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

Optimal. Leaf size=254 \[ -\frac{(b c-a d) \left (a^2 \left (-\left (-5 c^2 d^2+6 c^4+2 d^4\right )\right )+2 a b c d \left (4 c^2-d^2\right )-b^2 c^2 \left (c^2+2 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^3 f (c-d)^{5/2} (c+d)^{5/2}}+\frac{a^3 x}{c^3}+\frac{(b c-a d)^2 \left (5 a c^2-2 a d^2-3 b c d\right ) \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right )^2 (c \cos (e+f x)+d)}+\frac{(b c-a d)^2 \sin (e+f x) (a \cos (e+f x)+b)}{2 c f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2} \]

[Out]

(a^3*x)/c^3 - ((b*c - a*d)*(2*a*b*c*d*(4*c^2 - d^2) - b^2*c^2*(c^2 + 2*d^2) - a^2*(6*c^4 - 5*c^2*d^2 + 2*d^4))
*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^3*(c - d)^(5/2)*(c + d)^(5/2)*f) + ((b*c - a*d)^2*(b
+ a*Cos[e + f*x])*Sin[e + f*x])/(2*c*(c^2 - d^2)*f*(d + c*Cos[e + f*x])^2) + ((b*c - a*d)^2*(5*a*c^2 - 3*b*c*d
- 2*a*d^2)*Sin[e + f*x])/(2*c^2*(c^2 - d^2)^2*f*(d + c*Cos[e + f*x]))

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Rubi [A]  time = 1.13244, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3941, 2792, 3021, 2735, 2659, 208} \[ -\frac{(b c-a d) \left (a^2 \left (-\left (-5 c^2 d^2+6 c^4+2 d^4\right )\right )+2 a b c d \left (4 c^2-d^2\right )-b^2 c^2 \left (c^2+2 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^3 f (c-d)^{5/2} (c+d)^{5/2}}+\frac{a^3 x}{c^3}+\frac{(b c-a d)^2 \left (5 a c^2-2 a d^2-3 b c d\right ) \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right )^2 (c \cos (e+f x)+d)}+\frac{(b c-a d)^2 \sin (e+f x) (a \cos (e+f x)+b)}{2 c f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x)/c^3 - ((b*c - a*d)*(2*a*b*c*d*(4*c^2 - d^2) - b^2*c^2*(c^2 + 2*d^2) - a^2*(6*c^4 - 5*c^2*d^2 + 2*d^4))
*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^3*(c - d)^(5/2)*(c + d)^(5/2)*f) + ((b*c - a*d)^2*(b
+ a*Cos[e + f*x])*Sin[e + f*x])/(2*c*(c^2 - d^2)*f*(d + c*Cos[e + f*x])^2) + ((b*c - a*d)^2*(5*a*c^2 - 3*b*c*d
- 2*a*d^2)*Sin[e + f*x])/(2*c^2*(c^2 - d^2)^2*f*(d + c*Cos[e + f*x]))

Rule 3941

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Int[
((b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^(m + n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& NeQ[b*c - a*d, 0] && IntegerQ[m] && IntegerQ[n] && LeQ[-2, m + n, 0]

