3.521 \(\int \frac{a+b \sec (d+e x)}{(b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x))^2} \, dx\)

Optimal. Leaf size=230 \[ -\frac{\left (a^2-2 b^2\right ) \left (-a^2 b^2+2 a^4+b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b^4 e (a-b)^{5/2} (a+b)^{5/2}}-\frac{a \left (-11 a^2 b^2+6 a^4+11 b^4\right ) \tan (d+e x)}{6 b^3 e \left (a^2-b^2\right )^2 (a \sec (d+e x)+b)}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 e \left (a^2-b^2\right ) (a \sec (d+e x)+b)^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a x}{b^4} \]

[Out]

(a*x)/b^4 - ((a^2 - 2*b^2)*(2*a^4 - a^2*b^2 + b^4)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b
)^(5/2)*b^4*(a + b)^(5/2)*e) - (a*(3*a^2 - 5*b^2)*Tan[d + e*x])/(6*b^2*(a^2 - b^2)*e*(b + a*Sec[d + e*x])^2) -
(a*(6*a^4 - 11*a^2*b^2 + 11*b^4)*Tan[d + e*x])/(6*b^3*(a^2 - b^2)^2*e*(b + a*Sec[d + e*x])) - (a^4*Tan[d + e*
x])/(3*b*e*(a*b + a^2*Sec[d + e*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.832768, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.18, Rules used = {4172, 3923, 4060, 3919, 3831, 2659, 205} \[ -\frac{\left (a^2-2 b^2\right ) \left (-a^2 b^2+2 a^4+b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b^4 e (a-b)^{5/2} (a+b)^{5/2}}-\frac{a \left (-11 a^2 b^2+6 a^4+11 b^4\right ) \tan (d+e x)}{6 b^3 e \left (a^2-b^2\right )^2 (a \sec (d+e x)+b)}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 e \left (a^2-b^2\right ) (a \sec (d+e x)+b)^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a x}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/b^4 - ((a^2 - 2*b^2)*(2*a^4 - a^2*b^2 + b^4)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b
)^(5/2)*b^4*(a + b)^(5/2)*e) - (a*(3*a^2 - 5*b^2)*Tan[d + e*x])/(6*b^2*(a^2 - b^2)*e*(b + a*Sec[d + e*x])^2) -
(a*(6*a^4 - 11*a^2*b^2 + 11*b^4)*Tan[d + e*x])/(6*b^3*(a^2 - b^2)^2*e*(b + a*Sec[d + e*x])) - (a^4*Tan[d + e*
x])/(3*b*e*(a*b + a^2*Sec[d + e*x])^3)

Rule 4172

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rule 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4060

