3.303 \(\int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=54 \[ -\frac {\log (a+b \sec (c+d x))}{a^2 d}-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {1}{a d (a+b \sec (c+d x))} \]

[Out]

-ln(cos(d*x+c))/a^2/d-ln(a+b*sec(d*x+c))/a^2/d+1/a/d/(a+b*sec(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3885, 44} \[ -\frac {\log (a+b \sec (c+d x))}{a^2 d}-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {1}{a d (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[Cos[c + d*x]]/(a^2*d)) - Log[a + b*Sec[c + d*x]]/(a^2*d) + 1/(a*d*(a + b*Sec[c + d*x]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 54, normalized size = 1.00 \[ -\frac {b \log (a \cos (c+d x)+b)+a \cos (c+d x) \log (a \cos (c+d x)+b)+b}{a^2 d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b + b*Log[b + a*Cos[c + d*x]] + a*Cos[c + d*x]*Log[b + a*Cos[c + d*x]])/(a^2*d*(b + a*Cos[c + d*x])))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 46, normalized size = 0.85 \[ -\frac {{\left (a \cos \left (d x + c\right ) + b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + b}{a^{3} d \cos \left (d x + c\right ) + a^{2} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*cos(d*x + c) + b)*log(a*cos(d*x + c) + b) + b)/(a^3*d*cos(d*x + c) + a^2*b*d)

________________________________________________________________________________________

giac [B]  time = 1.41, size = 238, normalized size = 4.41 \[ -\frac {\frac {{\left (a - b\right )} \log \left ({\left | a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{3} - a^{2} b} - \frac {a^{2} - 2 \, a b - b^{2} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{{\left (a^{3} - a^{2} b\right )} {\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}} - \frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a - b)*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/
(a^3 - a^2*b) - (a^2 - 2*a*b - b^2 + a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*a*b*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((a^3 - a^2*b)*(a + b + a*(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))) - log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1
))/a^2)/d

________________________________________________________________________________________

maple [A]  time = 0.13, size = 54, normalized size = 1.00 \[ \frac {1}{a d \left (a +b \sec \left (d x +c \right )\right )}-\frac {\ln \left (a +b \sec \left (d x +c \right )\right )}{a^{2} d}+\frac {\ln \left (\sec \left (d x +c \right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+b*sec(d*x+c))^2,x)

[Out]

1/a/d/(a+b*sec(d*x+c))-ln(a+b*sec(d*x+c))/a^2/d+1/d/a^2*ln(sec(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 41, normalized size = 0.76 \[ -\frac {\frac {b}{a^{3} \cos \left (d x + c\right ) + a^{2} b} + \frac {\log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b/(a^3*cos(d*x + c) + a^2*b) + log(a*cos(d*x + c) + b)/a^2)/d

________________________________________________________________________________________

mupad [B]  time = 1.48, size = 257, normalized size = 4.76 \[ \frac {2\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )}{a^2\,d}-\frac {b\,\left (a+a\,\cos \left (c+d\,x\right )-2\,a\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )+2\,a\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )+\frac {2\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )\,\left (a^3\,d-a^3\,d\,\cos \left (c+d\,x\right )\right )}{a^2\,d}\right )}{a^2\,d\,\left (a-b\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(a/(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*cos(c + d*x))/2))))/(a^2*d)
- (b*(a + a*cos(c + d*x) - 2*a*atanh(a/(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*
cos(c + d*x))/2))) + 2*a*cos(c + d*x)*atanh(a/(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 +
b + (a*cos(c + d*x))/2))) + (2*atanh(a/(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*
cos(c + d*x))/2)))*(a^3*d - a^3*d*cos(c + d*x)))/(a^2*d)))/(a^2*d*(a - b)*(b + a*cos(c + d*x)))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {\tilde {\infty } x \tan {\relax (c )}}{\sec ^{2}{\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac {1}{2 b^{2} d \sec ^{2}{\left (c + d x \right )}} & \text {for}\: a = 0 \\\frac {\int \frac {\tan {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )} - 2 \cos {\left (c + d x \right )} \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} & \text {for}\: b = - a \cos {\left (c + d x \right )} \\\frac {x \tan {\relax (c )}}{\left (a + b \sec {\relax (c )}\right )^{2}} & \text {for}\: d = 0 \\\frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d} & \text {for}\: b = 0 \\- \frac {2 a \log {\left (\frac {a}{b} + \sec {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {2 a}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} - \frac {2 b \log {\left (\frac {a}{b} + \sec {\left (c + d x \right )} \right )} \sec {\left (c + d x \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(c)/sec(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-1/(2*b**2*d*sec(c + d*x)**2), Eq(a, 0)),
(Integral(tan(c + d*x)/(cos(c + d*x)**2*sec(c + d*x)**2 - 2*cos(c + d*x)*sec(c + d*x) + 1), x)/a**2, Eq(b, -a
*cos(c + d*x))), (x*tan(c)/(a + b*sec(c))**2, Eq(d, 0)), (log(tan(c + d*x)**2 + 1)/(2*a**2*d), Eq(b, 0)), (-2*
a*log(a/b + sec(c + d*x))/(2*a**3*d + 2*a**2*b*d*sec(c + d*x)) + a*log(tan(c + d*x)**2 + 1)/(2*a**3*d + 2*a**2
*b*d*sec(c + d*x)) + 2*a/(2*a**3*d + 2*a**2*b*d*sec(c + d*x)) - 2*b*log(a/b + sec(c + d*x))*sec(c + d*x)/(2*a*
*3*d + 2*a**2*b*d*sec(c + d*x)) + b*log(tan(c + d*x)**2 + 1)*sec(c + d*x)/(2*a**3*d + 2*a**2*b*d*sec(c + d*x))
, True))

________________________________________________________________________________________