3.302 \(\int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=74 \[ \frac {a^2-b^2}{a b^2 d (a+b \sec (c+d x))}+\frac {\left (a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^2 d}+\frac {\log (\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 74, 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 = {3885, 894} \[ \frac {a^2-b^2}{a b^2 d (a+b \sec (c+d x))}+\frac {\left (a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^2 d}+\frac {\log (\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {b^2-x^2}{x (a+x)^2} \, dx,x,b \sec (c+d x)\right )}{b^2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^2}{a^2 x}+\frac {a^2-b^2}{a (a+x)^2}+\frac {-a^2-b^2}{a^2 (a+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{b^2 d}\\ &=\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^2 d}+\frac {a^2-b^2}{a b^2 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 62, normalized size = 0.84 \[ -\frac {\frac {b-\frac {b^3}{a^2}}{a \cos (c+d x)+b}-\frac {\left (a^2+b^2\right ) \log (a \cos (c+d x)+b)}{a^2}+\log (\cos (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.51, size = 102, normalized size = 1.38 \[ -\frac {a^{2} b - b^{3} - {\left (a^{2} b + b^{3} + {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (a^{3} \cos \left (d x + c\right ) + a^{2} b\right )} \log \left (-\cos \left (d x + c\right )\right )}{a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.03, size = 313, normalized size = 4.23 \[ \frac {\frac {{\left (a^{3} - a^{2} b + a b^{2} - b^{3}\right )} \log \left ({\left | a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{3} b^{2} - a^{2} b^{3}} - \frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{2}} - \frac {a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + \frac {a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{{\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} a^{2} b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.53, size = 93, normalized size = 1.26 \[ -\frac {1}{d b \left (b +a \cos \left (d x +c \right )\right )}+\frac {b}{d \,a^{2} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\ln \left (b +a \cos \left (d x +c \right )\right )}{d \,b^{2}}+\frac {\ln \left (b +a \cos \left (d x +c \right )\right )}{d \,a^{2}}-\frac {\ln \left (\cos \left (d x +c \right )\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.39, size = 74, normalized size = 1.00 \[ -\frac {\frac {a^{2} - b^{2}}{a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}} + \frac {\log \left (\cos \left (d x + c\right )\right )}{b^{2}} - \frac {{\left (a^{2} + b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.65, size = 124, normalized size = 1.68 \[ \frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (\frac {1}{a^2}+\frac {1}{b^2}\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {2\,\left (a+b\right )}{a\,b\,d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + b/cos(c + d*x))^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**3/(a + b*sec(c + d*x))**2, x)

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