3.90 \(\int \frac {\tan ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=35 \[ \frac {2}{a^3 d (\cos (c+d x)+1)}+\frac {\log (\cos (c+d x)+1)}{a^3 d} \]

[Out]

2/a^3/d/(1+cos(d*x+c))+ln(1+cos(d*x+c))/a^3/d

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 43} \[ \frac {2}{a^3 d (\cos (c+d x)+1)}+\frac {\log (\cos (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + a*Sec[c + d*x])^3,x]

[Out]

2/(a^3*d*(1 + Cos[c + d*x])) + Log[1 + Cos[c + d*x]]/(a^3*d)

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3879

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

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {a-a x}{(a+a x)^2} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {2}{a (1+x)^2}-\frac {1}{a (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {2}{a^3 d (1+\cos (c+d x))}+\frac {\log (1+\cos (c+d x))}{a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 0.94 \[ \frac {\tan ^2\left (\frac {1}{2} (c+d x)\right )+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Log[Cos[(c + d*x)/2]] + Tan[(c + d*x)/2]^2)/(a^3*d)

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fricas [A]  time = 0.82, size = 42, normalized size = 1.20 \[ \frac {{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2}{a^{3} d \cos \left (d x + c\right ) + a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.90, size = 56, normalized size = 1.60 \[ -\frac {\frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} + \frac {\cos \left (d x + c\right ) - 1}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.63, size = 51, normalized size = 1.46 \[ -\frac {\ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}}-\frac {2}{d \,a^{3} \left (1+\sec \left (d x +c \right )\right )}+\frac {\ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+a*sec(d*x+c))^3,x)

[Out]

-1/d/a^3*ln(sec(d*x+c))-2/d/a^3/(1+sec(d*x+c))+1/d/a^3*ln(1+sec(d*x+c))

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maxima [A]  time = 0.62, size = 36, normalized size = 1.03 \[ \frac {\frac {2}{a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.17, size = 36, normalized size = 1.03 \[ -\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + a/cos(c + d*x))^3,x)

[Out]

-(log(tan(c/2 + (d*x)/2)^2 + 1) - tan(c/2 + (d*x)/2)^2)/(a^3*d)

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sympy [A]  time = 23.81, size = 457, normalized size = 13.06 \[ \begin {cases} - \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec ^{2}{\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sec {\left (c + d x \right )} + 1 \right )} \sec ^{2}{\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \log {\left (\sec {\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sec {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {\tan ^{2}{\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {2}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \tan ^{3}{\relax (c )}}{\left (a \sec {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+a*sec(d*x+c))**3,x)

[Out]

Piecewise((-log(tan(c + d*x)**2 + 1)*sec(c + d*x)**2/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**
3*d) - 2*log(tan(c + d*x)**2 + 1)*sec(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) -
log(tan(c + d*x)**2 + 1)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + 2*log(sec(c + d*x) +
1)*sec(c + d*x)**2/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + 4*log(sec(c + d*x) + 1)*se
c(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + 2*log(sec(c + d*x) + 1)/(2*a**3*d*s
ec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + tan(c + d*x)**2/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(
c + d*x) + 2*a**3*d) - 2*sec(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) - 2/(2*a**
3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*tan(c)**3/(a*sec(c) + a)**3, True))

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