)-1)^2+1/a/f*tan
(1/2*f*x+1/2*e)*d^4-1/3/a/f*d^4/(tan(1/2*f*x+1/2*e)+1)^3-3/2/a/f*ln(tan(1/2*f*x+1/2*e)+1)*d^4-5/2/a/f*d^4/(tan
(1/2*f*x+1/2*e)+1)+1/a/f*d^4/(tan(1/2*f*x+1/2*e)+1)^2-1/3/a/f*d^4/(tan(1/2*f*x+1/2*e)-1)^3+1/f*c^4/a*tan(1/2*f
*x+1/2*e)+3/2/a/f*ln(tan(1/2*f*x+1/2*e)-1)*d^4-4/a/f*tan(1/2*f*x+1/2*e)*c^3*d-6/a/f*d^2/(tan(1/2*f*x+1/2*e)+1)
*c^2
________________________________________________________________________________________
Maxima [B] time = 1.02911, size = 805, normalized size = 4.4 \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)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="maxima")
[Out]
1/6*(d^4*(2*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 16*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^5/(c
os(f*x + e) + 1)^5)/(a - 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - a
*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) - 9*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a + 9*log(sin(f*x + e)/(cos
(f*x + e) + 1) - 1)/a + 6*sin(f*x + e)/(a*(cos(f*x + e) + 1))) - 12*c*d^3*(2*(sin(f*x + e)/(cos(f*x + e) + 1)
- 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a - 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^4/(cos(
f*x + e) + 1)^4) - 3*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a + 3*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a
+ 2*sin(f*x + e)/(a*(cos(f*x + e) + 1))) - 36*c^2*d^2*(log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a - log(sin(f
*x + e)/(cos(f*x + e) + 1) - 1)/a - 2*sin(f*x + e)/((a - a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e)
+ 1)) - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 24*c^3*d*(log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a - log(sin(
f*x + e)/(cos(f*x + e) + 1) - 1)/a - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 6*c^4*sin(f*x + e)/(a*(cos(f*x + e
) + 1)))/f
________________________________________________________________________________________
Fricas [A] time = 0.51661, size = 693, normalized size = 3.79 \begin{align*} \frac{3 \,{\left ({\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{4} +{\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left ({\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{4} +{\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, d^{4} + 2 \,{\left (3 \, c^{4} - 12 \, c^{3} d + 36 \, c^{2} d^{2} - 24 \, c d^{3} + 8 \, d^{4}\right )} \cos \left (f x + e\right )^{3} +{\left (36 \, c^{2} d^{2} - 12 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{2} +{\left (12 \, c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \,{\left (a f \cos \left (f x + e\right )^{4} + a f \cos \left (f x + e\right )^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="fricas")
[Out]
1/12*(3*((8*c^3*d - 12*c^2*d^2 + 12*c*d^3 - 3*d^4)*cos(f*x + e)^4 + (8*c^3*d - 12*c^2*d^2 + 12*c*d^3 - 3*d^4)*
cos(f*x + e)^3)*log(sin(f*x + e) + 1) - 3*((8*c^3*d - 12*c^2*d^2 + 12*c*d^3 - 3*d^4)*cos(f*x + e)^4 + (8*c^3*d
- 12*c^2*d^2 + 12*c*d^3 - 3*d^4)*cos(f*x + e)^3)*log(-sin(f*x + e) + 1) + 2*(2*d^4 + 2*(3*c^4 - 12*c^3*d + 36
*c^2*d^2 - 24*c*d^3 + 8*d^4)*cos(f*x + e)^3 + (36*c^2*d^2 - 12*c*d^3 + 7*d^4)*cos(f*x + e)^2 + (12*c*d^3 - d^4
)*cos(f*x + e))*sin(f*x + e))/(a*f*cos(f*x + e)^4 + a*f*cos(f*x + e)^3)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{4} \sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{4 c d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 c^{2} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{4 c^{3} d \sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)*(c+d*sec(f*x+e))**4/(a+a*sec(f*x+e)),x)
[Out]
(Integral(c**4*sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(d**4*sec(e + f*x)**5/(sec(e + f*x) + 1), x) + In
tegral(4*c*d**3*sec(e + f*x)**4/(sec(e + f*x) + 1), x) + Integral(6*c**2*d**2*sec(e + f*x)**3/(sec(e + f*x) +
1), x) + Integral(4*c**3*d*sec(e + f*x)**2/(sec(e + f*x) + 1), x))/a
________________________________________________________________________________________
Giac [B] time = 1.35509, size = 487, normalized size = 2.66 \begin{align*} \frac{\frac{3 \,{\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{3 \,{\left (8 \, c^{3} d - 12 \, c^{2} d^{2} + 12 \, c d^{3} - 3 \, d^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} + \frac{6 \,{\left (c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a} - \frac{2 \,{\left (36 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 36 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 15 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 72 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 48 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 16 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 36 \, c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 12 \, c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 \, d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{3} a}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="giac")
[Out]
1/6*(3*(8*c^3*d - 12*c^2*d^2 + 12*c*d^3 - 3*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 3*(8*c^3*d - 12*c^2*d^
2 + 12*c*d^3 - 3*d^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a + 6*(c^4*tan(1/2*f*x + 1/2*e) - 4*c^3*d*tan(1/2*f*x
+ 1/2*e) + 6*c^2*d^2*tan(1/2*f*x + 1/2*e) - 4*c*d^3*tan(1/2*f*x + 1/2*e) + d^4*tan(1/2*f*x + 1/2*e))/a - 2*(3
6*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 - 36*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 15*d^4*tan(1/2*f*x + 1/2*e)^5 - 72*c^2*d^
2*tan(1/2*f*x + 1/2*e)^3 + 48*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 16*d^4*tan(1/2*f*x + 1/2*e)^3 + 36*c^2*d^2*tan(1/
2*f*x + 1/2*e) - 12*c*d^3*tan(1/2*f*x + 1/2*e) + 9*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a
))/f