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(a+b \sec (e+f x))^3}{(c+d \sec (e+f x))^3} \, dx &=\int \frac{(b+a \cos (e+f x))^3}{(d+c \cos (e+f x))^3} \, dx\\ &=\frac{(b c-a d)^2 (b+a \cos (e+f x)) \sin (e+f x)}{2 c \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac{\int \frac{5 a b^2 c^2-4 a^2 b c d-2 b^3 c d+a^3 d^2+\left (b^3 c^2-2 a^3 c d-4 a b^2 c d+a^2 b \left (6 c^2-d^2\right )\right ) \cos (e+f x)+2 a^3 \left (c^2-d^2\right ) \cos ^2(e+f x)}{(d+c \cos (e+f x))^2} \, dx}{2 c \left (c^2-d^2\right )}\\ &=\frac{(b c-a d)^2 (b+a \cos (e+f x)) \sin (e+f x)}{2 c \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac{(b c-a d)^2 \left (5 a c^2-3 b c d-2 a d^2\right ) \sin (e+f x)}{2 c^2 \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac{\int \frac{-c \left (9 a b^2 c^2 d-3 a^2 b c \left (2 c^2+d^2\right )-b^3 c \left (c^2+2 d^2\right )+a^3 \left (4 c^2 d-d^3\right )\right )+2 a^3 \left (c^2-d^2\right )^2 \cos (e+f x)}{d+c \cos (e+f x)} \, dx}{2 c^2 \left (c^2-d^2\right )^2}\\ &=\frac{a^3 x}{c^3}+\frac{(b c-a d)^2 (b+a \cos (e+f x)) \sin (e+f x)}{2 c \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac{(b c-a d)^2 \left (5 a c^2-3 b c d-2 a d^2\right ) \sin (e+f x)}{2 c^2 \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}-\frac{\left (9 a b^2 c^4 d-3 a^2 b c^3 \left (2 c^2+d^2\right )-b^3 c^3 \left (c^2+2 d^2\right )+a^3 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )\right ) \int \frac{1}{d+c \cos (e+f x)} \, dx}{2 c^3 \left (c^2-d^2\right )^2}\\ &=\frac{a^3 x}{c^3}+\frac{(b c-a d)^2 (b+a \cos (e+f x)) \sin (e+f x)}{2 c \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac{(b c-a d)^2 \left (5 a c^2-3 b c d-2 a d^2\right ) \sin (e+f x)}{2 c^2 \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}-\frac{\left (9 a b^2 c^4 d-3 a^2 b c^3 \left (2 c^2+d^2\right )-b^3 c^3 \left (c^2+2 d^2\right )+a^3 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{c^3 \left (c^2-d^2\right )^2 f}\\ &=\frac{a^3 x}{c^3}+\frac{(b c-a d) \left (6 a^2 c^4+b^2 c^4-8 a b c^3 d-5 a^2 c^2 d^2+2 b^2 c^2 d^2+2 a b c d^3+2 a^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^3 (c-d)^{5/2} (c+d)^{5/2} f}+\frac{(b c-a d)^2 (b+a \cos (e+f x)) \sin (e+f x)}{2 c \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac{(b c-a d)^2 \left (5 a c^2-3 b c d-2 a d^2\right ) \sin (e+f x)}{2 c^2 \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}\\ \end{align*}