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

Rule 3919

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 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 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{a+b \sec (d+e x)}{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^2} \, dx &=\left (16 a^4\right ) \int \frac{a+b \sec (d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^4} \, dx\\ &=-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{(2 a) \int \frac{12 a^3 \left (a^2-b^2\right )+12 a^2 b \left (a^2-b^2\right ) \sec (d+e x)-8 a^3 \left (a^2-b^2\right ) \sec ^2(d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\int \frac{96 a^5 \left (a^2-b^2\right )^2+32 a^4 b \left (a^2-3 b^2\right ) \left (a^2-b^2\right ) \sec (d+e x)-16 a^5 \left (3 a^2-5 b^2\right ) \left (a^2-b^2\right ) \sec ^2(d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^2} \, dx}{24 a^2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{a^2 \left (6 a^4-11 a^2 b^2+11 b^4\right ) \tan (d+e x)}{6 b^3 \left (a^2-b^2\right )^2 e \left (a b+a^2 \sec (d+e x)\right )}+\frac{\int \frac{384 a^7 \left (a^2-b^2\right )^3+192 a^6 b \left (a^6-2 a^4 b^2+3 a^2 b^4-2 b^6\right ) \sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{192 a^5 b^3 \left (a^2-b^2\right )^3}\\ &=\frac{a x}{b^4}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{a^2 \left (6 a^4-11 a^2 b^2+11 b^4\right ) \tan (d+e x)}{6 b^3 \left (a^2-b^2\right )^2 e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (a \left (a^2-2 b^2\right ) \left (2 a^4-a^2 b^2+b^4\right )\right ) \int \frac{\sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{b^4 \left (a^2-b^2\right )^2}\\ &=\frac{a x}{b^4}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{a^2 \left (6 a^4-11 a^2 b^2+11 b^4\right ) \tan (d+e x)}{6 b^3 \left (a^2-b^2\right )^2 e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (\left (a^2-2 b^2\right ) \left (2 a^4-a^2 b^2+b^4\right )\right ) \int \frac{1}{1+\frac{b \cos (d+e x)}{a}} \, dx}{2 a b^4 \left (a^2-b^2\right )^2}\\ &=\frac{a x}{b^4}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{a^2 \left (6 a^4-11 a^2 b^2+11 b^4\right ) \tan (d+e x)}{6 b^3 \left (a^2-b^2\right )^2 e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (\left (a^2-2 b^2\right ) \left (2 a^4-a^2 b^2+b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{b}{a}+\left (1-\frac{b}{a}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{a b^4 \left (a^2-b^2\right )^2 e}\\ &=\frac{a x}{b^4}-\frac{\left (a^2-2 b^2\right ) \left (2 a^4-a^2 b^2+b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} e}-\frac{a \left (3 a^2-5 b^2\right ) \tan (d+e x)}{6 b^2 \left (a^2-b^2\right ) e (b+a \sec (d+e x))^2}-\frac{a^4 \tan (d+e x)}{3 b e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{a^2 \left (6 a^4-11 a^2 b^2+11 b^4\right ) \tan (d+e x)}{6 b^3 \left (a^2-b^2\right )^2 e \left (a b+a^2 \sec (d+e x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.57569, size = 276, normalized size = 1.2 \[ \frac{\sec ^3(d+e x) (a+b \cos (d+e x)) (a+b \sec (d+e x)) \left (\frac{a^2 b \left (7 a^2-9 b^2\right ) \sin (d+e x) (a+b \cos (d+e x))}{(a-b) (a+b)}-\frac{a b \left (-23 a^2 b^2+11 a^4+18 b^4\right ) \sin (d+e x) (a+b \cos (d+e x))^2}{(a-b)^2 (a+b)^2}+\frac{6 \left (5 a^4 b^2-3 a^2 b^4-2 a^6+2 b^6\right ) (a+b \cos (d+e x))^3 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}-2 a^3 b \sin (d+e x)+6 a (d+e x) (a+b \cos (d+e x))^3\right )}{6 b^4 e (a \cos (d+e x)+b) (a \sec (d+e x)+b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Cos[d + e*x])*Sec[d + e*x]^3*(a + b*Sec[d + e*x])*(6*a*(d + e*x)*(a + b*Cos[d + e*x])^3 + (6*(-2*a^6 +
5*a^4*b^2 - 3*a^2*b^4 + 2*b^6)*ArcTanh[((-a + b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]]*(a + b*Cos[d + e*x])^3)/
(-a^2 + b^2)^(5/2) - 2*a^3*b*Sin[d + e*x] + (a^2*b*(7*a^2 - 9*b^2)*(a + b*Cos[d + e*x])*Sin[d + e*x])/((a - b)
*(a + b)) - (a*b*(11*a^4 - 23*a^2*b^2 + 18*b^4)*(a + b*Cos[d + e*x])^2*Sin[d + e*x])/((a - b)^2*(a + b)^2)))/(
6*b^4*e*(b + a*Cos[d + e*x])*(b + a*Sec[d + e*x])^4)

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Maple [B]  time = 0.113, size = 1118, normalized size = 4.9 \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(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x)

[Out]

