3.87 \(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=60 \[ \frac {(2 A-B) \sin (c+d x)}{a d}-\frac {(A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac {x (A-B)}{a} \]

[Out]

-(A-B)*x/a+(2*A-B)*sin(d*x+c)/a/d-(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4020, 3787, 2637, 8} \[ \frac {(2 A-B) \sin (c+d x)}{a d}-\frac {(A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac {x (A-B)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - B)*x)/a) + ((2*A - B)*Sin[c + d*x])/(a*d) - ((A - B)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx &=-\frac {(A-B) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \cos (c+d x) (a (2 A-B)-a (A-B) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(A-B) \int 1 \, dx}{a}+\frac {(2 A-B) \int \cos (c+d x) \, dx}{a}\\ &=-\frac {(A-B) x}{a}+\frac {(2 A-B) \sin (c+d x)}{a d}-\frac {(A-B) \sin (c+d x)}{d (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 76, normalized size = 1.27 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) (d x (B-A)+A \sin (c+d x))+(A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((A - B)*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*((-A + B)*d*x + A*Sin[c + d*x])))/(a*d*(
1 + Cos[c + d*x]))

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fricas [A]  time = 0.44, size = 63, normalized size = 1.05 \[ -\frac {{\left (A - B\right )} d x \cos \left (d x + c\right ) + {\left (A - B\right )} d x - {\left (A \cos \left (d x + c\right ) + 2 \, A - B\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((A - B)*d*x*cos(d*x + c) + (A - B)*d*x - (A*cos(d*x + c) + 2*A - B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 0.23, size = 79, normalized size = 1.32 \[ -\frac {\frac {{\left (d x + c\right )} {\left (A - B\right )}}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d*x + c)*(A - B)/a - (A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a - 2*A*tan(1/2*d*x + 1/2*c)/((tan(1
/2*d*x + 1/2*c)^2 + 1)*a))/d

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maple [A]  time = 1.09, size = 108, normalized size = 1.80 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.43, size = 143, normalized size = 2.38 \[ -\frac {A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.00, size = 65, normalized size = 1.08 \[ \frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {x\,\left (A-B\right )}{a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*cos(c + d*x)/(sec(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)/(sec(c + d*x) + 1), x))
/a

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