3.525 \(\int \frac{a+b \sec (d+e x)}{(b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=330 \[ \frac{x \left (a^2 \sec (d+e x)+a b\right )^3}{a^2 b^3 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\left (-3 a^2 b^2+2 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (a \sec (d+e x)+b)^3}{b^3 e (a-b)^{3/2} (a+b)^{3/2} \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )}{2 b e \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{2 b^2 e \left (a^2-b^2\right ) \left (a^2 b+a^3 \sec (d+e x)\right ) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.566333, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4174, 3923, 4060, 3919, 3831, 2659, 205} \[ \frac{x \left (a^2 \sec (d+e x)+a b\right )^3}{a^2 b^3 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\left (-3 a^2 b^2+2 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (a \sec (d+e x)+b)^3}{b^3 e (a-b)^{3/2} (a+b)^{3/2} \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )}{2 b e \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{2 b^2 e \left (a^2-b^2\right ) \left (a^2 b+a^3 \sec (d+e x)\right ) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}} \]

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)^(3/2),x]

[Out]

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

Rule 4174

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Sec[d + e*x] + c*Sec[d + e*x]^2)^n/(b + 2*c*Sec[d + e*x])^(2*n), Int[(A +
B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; FreeQ[{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 )^{3/2}} \, dx &=\frac{\left (2 a b+2 a^2 \sec (d+e x)\right )^3 \int \frac{a+b \sec (d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3} \, dx}{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}+\frac{\left (2 a b+2 a^2 \sec (d+e x)\right )^3 \int \frac{8 a^3 \left (a^2-b^2\right )+8 a^2 b \left (a^2-b^2\right ) \sec (d+e x)-4 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 )^2} \, dx}{16 a^3 b \left (a^2-b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{2 a b^2 \left (a^2-b^2\right ) e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}+\frac{\left (2 a b+2 a^2 \sec (d+e x)\right )^3 \int \frac{32 a^5 \left (a^2-b^2\right )^2+16 a^4 b \left (a^2-2 b^2\right ) \left (a^2-b^2\right ) \sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{128 a^6 b^2 \left (a^2-b^2\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )^3}{a^2 b^3 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{2 a b^2 \left (a^2-b^2\right ) e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (\left (-32 a^5 b^2 \left (a^2-2 b^2\right ) \left (a^2-b^2\right )+64 a^7 \left (a^2-b^2\right )^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )^3\right ) \int \frac{\sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{256 a^7 b^3 \left (a^2-b^2\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )^3}{a^2 b^3 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{2 a b^2 \left (a^2-b^2\right ) e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (\left (-32 a^5 b^2 \left (a^2-2 b^2\right ) \left (a^2-b^2\right )+64 a^7 \left (a^2-b^2\right )^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )^3\right ) \int \frac{1}{1+\frac{b \cos (d+e x)}{a}} \, dx}{512 a^9 b^3 \left (a^2-b^2\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=\frac{x \left (a b+a^2 \sec (d+e x)\right )^3}{a^2 b^3 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{2 a b^2 \left (a^2-b^2\right ) e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (\left (-32 a^5 b^2 \left (a^2-2 b^2\right ) \left (a^2-b^2\right )+64 a^7 \left (a^2-b^2\right )^2\right ) \left (2 a b+2 a^2 \sec (d+e x)\right )^3\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 )}{256 a^9 b^3 \left (a^2-b^2\right )^2 e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ &=-\frac{\left (2 a^4-3 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right ) (b+a \sec (d+e x))^3}{(a-b)^{3/2} b^3 (a+b)^{3/2} e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}+\frac{x \left (a b+a^2 \sec (d+e x)\right )^3}{a^2 b^3 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (a b+a^2 \sec (d+e x)\right ) \tan (d+e x)}{2 b e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}-\frac{\left (2 a^2-3 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{2 a b^2 \left (a^2-b^2\right ) e \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.01106, size = 216, normalized size = 0.65 \[ \frac{\sec ^2(d+e x) (a+b \cos (d+e x)) (a+b \sec (d+e x)) \left (\frac{a b \left (3 a^2-4 b^2\right ) \sin (d+e x) (a+b \cos (d+e x))}{(b-a) (a+b)}+\frac{2 \left (-3 a^2 b^2+2 a^4+2 b^4\right ) (a+b \cos (d+e x))^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+a^2 b \sin (d+e x)+2 a (d+e x) (a+b \cos (d+e x))^2\right )}{2 b^3 e (a \cos (d+e x)+b) \left ((a \sec (d+e x)+b)^2\right )^{3/2}} \]

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)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.21, size = 756, normalized size = 2.3 \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)^(3/2),x)

[Out]

