\(\int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [293]

Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f} \]

[Out]

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/f+(a+b*tan(f*x+e)^2)^(1/2)/f-1/3*(a+b)*(a+b*tan(f*x
+e)^2)^(3/2)/b^2/f+1/5*(a+b*tan(f*x+e)^2)^(5/2)/b^2/f

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 457, 90, 52, 65, 214} \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f} \]

[In]

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

[Out]

-((Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/f) + Sqrt[a + b*Tan[e + f*x]^2]/f - ((a + b)*(
a + b*Tan[e + f*x]^2)^(3/2))/(3*b^2*f) + (a + b*Tan[e + f*x]^2)^(5/2)/(5*b^2*f)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {-15 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\frac {\sqrt {a+b \tan ^2(e+f x)} \left (-2 a^2-5 a b+15 b^2+(a-5 b) b \tan ^2(e+f x)+3 b^2 \tan ^4(e+f x)\right )}{b^2}}{15 f} \]

[In]

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

[Out]

(-15*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + (Sqrt[a + b*Tan[e + f*x]^2]*(-2*a^2 - 5*a*b
+ 15*b^2 + (a - 5*b)*b*Tan[e + f*x]^2 + 3*b^2*Tan[e + f*x]^4))/b^2)/(15*f)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\)	\(157\)
default	\(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\)	\(157\)

[In]

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

[Out]

1/f*(1/5*tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^(3/2)/b-2/15*a/b^2*(a+b*tan(f*x+e)^2)^(3/2)-1/3*(a+b*tan(f*x+e)^2)^(3
/2)/b+b*(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)))+a/(-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 = 0.32 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.74 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, b^{2} f}, \frac {15 \, \sqrt {-a + b} b^{2} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, b^{2} f}\right ] \]

[In]

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

[Out]

[1/60*(15*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*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a*b + 15*b^2)*sqrt(b*tan(f*x + e)^2 + a
))/(b^2*f), 1/30*(15*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*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a*b + 15*b^2)*sqrt(b*tan(f*x + e)^
2 + a))/(b^2*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 19.81 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}}{5\,b^2\,f}-\left (\frac {2\,a}{3\,b^2\,f}-\frac {a-b}{3\,b^2\,f}\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}-\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\left (a-b\right )-\frac {a^2}{b^2\,f}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}}{\sqrt {a-b}}\right )\,\sqrt {a-b}\,1{}\mathrm {i}}{f} \]

[In]

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

[Out]

(atan(((a + b*tan(e + f*x)^2)^(1/2)*1i)/(a - b)^(1/2))*(a - b)^(1/2)*1i)/f - ((2*a)/(3*b^2*f) - (a - b)/(3*b^2
*f))*(a + b*tan(e + f*x)^2)^(3/2) - (a + b*tan(e + f*x)^2)^(1/2)*(((2*a)/(b^2*f) - (a - b)/(b^2*f))*(a - b) -
a^2/(b^2*f)) + (a + b*tan(e + f*x)^2)^(5/2)/(5*b^2*f)