3.931 \(\int \frac{a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=75 \[ \frac{x (b B-a C)}{a}-\frac{2 b (b B-2 a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

[Out]

((b*B - a*C)*x)/a - (2*b*(b*B - 2*a*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqr
t[a + b]*d)

________________________________________________________________________________________

Rubi [A]  time = 0.150927, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.104, Rules used = {24, 3919, 3831, 2659, 208} \[ \frac{x (b B-a C)}{a}-\frac{2 b (b B-2 a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - a*C)*x)/a - (2*b*(b*B - 2*a*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqr
t[a + b]*d)

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[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 b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac{\int \frac{b^2 (b B-a C)+b^3 C \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^2}\\ &=\frac{(b B-a C) x}{a}-\frac{(b (b B-2 a C)) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a}\\ &=\frac{(b B-a C) x}{a}-\frac{(b B-2 a C) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a}\\ &=\frac{(b B-a C) x}{a}-\frac{(2 (b B-2 a C)) \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 d}\\ &=\frac{(b B-a C) x}{a}-\frac{2 b (b B-2 a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a \sqrt{a-b} \sqrt{a+b} d}\\ \end{align*}

Mathematica [A]  time = 0.206372, size = 76, normalized size = 1.01 \[ \frac{\frac{2 b (b B-2 a C) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+(c+d x) (b B-a C)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.089, size = 133, normalized size = 1.8 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) Bb}{ad}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d}}-2\,{\frac{B{b}^{2}}{ad\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{Cb}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \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((B*a*b-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)

[Out]

2/d/a*arctan(tan(1/2*d*x+1/2*c))*B*b-2/d*arctan(tan(1/2*d*x+1/2*c))*C-2/d*b^2/a/((a+b)*(a-b))^(1/2)*arctanh((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+4/d*b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b
)*(a-b))^(1/2))*C

________________________________________________________________________________________

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*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.533034, size = 613, normalized size = 8.17 \begin{align*} \left [-\frac{2 \,{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x +{\left (2 \, C a b - B b^{2}\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d}, -\frac{{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x -{\left (2 \, C a b - B b^{2}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x + (2*C*a*b - B*b^2)*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)))/((a^3 - a*b^2)*d), -((C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x - (2*C*a*b
- B*b^2)*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))))/((a^3 -
a*b^2)*d)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.29049, size = 153, normalized size = 2.04 \begin{align*} -\frac{\frac{{\left (C a - B b\right )}{\left (d x + c\right )}}{a} - \frac{2 \,{\left (2 \, C a b - B b^{2}\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 )}}{\sqrt{-a^{2} + b^{2}} a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((C*a - B*b)*(d*x + c)/a - 2*(2*C*a*b - 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)*a))/d