2/e*a/b^4*arctan(tan(1/2*d+1/2*e*x))-2/e/b^3/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^5/(a^2+2*
a*b+b^2)*tan(1/2*d+1/2*e*x)^5+1/e/b^2/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^4/(a^2+2*a*b+b^2
)*tan(1/2*d+1/2*e*x)^5+4/e/b/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^3/(a^2+2*a*b+b^2)*tan(1/2
*d+1/2*e*x)^5-3/e/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^2/(a^2+2*a*b+b^2)*tan(1/2*d+1/2*e*x)
^5-6/e*b/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a/(a^2+2*a*b+b^2)*tan(1/2*d+1/2*e*x)^5-4/e/b^3/
(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^5/(a-b)/(a+b)*tan(1/2*d+1/2*e*x)^3+32/3/e/b/(a*tan(1/2
*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^3/(a-b)/(a+b)*tan(1/2*d+1/2*e*x)^3-12/e*b/(a*tan(1/2*d+1/2*e*x)^
2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a/(a-b)/(a+b)*tan(1/2*d+1/2*e*x)^3-2/e/b^3/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d
+1/2*e*x)^2+a+b)^3*a^5/(a^2-2*a*b+b^2)*tan(1/2*d+1/2*e*x)-1/e/b^2/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)
^2+a+b)^3*a^4/(a^2-2*a*b+b^2)*tan(1/2*d+1/2*e*x)+4/e/b/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a
^3/(a^2-2*a*b+b^2)*tan(1/2*d+1/2*e*x)+3/e/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a^2/(a^2-2*a*b
+b^2)*tan(1/2*d+1/2*e*x)-6/e*b/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+a+b)^3*a/(a^2-2*a*b+b^2)*tan(1/2
*d+1/2*e*x)-2/e/b^4/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d+1/2*e*x)*(a-b)/((a-b)*(a+b))^(1/2
))*a^6+5/e/b^2/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d+1/2*e*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^
4-3/e/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d+1/2*e*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^2+2/e*b^2
/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d+1/2*e*x)*(a-b)/((a-b)*(a+b))^(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((a+b*sec(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.05475, size = 2871, normalized size = 12.48 \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(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="fricas")

[Out]

[1/12*(12*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*e*x*cos(e*x + d)^3 + 36*(a^8*b^2 - 3*a^6*b^4 + 3*a^4*b^6 -
a^2*b^8)*e*x*cos(e*x + d)^2 + 36*(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*e*x*cos(e*x + d) + 12*(a^10 - 3*a^
8*b^2 + 3*a^6*b^4 - a^4*b^6)*e*x + 3*(2*a^9 - 5*a^7*b^2 + 3*a^5*b^4 - 2*a^3*b^6 + (2*a^6*b^3 - 5*a^4*b^5 + 3*a
^2*b^7 - 2*b^9)*cos(e*x + d)^3 + 3*(2*a^7*b^2 - 5*a^5*b^4 + 3*a^3*b^6 - 2*a*b^8)*cos(e*x + d)^2 + 3*(2*a^8*b -
5*a^6*b^3 + 3*a^4*b^5 - 2*a^2*b^7)*cos(e*x + d))*sqrt(-a^2 + b^2)*log((2*a*b*cos(e*x + d) + (2*a^2 - b^2)*cos
(e*x + d)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(e*x + d) + b)*sin(e*x + d) - a^2 + 2*b^2)/(b^2*cos(e*x + d)^2 + 2*a*b*
cos(e*x + d) + a^2)) - 2*(6*a^9*b - 17*a^7*b^3 + 22*a^5*b^5 - 11*a^3*b^7 + (11*a^7*b^3 - 34*a^5*b^5 + 41*a^3*b
^7 - 18*a*b^9)*cos(e*x + d)^2 + 3*(5*a^8*b^2 - 15*a^6*b^4 + 19*a^4*b^6 - 9*a^2*b^8)*cos(e*x + d))*sin(e*x + d)
)/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*e*cos(e*x + d)^3 + 3*(a^7*b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*
e*cos(e*x + d)^2 + 3*(a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*e*cos(e*x + d) + (a^9*b^4 - 3*a^7*b^6 + 3*a^
5*b^8 - a^3*b^10)*e), 1/6*(6*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*e*x*cos(e*x + d)^3 + 18*(a^8*b^2 - 3*a^
6*b^4 + 3*a^4*b^6 - a^2*b^8)*e*x*cos(e*x + d)^2 + 18*(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*e*x*cos(e*x + d
) + 6*(a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*e*x - 3*(2*a^9 - 5*a^7*b^2 + 3*a^5*b^4 - 2*a^3*b^6 + (2*a^6*b^3
- 5*a^4*b^5 + 3*a^2*b^7 - 2*b^9)*cos(e*x + d)^3 + 3*(2*a^7*b^2 - 5*a^5*b^4 + 3*a^3*b^6 - 2*a*b^8)*cos(e*x + d
)^2 + 3*(2*a^8*b - 5*a^6*b^3 + 3*a^4*b^5 - 2*a^2*b^7)*cos(e*x + d))*sqrt(a^2 - b^2)*arctan(-(a*cos(e*x + d) +
b)/(sqrt(a^2 - b^2)*sin(e*x + d))) - (6*a^9*b - 17*a^7*b^3 + 22*a^5*b^5 - 11*a^3*b^7 + (11*a^7*b^3 - 34*a^5*b^
5 + 41*a^3*b^7 - 18*a*b^9)*cos(e*x + d)^2 + 3*(5*a^8*b^2 - 15*a^6*b^4 + 19*a^4*b^6 - 9*a^2*b^8)*cos(e*x + d))*
sin(e*x + d))/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*e*cos(e*x + d)^3 + 3*(a^7*b^6 - 3*a^5*b^8 + 3*a^3*b^1
0 - a*b^12)*e*cos(e*x + d)^2 + 3*(a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*e*cos(e*x + d) + (a^9*b^4 - 3*a^
7*b^6 + 3*a^5*b^8 - a^3*b^10)*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \sec{\left (d + e x \right )}}{\left (a \sec{\left (d + e x \right )} + b\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))/(b**2+2*a*b*sec(e*x+d)+a**2*sec(e*x+d)**2)**2,x)

