3.708 \(\int \frac{a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=162 \[ -\frac{2 b \left (-2 a^2 b^2+4 a^4+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b^2 \left (4 a^2-b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{x}{a^2} \]

[Out]

x/a^2 - (2*b*(4*a^4 - 2*a^2*b^2 + b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)
*(a + b)^(5/2)*d) + (b^2*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (b^2*(4*a^2 - b^2)*Tan[c + d*x
])/(a*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.339561, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {4042, 3923, 4060, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (-2 a^2 b^2+4 a^4+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b^2 \left (4 a^2-b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^4,x]

[Out]

x/a^2 - (2*b*(4*a^4 - 2*a^2*b^2 + b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)
*(a + b)^(5/2)*d) + (b^2*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (b^2*(4*a^2 - b^2)*Tan[c + d*x
])/(a*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4042

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

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

Mathematica [A]  time = 0.775569, size = 223, normalized size = 1.38 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) (a-b \sec (c+d x)) \left (\frac{a b^2 \left (5 a^2-2 b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}+\frac{2 b \left (-2 a^2 b^2+4 a^4+b^4\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^3 \sin (c+d x)}{(b-a) (a+b)}+(c+d x) (a \cos (c+d x)+b)^2\right )}{a^2 d (a \cos (c+d x)-b) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^4,x]

[Out]

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

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Maple [B]  time = 0.102, size = 659, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x)

[Out]

2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-10/d*b^2*a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2
+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-2/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*
b+b^2)*tan(1/2*d*x+1/2*c)^3+2/d*b^4/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b
^2)*tan(1/2*d*x+1/2*c)^3+10/d*b^2*a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2
)*tan(1/2*d*x+1/2*c)-2/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1
/2*d*x+1/2*c)-2/d*b^4/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*
x+1/2*c)-8/d*b*a^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2
))+4/d*b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-2/d*b
^5/a^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((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^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.684094, size = 1925, normalized size = 11.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2-b**2*sec(d*x+c)**2)/(a+b*sec(d*x+c))**4,x)

[Out]

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

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Giac [B]  time = 1.30622, size = 428, normalized size = 2.64 \begin{align*} \frac{\frac{2 \,{\left (4 \, a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{d x + c}{a^{2}} - \frac{2 \,{\left (5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

(2*(4*a^4*b - 2*a^2*b^3 + b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*
c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - 2*a^4*b^2 + a^2*b^4)*sqrt(-a^2 + b^2)) + (d*x + c)/a^2
- 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^4*tan(1/2*d*x + 1/2*c)^3 + b
^5*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*b^2*tan(1/2*d*x + 1/2*c) - 4*a^2*b^3*tan(1/2*d*x + 1/2*c) + 2*a*b^4*tan(1/2*
d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c))/((a^5 - 2*a^3*b^2 + a*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x
+ 1/2*c)^2 - a - b)^2))/d