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

Optimal. Leaf size=43 \[ \frac {(A-B) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3998, 3770, 3794} \[ \frac {(A-B) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rubi steps

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

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Mathematica [B]  time = 0.27, size = 109, normalized size = 2.53 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left ((A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+B \cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 74, normalized size = 1.72 \[ \frac {{\left (B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A - B\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 70, normalized size = 1.63 \[ \frac {\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.68, size = 78, normalized size = 1.81 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(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*ln(tan(1/2*d*x+1/2*c)+1)*B-1/a/d*B*tan(1/2*d*x+1/2*c)-1/a/d*ln(tan(1/2*d*x+1/
2*c)-1)*B

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maxima [B]  time = 0.35, size = 99, normalized size = 2.30 \[ \frac {B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {A \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*
(cos(d*x + c) + 1))) + A*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 1.93, size = 41, normalized size = 0.95 \[ \frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*atanh(tan(c/2 + (d*x)/2)))/(a*d) + (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 \[ \frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a

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