-1/2/e/((a-b)*(a+b))^(1/2)/(a^2-b^2)/b^3*(b*cos(e*x+d)+a)*(-4*cos(e*x+d)^2*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x
+d)/((a-b)*(a+b))^(1/2))*a^4*b^2+6*cos(e*x+d)^2*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*a^
2*b^4-4*cos(e*x+d)^2*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*b^6-2*cos(e*x+d)^2*((a-b)*(a+
b))^(1/2)*(e*x+d)*a^3*b^2+2*cos(e*x+d)^2*((a-b)*(a+b))^(1/2)*(e*x+d)*a*b^4+3*sin(e*x+d)*cos(e*x+d)*((a-b)*(a+b
))^(1/2)*a^3*b^2-4*sin(e*x+d)*cos(e*x+d)*((a-b)*(a+b))^(1/2)*a*b^4-8*cos(e*x+d)*arctan((cos(e*x+d)-1)*(a-b)/si
n(e*x+d)/((a-b)*(a+b))^(1/2))*a^5*b+12*cos(e*x+d)*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*
a^3*b^3-8*cos(e*x+d)*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*a*b^5-4*cos(e*x+d)*((a-b)*(a+
b))^(1/2)*(e*x+d)*a^4*b+4*cos(e*x+d)*((a-b)*(a+b))^(1/2)*(e*x+d)*a^2*b^3+2*((a-b)*(a+b))^(1/2)*a^4*b*sin(e*x+d
)-3*((a-b)*(a+b))^(1/2)*a^2*b^3*sin(e*x+d)-4*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*a^6+6
*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/((a-b)*(a+b))^(1/2))*a^4*b^2-4*arctan((cos(e*x+d)-1)*(a-b)/sin(e*x+d)/
((a-b)*(a+b))^(1/2))*a^2*b^4-2*(e*x+d)*((a-b)*(a+b))^(1/2)*a^5+2*((a-b)*(a+b))^(1/2)*(e*x+d)*a^3*b^2)/cos(e*x+
d)^3/((b*cos(e*x+d)+a)^2/cos(e*x+d)^2)^(3/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.61807, size = 1739, normalized size = 5.27 \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)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(4*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*e*x*cos(e*x + d)^2 + 8*(a^6*b - 2*a^4*b^3 + a^2*b^5)*e*x*cos(e*x + d) +
4*(a^7 - 2*a^5*b^2 + a^3*b^4)*e*x + (2*a^6 - 3*a^4*b^2 + 2*a^2*b^4 + (2*a^4*b^2 - 3*a^2*b^4 + 2*b^6)*cos(e*x +
d)^2 + 2*(2*a^5*b - 3*a^3*b^3 + 2*a*b^5)*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*(2*a^6*b - 5*a^4*b^3 + 3*a^2*b^5 + (3*a^5*b^2 - 7*a^3*b^4 + 4*a*b^6)*cos(e*x +
d))*sin(e*x + d))/((a^4*b^5 - 2*a^2*b^7 + b^9)*e*cos(e*x + d)^2 + 2*(a^5*b^4 - 2*a^3*b^6 + a*b^8)*e*cos(e*x +
d) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*e), 1/2*(2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*e*x*cos(e*x + d)^2 + 4*(a^6*b -
2*a^4*b^3 + a^2*b^5)*e*x*cos(e*x + d) + 2*(a^7 - 2*a^5*b^2 + a^3*b^4)*e*x - (2*a^6 - 3*a^4*b^2 + 2*a^2*b^4 + (
2*a^4*b^2 - 3*a^2*b^4 + 2*b^6)*cos(e*x + d)^2 + 2*(2*a^5*b - 3*a^3*b^3 + 2*a*b^5)*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))) - (2*a^6*b - 5*a^4*b^3 + 3*a^2*b^5 + (3*a^5*b^2
- 7*a^3*b^4 + 4*a*b^6)*cos(e*x + d))*sin(e*x + d))/((a^4*b^5 - 2*a^2*b^7 + b^9)*e*cos(e*x + d)^2 + 2*(a^5*b^4
- 2*a^3*b^6 + a*b^8)*e*cos(e*x + d) + (a^6*b^3 - 2*a^4*b^5 + a^2*b^7)*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 (\left (a \sec{\left (d + e x \right )} + b\right )^{2}\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral((a + b*sec(d + e*x))/((a*sec(d + e*x) + b)**2)**(3/2), x)

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Giac [A]  time = 1.88619, size = 768, normalized size = 2.33 \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)^(3/2),x, algorithm="giac")

[Out]

((2*a^4 - 3*a^2*b^2 + 2*b^4)*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^2*b
^3*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1/2*x*e + 1/2*d)^2 - a - b) - b^5*sgn(a*t
an(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1/2*x*e + 1/2*d)^2 - a - b))*sqrt(a^2 - b^2)) + (2*
a^4*tan(1/2*x*e + 1/2*d)^3 - 3*a^3*b*tan(1/2*x*e + 1/2*d)^3 - 3*a^2*b^2*tan(1/2*x*e + 1/2*d)^3 + 4*a*b^3*tan(1
/2*x*e + 1/2*d)^3 + 2*a^4*tan(1/2*x*e + 1/2*d) + 3*a^3*b*tan(1/2*x*e + 1/2*d) - 3*a^2*b^2*tan(1/2*x*e + 1/2*d)
- 4*a*b^3*tan(1/2*x*e + 1/2*d))/((a^2*b^2*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1
/2*x*e + 1/2*d)^2 - a - b) - b^4*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1/2*x*e + 1
/2*d)^2 - a - b))*(a*tan(1/2*x*e + 1/2*d)^2 - b*tan(1/2*x*e + 1/2*d)^2 + a + b)^2) - (x*e - 2*pi*floor(1/2*(x*
e + d)/pi + 1/2) + d)*a/(b^3*sgn(a*tan(1/2*x*e + 1/2*d)^4 - b*tan(1/2*x*e + 1/2*d)^4 + 2*b*tan(1/2*x*e + 1/2*d
)^2 - a - b)))*e^(-1)