\(\int \frac {\tan ^{\frac {5}{2}}(a+b \log (c x^n))}{x} \, dx\) [180]

Optimal result

Integrand size = 19, antiderivative size = 201 \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\log \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n} \]

[Out]

-1/2*arctan(-1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2))/b/n*2^(1/2)-1/2*arctan(1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2))/b/
n*2^(1/2)-1/4*ln(1-2^(1/2)*tan(a+b*ln(c*x^n))^(1/2)+tan(a+b*ln(c*x^n)))/b/n*2^(1/2)+1/4*ln(1+2^(1/2)*tan(a+b*l
n(c*x^n))^(1/2)+tan(a+b*ln(c*x^n)))/b/n*2^(1/2)+2/3*tan(a+b*ln(c*x^n))^(3/2)/b/n

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\arctan \left (\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{\sqrt {2} b n}+\frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n}+\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n} \]

[In]

Int[Tan[a + b*Log[c*x^n]]^(5/2)/x,x]

[Out]

ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]]/(Sqrt[2]*b*n) - ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]
]/(Sqrt[2]*b*n) - Log[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n) + Log[1
+ Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n) + (2*Tan[a + b*Log[c*x^n]]^(3/
2))/(3*b*n)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \tan ^{\frac {5}{2}}(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = \frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {\text {Subst}\left (\int \sqrt {\tan (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = \frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \\ & = \frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {2 \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \\ & = \frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}+\frac {\text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {\text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \\ & = \frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {\text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{2 b n}-\frac {\text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{2 b n}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{2 \sqrt {2} b n}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{2 \sqrt {2} b n} \\ & = -\frac {\log \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n} \\ & = \frac {\arctan \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\log \left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{2 \sqrt {2} b n}+\frac {2 \tan ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {-3 \arctan \left (\sqrt [4]{-\tan ^2\left (a+b \log \left (c x^n\right )\right )}\right ) \sqrt [4]{-\tan \left (a+b \log \left (c x^n\right )\right )}+3 \text {arctanh}\left (\sqrt [4]{-\tan ^2\left (a+b \log \left (c x^n\right )\right )}\right ) \sqrt [4]{-\tan \left (a+b \log \left (c x^n\right )\right )}+2 \tan ^{\frac {7}{4}}\left (a+b \log \left (c x^n\right )\right )}{3 b n \sqrt [4]{\tan \left (a+b \log \left (c x^n\right )\right )}} \]

[In]

Integrate[Tan[a + b*Log[c*x^n]]^(5/2)/x,x]

[Out]

(-3*ArcTan[(-Tan[a + b*Log[c*x^n]]^2)^(1/4)]*(-Tan[a + b*Log[c*x^n]])^(1/4) + 3*ArcTanh[(-Tan[a + b*Log[c*x^n]
]^2)^(1/4)]*(-Tan[a + b*Log[c*x^n]])^(1/4) + 2*Tan[a + b*Log[c*x^n]]^(7/4))/(3*b*n*Tan[a + b*Log[c*x^n]]^(1/4)
)

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {\frac {2 {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}{1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )\right )}{4}}{n b}\)	\(139\)
default	\(\frac {\frac {2 {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}{1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )\right )}{4}}{n b}\)	\(139\)

[In]

int(tan(a+b*ln(c*x^n))^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

1/n/b*(2/3*tan(a+b*ln(c*x^n))^(3/2)-1/4*2^(1/2)*(ln((1-2^(1/2)*tan(a+b*ln(c*x^n))^(1/2)+tan(a+b*ln(c*x^n)))/(1
+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2)+tan(a+b*ln(c*x^n))))+2*arctan(1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2))+2*arctan(-
1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.51 \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {3 \, {\left (b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}} \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}}\right )} \log \left (b^{3} n^{3} \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {3}{4}} + \sqrt {\frac {\sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1}}\right ) + 3 \, {\left (-i \, b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}} \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) - i \, b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}}\right )} \log \left (i \, b^{3} n^{3} \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {3}{4}} + \sqrt {\frac {\sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1}}\right ) + 3 \, {\left (i \, b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}} \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + i \, b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}}\right )} \log \left (-i \, b^{3} n^{3} \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {3}{4}} + \sqrt {\frac {\sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1}}\right ) - 3 \, {\left (b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}} \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + b n \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {1}{4}}\right )} \log \left (-b^{3} n^{3} \left (-\frac {1}{b^{4} n^{4}}\right )^{\frac {3}{4}} + \sqrt {\frac {\sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1}}\right ) - 4 \, \sqrt {\frac {\sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1}} \sin \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right )}{6 \, {\left (b n \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + b n\right )}} \]

[In]

integrate(tan(a+b*log(c*x^n))^(5/2)/x,x, algorithm="fricas")

[Out]

-1/6*(3*(b*n*(-1/(b^4*n^4))^(1/4)*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + b*n*(-1/(b^4*n^4))^(1/4))*log(b^3*n^3
*(-1/(b^4*n^4))^(3/4) + sqrt(sin(2*b*n*log(x) + 2*b*log(c) + 2*a)/(cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + 1)))
+ 3*(-I*b*n*(-1/(b^4*n^4))^(1/4)*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) - I*b*n*(-1/(b^4*n^4))^(1/4))*log(I*b^3
*n^3*(-1/(b^4*n^4))^(3/4) + sqrt(sin(2*b*n*log(x) + 2*b*log(c) + 2*a)/(cos(2*b*n*log(x) + 2*b*log(c) + 2*a) +
1))) + 3*(I*b*n*(-1/(b^4*n^4))^(1/4)*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + I*b*n*(-1/(b^4*n^4))^(1/4))*log(-I
*b^3*n^3*(-1/(b^4*n^4))^(3/4) + sqrt(sin(2*b*n*log(x) + 2*b*log(c) + 2*a)/(cos(2*b*n*log(x) + 2*b*log(c) + 2*a
) + 1))) - 3*(b*n*(-1/(b^4*n^4))^(1/4)*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + b*n*(-1/(b^4*n^4))^(1/4))*log(-b
^3*n^3*(-1/(b^4*n^4))^(3/4) + sqrt(sin(2*b*n*log(x) + 2*b*log(c) + 2*a)/(cos(2*b*n*log(x) + 2*b*log(c) + 2*a)
+ 1))) - 4*sqrt(sin(2*b*n*log(x) + 2*b*log(c) + 2*a)/(cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + 1))*sin(2*b*n*log
(x) + 2*b*log(c) + 2*a))/(b*n*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + b*n)

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \]

[In]

integrate(tan(a+b*ln(c*x**n))**(5/2)/x,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {\tan \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}{x} \,d x } \]

[In]

integrate(tan(a+b*log(c*x^n))^(5/2)/x,x, algorithm="maxima")

[Out]

integrate(tan(b*log(c*x^n) + a)^(5/2)/x, x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \]

[In]

integrate(tan(a+b*log(c*x^n))^(5/2)/x,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 30.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.39 \[ \int \frac {\tan ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {2\,{\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2}}{3\,b\,n}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n} \]

[In]

int(tan(a + b*log(c*x^n))^(5/2)/x,x)

[Out]

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