Mathematica [B]  time = 2.25181, size = 517, normalized size = 2.04 \[ \frac{\frac{-18 a^2 b c^4 d^2 \sin (e+f x)+3 a^2 b c^3 d^3 \sin (2 (e+f x))-12 a^2 b c^5 d \sin (2 (e+f x))+6 a^3 c^4 d^2 \sin (2 (e+f x))+10 a^3 c^3 d^3 \sin (e+f x)-3 a^3 c^2 d^4 \sin (2 (e+f x))+8 a^3 c d \left (c^2-d^2\right )^2 (e+f x) \cos (e+f x)+2 a^3 \left (c^3-c d^2\right )^2 (e+f x) \cos (2 (e+f x))-6 a^3 c^2 d^4 e-6 a^3 c^2 d^4 f x+2 a^3 c^6 e+2 a^3 c^6 f x-4 a^3 c d^5 \sin (e+f x)+4 a^3 d^6 e+4 a^3 d^6 f x+3 a b^2 c^4 d^2 \sin (2 (e+f x))+12 a b^2 c^3 d^3 \sin (e+f x)+6 a b^2 c^5 d \sin (e+f x)+6 a b^2 c^6 \sin (2 (e+f x))-8 b^3 c^4 d^2 \sin (e+f x)-3 b^3 c^5 d \sin (2 (e+f x))+2 b^3 c^6 \sin (e+f x)}{\left (c^2-d^2\right )^2 (c \cos (e+f x)+d)^2}-\frac{4 \left (3 a^2 b c^3 \left (2 c^2+d^2\right )+a^3 \left (5 c^2 d^3-6 c^4 d-2 d^5\right )-9 a b^2 c^4 d+b^3 c^3 \left (c^2+2 d^2\right )\right ) \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}}{4 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*(-9*a*b^2*c^4*d + 3*a^2*b*c^3*(2*c^2 + d^2) + b^3*c^3*(c^2 + 2*d^2) + a^3*(-6*c^4*d + 5*c^2*d^3 - 2*d^5))
*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(5/2) + (2*a^3*c^6*e - 6*a^3*c^2*d^4*e + 4*
a^3*d^6*e + 2*a^3*c^6*f*x - 6*a^3*c^2*d^4*f*x + 4*a^3*d^6*f*x + 8*a^3*c*d*(c^2 - d^2)^2*(e + f*x)*Cos[e + f*x]
+ 2*a^3*(c^3 - c*d^2)^2*(e + f*x)*Cos[2*(e + f*x)] + 2*b^3*c^6*Sin[e + f*x] + 6*a*b^2*c^5*d*Sin[e + f*x] - 18
*a^2*b*c^4*d^2*Sin[e + f*x] - 8*b^3*c^4*d^2*Sin[e + f*x] + 10*a^3*c^3*d^3*Sin[e + f*x] + 12*a*b^2*c^3*d^3*Sin[
e + f*x] - 4*a^3*c*d^5*Sin[e + f*x] + 6*a*b^2*c^6*Sin[2*(e + f*x)] - 12*a^2*b*c^5*d*Sin[2*(e + f*x)] - 3*b^3*c
^5*d*Sin[2*(e + f*x)] + 6*a^3*c^4*d^2*Sin[2*(e + f*x)] + 3*a*b^2*c^4*d^2*Sin[2*(e + f*x)] + 3*a^2*b*c^3*d^3*Si
n[2*(e + f*x)] - 3*a^3*c^2*d^4*Sin[2*(e + f*x)])/((c^2 - d^2)^2*(d + c*Cos[e + f*x])^2))/(4*c^3*f)

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Maple [B]  time = 0.111, size = 2031, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e))^3/(c+d*sec(f*x+e))^3,x)

[Out]

-9/f*c/(c^4-2*c^2*d^2+d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a*b^2*d-1
/f/c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a^3*d^3+
2/f/c^2/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a^3*d
^4-6/f*c^2/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a*
b^2+4/f*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*b^3
*d-1/f/c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a^3*d^3-2/f/c^
2/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a^3*d^4+6/f*c^2/(tan(
1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a*b^2-4/f*c/(tan(1/2*f*x+1/2
*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*b^3*d+3/f/(tan(1/2*f*x+1/2*e)^2*c-tan(1
/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a^2*b*d^2-6/f/(tan(1/2*f*x+1/2*e)^2*c-tan(
1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a*b^2*d^2+2/f*a^3/c^3*arctan(tan(1/2*f*x+
1/2*e))+3/f/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a^2*b*d^2+6
/f/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a*b^2*d^2+2/f/(c^4-2
*c^2*d^2+d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*b^3*d^2+1/f*c^2/(c^4-2
*c^2*d^2+d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*b^3+5/f/c/(c^4-2*c^2*d
^2+d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a^3*d^3-6/f*c/(c^4-2*c^2*d^2
+d^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a^3*d-2/f/c^3/(c^4-2*c^2*d^2+d
^4)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a^3*d^5+1/f*c^2/(tan(1/2*f*x+1/2
*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*b^3+1/f*c^2/(tan(1/2*f*x+1/2*
e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*b^3+6/f*c^2/(c^4-2*c^2*d^2+d^4)/((c+d)*(
c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a^2*b+3/f/(c^4-2*c^2*d^2+d^4)/((c+d)*(c-d))^
(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2))*a^2*b*d^2-6/f/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+
1/2*e)^2*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a^3*d^2+6/f/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1
/2*e)^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a^3*d^2-3/f*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d
-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a*b^2*d+12/f*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2
*d-c-d)^2/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*a^2*b*d-12/f*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)
^2*d-c-d)^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a^2*b*d-3/f*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)
^2/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e)*a*b^2*d

