\(\int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [320]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f} \]

[Out]

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/f/(a-b)^(1/2)-(a+b)*(a+b*tan(f*x+e)^2)^(1/2)/b^2/f+1/3*(a+b*tan
(f*x+e)^2)^(3/2)/b^2/f

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3751, 457, 90, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \]

[In]

Int[Tan[e + f*x]^5/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-(ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]/(Sqrt[a - b]*f)) - ((a + b)*Sqrt[a + b*Tan[e + f*x]^2])/(b^2
*f) + (a + b*Tan[e + f*x]^2)^(3/2)/(3*b^2*f)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {-a-b}{b \sqrt {a+b x}}+\frac {1}{(1+x) \sqrt {a+b x}}+\frac {\sqrt {a+b x}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{b f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \left (2 a+3 b-b \tan ^2(e+f x)\right ) \sqrt {a+b \tan ^2(e+f x)}}{3 b^2}}{2 f} \]

[In]

Integrate[Tan[e + f*x]^5/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-1/2*((2*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b] + (2*(2*a + 3*b - b*Tan[e + f*x]^2)*Sqrt
[a + b*Tan[e + f*x]^2])/(3*b^2))/f

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b^{2}}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\)	\(103\)
default	\(\frac {\frac {\tan \left (f x +e \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b^{2}}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\)	\(103\)

[In]

int(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f*(1/3*tan(f*x+e)^2/b*(a+b*tan(f*x+e)^2)^(1/2)-2/3*a/b^2*(a+b*tan(f*x+e)^2)^(1/2)-1/b*(a+b*tan(f*x+e)^2)^(1/
2)+1/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.31 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {3 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {3 \, \sqrt {-a + b} b^{2} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \]

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*sqrt(a - b)*b^2*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2
*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1))
+ 4*((a*b - b^2)*tan(f*x + e)^2 - 2*a^2 - a*b + 3*b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a*b^2 - b^3)*f), 1/6*(3*
sqrt(-a + b)*b^2*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + 2*((a*b - b^
2)*tan(f*x + e)^2 - 2*a^2 - a*b + 3*b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a*b^2 - b^3)*f)]

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

integrate(tan(f*x+e)**5/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(tan(e + f*x)**5/sqrt(a + b*tan(e + f*x)**2), x)

Maxima [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(f*x + e)^5/sqrt(b*tan(f*x + e)^2 + a), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 12.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{3\,b^2\,f}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )}{f\,\sqrt {a-b}}-\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \]

[In]

int(tan(e + f*x)^5/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

(a + b*tan(e + f*x)^2)^(3/2)/(3*b^2*f) - atanh((a + b*tan(e + f*x)^2)^(1/2)/(a - b)^(1/2))/(f*(a - b)^(1/2)) -
((2*a)/(b^2*f) - (a - b)/(b^2*f))*(a + b*tan(e + f*x)^2)^(1/2)