\(\int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [294]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-1)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {\sqrt {a+b \tan ^2(e+f x)}}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b f} \]

[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*tan(f*x+e)^2)^
(3/2)/b/f

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 457, 81, 52, 65, 214} \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b f}-\frac {\sqrt {a+b \tan ^2(e+f x)}}{f} \]

[In]

Int[Tan[e + f*x]^3*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*Tan[e
+ f*x]^2)^(3/2)/(3*b*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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93 \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {3 \sqrt {a-b} b \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\sqrt {a+b \tan ^2(e+f x)} \left (a-3 b+b \tan ^2(e+f x)\right )}{3 b f} \]

[In]

Integrate[Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30

method result size
derivativedivides \(\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b f}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f}+\frac {b \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}-\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) \(114\)
default \(\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b f}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f}+\frac {b \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}-\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) \(114\)

[In]

int((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x,method=_RETURNVERBOSE)

[Out]

1/3*(a+b*tan(f*x+e)^2)^(3/2)/b/f-(a+b*tan(f*x+e)^2)^(1/2)/f+1/f*b/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)
/(-a+b)^(1/2))-1/f*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 = 254, normalized size of antiderivative = 2.89 \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, \sqrt {a - b} b \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (b \tan \left (f x + e\right )^{2} + a - 3 \, b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, b f}, -\frac {3 \, \sqrt {-a + b} b \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) - 2 \, {\left (b \tan \left (f x + e\right )^{2} + a - 3 \, b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, b f}\right ] \]

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx \]

[In]

integrate((a+b*tan(f*x+e)**2)**(1/2)*tan(f*x+e)**3,x)

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*tan(e + f*x)**3, x)

Maxima [F]

\[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{3} \,d x } \]

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*tan(f*x + e)^3, x)

Giac [F(-1)]

Timed out. \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.86 \[ \int \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )\,\sqrt {a-b}}{f}-\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{f}+\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{3\,b\,f} \]

[In]

int(tan(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

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