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.928389, size = 3343, normalized size = 13.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^3*c^8 - 3*a^3*c^6*d^2 + 3*a^3*c^4*d^4 - a^3*c^2*d^6)*f*x*cos(f*x + e)^2 + 8*(a^3*c^7*d - 3*a^3*c^5*
d^3 + 3*a^3*c^3*d^5 - a^3*c*d^7)*f*x*cos(f*x + e) + 4*(a^3*c^6*d^2 - 3*a^3*c^4*d^4 + 3*a^3*c^2*d^6 - a^3*d^8)*
f*x - (5*a^3*c^2*d^5 - 2*a^3*d^7 + (6*a^2*b + b^3)*c^5*d^2 - 3*(2*a^3 + 3*a*b^2)*c^4*d^3 + (3*a^2*b + 2*b^3)*c
^3*d^4 + (5*a^3*c^4*d^3 - 2*a^3*c^2*d^5 + (6*a^2*b + b^3)*c^7 - 3*(2*a^3 + 3*a*b^2)*c^6*d + (3*a^2*b + 2*b^3)*
c^5*d^2)*cos(f*x + e)^2 + 2*(5*a^3*c^3*d^4 - 2*a^3*c*d^6 + (6*a^2*b + b^3)*c^6*d - 3*(2*a^3 + 3*a*b^2)*c^5*d^2
+ (3*a^2*b + 2*b^3)*c^4*d^3)*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*(b^3*c^8 + 3*a*b^2*c^7*d + 2*a^3*c*d^7 - (9*a^2*b + 5*b^3)*c^6*d^2 + (5*a^3 + 3*a*b^2)*c^5*d^
3 + (9*a^2*b + 4*b^3)*c^4*d^4 - (7*a^3 + 6*a*b^2)*c^3*d^5 + 3*(2*a*b^2*c^8 - a^2*b*c^3*d^5 + a^3*c^2*d^6 - (4*
a^2*b + b^3)*c^7*d + (2*a^3 - a*b^2)*c^6*d^2 + (5*a^2*b + b^3)*c^5*d^3 - (3*a^3 + a*b^2)*c^4*d^4)*cos(f*x + e)
)*sin(f*x + e))/((c^11 - 3*c^9*d^2 + 3*c^7*d^4 - c^5*d^6)*f*cos(f*x + e)^2 + 2*(c^10*d - 3*c^8*d^3 + 3*c^6*d^5
- c^4*d^7)*f*cos(f*x + e) + (c^9*d^2 - 3*c^7*d^4 + 3*c^5*d^6 - c^3*d^8)*f), 1/2*(2*(a^3*c^8 - 3*a^3*c^6*d^2 +
3*a^3*c^4*d^4 - a^3*c^2*d^6)*f*x*cos(f*x + e)^2 + 4*(a^3*c^7*d - 3*a^3*c^5*d^3 + 3*a^3*c^3*d^5 - a^3*c*d^7)*f
*x*cos(f*x + e) + 2*(a^3*c^6*d^2 - 3*a^3*c^4*d^4 + 3*a^3*c^2*d^6 - a^3*d^8)*f*x + (5*a^3*c^2*d^5 - 2*a^3*d^7 +
(6*a^2*b + b^3)*c^5*d^2 - 3*(2*a^3 + 3*a*b^2)*c^4*d^3 + (3*a^2*b + 2*b^3)*c^3*d^4 + (5*a^3*c^4*d^3 - 2*a^3*c^
2*d^5 + (6*a^2*b + b^3)*c^7 - 3*(2*a^3 + 3*a*b^2)*c^6*d + (3*a^2*b + 2*b^3)*c^5*d^2)*cos(f*x + e)^2 + 2*(5*a^3
*c^3*d^4 - 2*a^3*c*d^6 + (6*a^2*b + b^3)*c^6*d - 3*(2*a^3 + 3*a*b^2)*c^5*d^2 + (3*a^2*b + 2*b^3)*c^4*d^3)*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))) + (b^3*c^
8 + 3*a*b^2*c^7*d + 2*a^3*c*d^7 - (9*a^2*b + 5*b^3)*c^6*d^2 + (5*a^3 + 3*a*b^2)*c^5*d^3 + (9*a^2*b + 4*b^3)*c^
4*d^4 - (7*a^3 + 6*a*b^2)*c^3*d^5 + 3*(2*a*b^2*c^8 - a^2*b*c^3*d^5 + a^3*c^2*d^6 - (4*a^2*b + b^3)*c^7*d + (2*
a^3 - a*b^2)*c^6*d^2 + (5*a^2*b + b^3)*c^5*d^3 - (3*a^3 + a*b^2)*c^4*d^4)*cos(f*x + e))*sin(f*x + e))/((c^11 -
3*c^9*d^2 + 3*c^7*d^4 - c^5*d^6)*f*cos(f*x + e)^2 + 2*(c^10*d - 3*c^8*d^3 + 3*c^6*d^5 - c^4*d^7)*f*cos(f*x +
e) + (c^9*d^2 - 3*c^7*d^4 + 3*c^5*d^6 - c^3*d^8)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sec{\left (e + f x \right )}\right )^{3}}{\left (c + d \sec{\left (e + f x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(e + f*x))**3/(c + d*sec(e + f*x))**3, x)

