3.344 \(\int \frac{\frac{b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{2 B \sqrt{a-b} \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d}+\frac{b B x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.11474, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3919, 3831, 2659, 208} \[ \frac{2 B \sqrt{a-b} \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d}+\frac{b B x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.144292, size = 61, normalized size = 1. \[ \frac{B \left (b (c+d x)-2 \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.077, size = 116, normalized size = 1.9 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) Bb}{d{a}^{2}}}+2\,{\frac{B}{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 ) }-2\,{\frac{B{b}^{2}}{d{a}^{2}\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*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.536089, size = 454, normalized size = 7.44 \begin{align*} \left [\frac{2 \, B b d x + \sqrt{a^{2} - b^{2}} B \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 \, a^{2} d}, \frac{B b d x + \sqrt{-a^{2} + b^{2}} B \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 )}{a^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*B*b*d*x + sqrt(a^2 - b^2)*B*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^2*d), (
B*b*d*x + sqrt(-a^2 + b^2)*B*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))))/(a^2*d
)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.31344, size = 143, normalized size = 2.34 \begin{align*} \frac{\frac{{\left (d x + c\right )} B b}{a^{2}} + \frac{2 \,{\left (B a^{2} - 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^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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