[Out]

Integral((a + b*sec(d + e*x))/(a*sec(d + e*x) + b)**4, x)

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Giac [B]  time = 1.33189, size = 662, normalized size = 2.88 \begin{align*} \frac{1}{3} \,{\left (\frac{3 \,{\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - 2 \, b^{6}\right )}{\left (\pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \,{\left (x e + d\right )} a}{b^{4}} - \frac{6 \, a^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} - 15 \, a^{6} b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} + 30 \, a^{4} b^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} - 12 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} - 27 \, a^{2} b^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} + 18 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{5} + 12 \, a^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} - 44 \, a^{5} b^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 68 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} - 36 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 6 \, a^{7} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 15 \, a^{6} b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 30 \, a^{4} b^{3} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 12 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 27 \, a^{2} b^{5} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 18 \, a b^{6} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )}{\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + a + b\right )}^{3}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))/(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="giac")

[Out]

1/3*(3*(2*a^6 - 5*a^4*b^2 + 3*a^2*b^4 - 2*b^6)*(pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*
tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/2*d))/sqrt(a^2 - b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(a^2 - b^2))
+ 3*(x*e + d)*a/b^4 - (6*a^7*tan(1/2*x*e + 1/2*d)^5 - 15*a^6*b*tan(1/2*x*e + 1/2*d)^5 + 30*a^4*b^3*tan(1/2*x*
e + 1/2*d)^5 - 12*a^3*b^4*tan(1/2*x*e + 1/2*d)^5 - 27*a^2*b^5*tan(1/2*x*e + 1/2*d)^5 + 18*a*b^6*tan(1/2*x*e +
1/2*d)^5 + 12*a^7*tan(1/2*x*e + 1/2*d)^3 - 44*a^5*b^2*tan(1/2*x*e + 1/2*d)^3 + 68*a^3*b^4*tan(1/2*x*e + 1/2*d)
^3 - 36*a*b^6*tan(1/2*x*e + 1/2*d)^3 + 6*a^7*tan(1/2*x*e + 1/2*d) + 15*a^6*b*tan(1/2*x*e + 1/2*d) - 30*a^4*b^3
*tan(1/2*x*e + 1/2*d) - 12*a^3*b^4*tan(1/2*x*e + 1/2*d) + 27*a^2*b^5*tan(1/2*x*e + 1/2*d) + 18*a*b^6*tan(1/2*x
*e + 1/2*d))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*x*e + 1/2*d)^2 - b*tan(1/2*x*e + 1/2*d)^2 + a + b)^3))*e^
(-1)