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Giac [B]  time = 1.56538, size = 1150, normalized size = 4.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f*x + e)*a^3/c^3 + (6*a^2*b*c^5 + b^3*c^5 - 6*a^3*c^4*d - 9*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 + 2*b^3*c^3*d^2 +
5*a^3*c^2*d^3 - 2*a^3*d^5)*(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^7 - 2*c^5*d^2 + c^3*d^4)*sqrt(-c^2 + d^2)) - (6*a*b^2*c^5*ta
n(1/2*f*x + 1/2*e)^3 - b^3*c^5*tan(1/2*f*x + 1/2*e)^3 - 12*a^2*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^2*c^4*d*
tan(1/2*f*x + 1/2*e)^3 - 3*b^3*c^4*d*tan(1/2*f*x + 1/2*e)^3 + 6*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 9*a^2*b*c
^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 3*a*b^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 4*b^3*c^3*d^2*tan(1/2*f*x + 1/2*e)^3
- 5*a^3*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 + 3*a^2*b*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*a*b^2*c^2*d^3*tan(1/2*f*x
+ 1/2*e)^3 - 3*a^3*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*a^3*d^5*tan(1/2*f*x + 1/2*e)^3 - 6*a*b^2*c^5*tan(1/2*f*x +
1/2*e) - b^3*c^5*tan(1/2*f*x + 1/2*e) + 12*a^2*b*c^4*d*tan(1/2*f*x + 1/2*e) - 3*a*b^2*c^4*d*tan(1/2*f*x + 1/2
*e) + 3*b^3*c^4*d*tan(1/2*f*x + 1/2*e) - 6*a^3*c^3*d^2*tan(1/2*f*x + 1/2*e) + 9*a^2*b*c^3*d^2*tan(1/2*f*x + 1/
2*e) - 3*a*b^2*c^3*d^2*tan(1/2*f*x + 1/2*e) + 4*b^3*c^3*d^2*tan(1/2*f*x + 1/2*e) - 5*a^3*c^2*d^3*tan(1/2*f*x +
1/2*e) - 3*a^2*b*c^2*d^3*tan(1/2*f*x + 1/2*e) - 6*a*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e) + 3*a^3*c*d^4*tan(1/2*f*
x + 1/2*e) + 2*a^3*d^5*tan(1/2*f*x + 1/2*e))/((c^6 - 2*c^4*d^2 + c^2*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/
2*f*x + 1/2*e)^2 - c - d)^2))/f