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

Optimal. Leaf size=86 \[ -\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d}+\frac{a x}{b^2}-\frac{a \tan (c+d x)}{b d (a \sec (c+d x)+b)} \]

[Out]

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

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Rubi [A]  time = 0.180483, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3923, 3919, 3831, 2659, 205} \[ -\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d}+\frac{a x}{b^2}-\frac{a \tan (c+d x)}{b d (a \sec (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 (c+d x)}{(b+a \sec (c+d x))^2} \, dx &=-\frac{a \tan (c+d x)}{b d (b+a \sec (c+d x))}+\frac{\int \frac{a \left (a^2-b^2\right )+b \left (a^2-b^2\right ) \sec (c+d x)}{b+a \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{a x}{b^2}-\frac{a \tan (c+d x)}{b d (b+a \sec (c+d x))}-\frac{\left (a^2-b^2\right ) \int \frac{\sec (c+d x)}{b+a \sec (c+d x)} \, dx}{b^2}\\ &=\frac{a x}{b^2}-\frac{a \tan (c+d x)}{b d (b+a \sec (c+d x))}-\frac{\left (a^2-b^2\right ) \int \frac{1}{1+\frac{b \cos (c+d x)}{a}} \, dx}{a b^2}\\ &=\frac{a x}{b^2}-\frac{a \tan (c+d x)}{b d (b+a \sec (c+d x))}-\frac{\left (2 \left (a^2-b^2\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} (c+d x)\right )\right )}{a b^2 d}\\ &=\frac{a x}{b^2}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d}-\frac{a \tan (c+d x)}{b d (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.35524, size = 97, normalized size = 1.13 \[ \frac{2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )+\frac{a (a c+a d x-b \sin (c+d x)+b (c+d x) \cos (c+d x))}{a+b \cos (c+d x)}}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.092, size = 163, normalized size = 1.9 \begin{align*} 2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{2}}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) a}{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-2\,{\frac{{a}^{2}}{d{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{1}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a/b^2*arctan(tan(1/2*d*x+1/2*c))-2/d/b*tan(1/2*d*x+1/2*c)*a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b
+a+b)-2/d/b^2/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^2+2/d/((a+b)*(a-b))^(
1/2)*arctan((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+b*sec(d*x+c))/(b+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.550893, size = 670, normalized size = 7.79 \begin{align*} \left [\frac{2 \, a b d x \cos \left (d x + c\right ) + 2 \, a^{2} d x - 2 \, a b \sin \left (d x + c\right ) + \sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \,{\left (b^{3} d \cos \left (d x + c\right ) + a b^{2} d\right )}}, \frac{a b d x \cos \left (d x + c\right ) + a^{2} d x - a b \sin \left (d x + c\right ) - \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{b^{3} d \cos \left (d x + c\right ) + a b^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*b*d*x*cos(d*x + c) + 2*a^2*d*x - 2*a*b*sin(d*x + c) + sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)*log((2*a
*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 +
2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)))/(b^3*d*cos(d*x + c) + a*b^2*d), (a*b*d*x*cos(d*x + c)
+ a^2*d*x - a*b*sin(d*x + c) - sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 -
b^2)*sin(d*x + c))))/(b^3*d*cos(d*x + c) + a*b^2*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 \sec{\left (c + d x \right )} + b\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37274, size = 188, normalized size = 2.19 \begin{align*} \frac{\frac{{\left (d x + c\right )} a}{b^{2}} - \frac{2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} b} - \frac{2 \,{\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 )} \sqrt{a^{2} - b^{2}}}{b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*a/b^2 - 2*a*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*b)
- 2*(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)))*sqrt(a^2 - b^2)/b^2)/d