\(\int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [293]

Optimal result

Integrand size = 32, antiderivative size = 70 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d}+\frac {B \text {arctanh}(\sin (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {21, 2826, 3855, 2738, 211} \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

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 2826

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = B \int \frac {\sec (c+d x)}{a+b \cos (c+d x)} \, dx \\ & = \frac {B \int \sec (c+d x) \, dx}{a}-\frac {(b B) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{a d}-\frac {(2 b B) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \\ & = -\frac {2 b B \arctan \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}+\frac {B \text {arctanh}(\sin (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.47 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (\frac {2 b \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {2 B \left (-\frac {b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a}\right )}{d}\)	\(86\)
default	\(\frac {2 B \left (-\frac {b \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a}\right )}{d}\)	\(86\)
risch	\(-\frac {b B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}\)	\(192\)

[In]

int((B*a+b*B*cos(d*x+c))*sec(d*x+c)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 292, normalized size of antiderivative = 4.17 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} B b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, -\frac {2 \, \sqrt {a^{2} - b^{2}} B b \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}\right ] \]

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*B*b*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*a^2 - B*b^2)*
log(sin(d*x + c) + 1) + (B*a^2 - B*b^2)*log(-sin(d*x + c) + 1))/((a^3 - a*b^2)*d), -1/2*(2*sqrt(a^2 - b^2)*B*b
*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (B*a^2 - B*b^2)*log(sin(d*x + c) + 1) + (B*a^2
- B*b^2)*log(-sin(d*x + c) + 1))/((a^3 - a*b^2)*d)]

Sympy [F]

\[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=B \int \frac {\sec {\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.74 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B b}{\sqrt {a^{2} - b^{2}} a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.44 \[ \int \frac {(a B+b B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,B\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {2\,B\,b\,\mathrm {atanh}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )}{a\,d\,\sqrt {b^2-a^2}} \]

[In]

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

[Out]

(2*B*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d) + (2*B*b*atanh((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d
*x)/2))/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))))/(a*d*(b^2 - a^2)^(1/2))