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

Optimal. Leaf size=79 \[ -\frac{a^3 \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}+\frac{a \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{b x}{a^2+b^2}+\frac{\tan (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.129113, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3566, 3626, 3617, 31, 3475} \[ -\frac{a^3 \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )}+\frac{a \log (\cos (c+d x))}{d \left (a^2+b^2\right )}-\frac{b x}{a^2+b^2}+\frac{\tan (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rubi steps

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

Mathematica [C]  time = 0.386194, size = 91, normalized size = 1.15 \[ -\frac{\frac{2 a^3 \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )}+\frac{\log (-\tan (c+d x)+i)}{a+i b}+\frac{\log (\tan (c+d x)+i)}{a-i b}-\frac{2 \tan (c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 94, normalized size = 1.2 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{bd}}-{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{b\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ){b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(d*x+c)/b/d-1/2/d/(a^2+b^2)*a*ln(1+tan(d*x+c)^2)-1/d/(a^2+b^2)*b*arctan(tan(d*x+c))-a^3*ln(a+b*tan(d*x+c))/
b^2/(a^2+b^2)/d

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Maxima [A]  time = 1.54494, size = 115, normalized size = 1.46 \begin{align*} -\frac{\frac{2 \, a^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} + \frac{a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3*log(b*tan(d*x + c) + a)/(a^2*b^2 + b^4) + 2*(d*x + c)*b/(a^2 + b^2) + a*log(tan(d*x + c)^2 + 1)/(a
^2 + b^2) - 2*tan(d*x + c)/b)/d

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Fricas [A]  time = 2.21689, size = 261, normalized size = 3.3 \begin{align*} -\frac{2 \, b^{3} d x + a^{3} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (a^{3} + a b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^3*d*x + a^3*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - (a^3 + a*b^2
)*log(1/(tan(d*x + c)^2 + 1)) - 2*(a^2*b + b^3)*tan(d*x + c))/((a^2*b^2 + b^4)*d)

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Sympy [A]  time = 7.44149, size = 554, normalized size = 7.01 \begin{align*} \begin{cases} \tilde{\infty } x \tan ^{2}{\left (c \right )} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{- \frac{\log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{\tan ^{2}{\left (c + d x \right )}}{2 d}}{a} & \text{for}\: b = 0 \\\frac{3 d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{3 i d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{\log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{2 \tan ^{2}{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{3}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = - i b \\- \frac{3 d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{3 i d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{\log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{2 \tan ^{2}{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{3}{2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = i b \\\frac{x \tan ^{3}{\left (c \right )}}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\- \frac{2 a^{3} \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} b^{2} d + 2 b^{4} d} + \frac{2 a^{2} b \tan{\left (c + d x \right )}}{2 a^{2} b^{2} d + 2 b^{4} d} - \frac{a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b^{2} d + 2 b^{4} d} - \frac{2 b^{3} d x}{2 a^{2} b^{2} d + 2 b^{4} d} + \frac{2 b^{3} \tan{\left (c + d x \right )}}{2 a^{2} b^{2} d + 2 b^{4} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-log(tan(c + d*x)**2 + 1)/(2*d) + tan(c + d*x)*
*2/(2*d))/a, Eq(b, 0)), (3*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I*d*x/(-2*b*d*tan(c + d*x) + 2
*I*b*d) - I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - log(tan(c + d*x)**2 + 1)/(
-2*b*d*tan(c + d*x) + 2*I*b*d) - 2*tan(c + d*x)**2/(-2*b*d*tan(c + d*x) + 2*I*b*d) - 3/(-2*b*d*tan(c + d*x) +
2*I*b*d), Eq(a, -I*b)), (-3*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I*d*x/(2*b*d*tan(c + d*x) + 2*
I*b*d) - I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + log(tan(c + d*x)**2 + 1)/(2*
b*d*tan(c + d*x) + 2*I*b*d) + 2*tan(c + d*x)**2/(2*b*d*tan(c + d*x) + 2*I*b*d) + 3/(2*b*d*tan(c + d*x) + 2*I*b
*d), Eq(a, I*b)), (x*tan(c)**3/(a + b*tan(c)), Eq(d, 0)), (-2*a**3*log(a/b + tan(c + d*x))/(2*a**2*b**2*d + 2*
b**4*d) + 2*a**2*b*tan(c + d*x)/(2*a**2*b**2*d + 2*b**4*d) - a*b**2*log(tan(c + d*x)**2 + 1)/(2*a**2*b**2*d +
2*b**4*d) - 2*b**3*d*x/(2*a**2*b**2*d + 2*b**4*d) + 2*b**3*tan(c + d*x)/(2*a**2*b**2*d + 2*b**4*d), True))

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Giac [A]  time = 1.73415, size = 116, normalized size = 1.47 \begin{align*} -\frac{\frac{2 \, a^{3} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (d x + c\right )} b}{a^{2} + b^{2}} + \frac{a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, \tan \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3*log(abs(b*tan(d*x + c) + a))/(a^2*b^2 + b^4) + 2*(d*x + c)*b/(a^2 + b^2) + a*log(tan(d*x + c)^2 +
1)/(a^2 + b^2) - 2*tan(d*x + c)/b)/d