\(\int (b \tan ^p(c+d x))^{\frac {1}{p}} \, dx\) [52]

Optimal result

Integrand size = 14, antiderivative size = 32 \[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=-\frac {\cot (c+d x) \log (\cos (c+d x)) \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}}}{d} \]

[Out]

-cot(d*x+c)*ln(cos(d*x+c))*(b*tan(d*x+c)^p)^(1/p)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3740, 3556} \[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=-\frac {\cot (c+d x) \log (\cos (c+d x)) \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}}}{d} \]

[In]

Int[(b*Tan[c + d*x]^p)^p^(-1),x]

[Out]

-((Cot[c + d*x]*Log[Cos[c + d*x]]*(b*Tan[c + d*x]^p)^p^(-1))/d)

Rule 3556

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

Rule 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps \begin{align*} \text {integral}& = \left (\cot (c+d x) \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}}\right ) \int \tan (c+d x) \, dx \\ & = -\frac {\cot (c+d x) \log (\cos (c+d x)) \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=-\frac {\cot (c+d x) \log (\cos (c+d x)) \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}}}{d} \]

[In]

Integrate[(b*Tan[c + d*x]^p)^p^(-1),x]

[Out]

-((Cot[c + d*x]*Log[Cos[c + d*x]]*(b*Tan[c + d*x]^p)^p^(-1))/d)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 6.38 (sec) , antiderivative size = 5979, normalized size of antiderivative = 186.84

method	result	size
risch	\(\text {Expression too large to display}\)	\(5979\)

[In]

int((b*tan(d*x+c)^p)^(1/p),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=-\frac {b^{\left (\frac {1}{p}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

[In]

integrate((b*tan(d*x+c)^p)^(1/p),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=\int \left (b \tan ^{p}{\left (c + d x \right )}\right )^{\frac {1}{p}}\, dx \]

[In]

integrate((b*tan(d*x+c)**p)**(1/p),x)

[Out]

Integral((b*tan(c + d*x)**p)**(1/p), x)

Maxima [F]

\[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=\int { \left (b \tan \left (d x + c\right )^{p}\right )^{\left (\frac {1}{p}\right )} \,d x } \]

[In]

integrate((b*tan(d*x+c)^p)^(1/p),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c)^p)^(1/p), x)

Giac [F]

\[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=\int { \left (b \tan \left (d x + c\right )^{p}\right )^{\left (\frac {1}{p}\right )} \,d x } \]

[In]

integrate((b*tan(d*x+c)^p)^(1/p),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c)^p)^(1/p), x)

Mupad [F(-1)]

Timed out. \[ \int \left (b \tan ^p(c+d x)\right )^{\frac {1}{p}} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^p\right )}^{1/p} \,d x \]

[In]

int((b*tan(c + d*x)^p)^(1/p),x)

[Out]

int((b*tan(c + d*x)^p)^(1/p), x)