3.401 \(\int \frac{\tan ^3(e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\)

Optimal. Leaf size=187 \[ \frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f}+\frac{2 \tan (e+f x) \sqrt{\tan (e+f x)+1}}{3 f}-\frac{4 \sqrt{\tan (e+f x)+1}}{3 f}+\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f} \]

[Out]

(Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan
[e + f*x]])])/(2*f) + (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*S
qrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*f) - (4*Sqrt[1 + Tan[e + f*x]])/(3*f) + (2*Tan[e + f*x]*Sqrt[1 + Tan[e +
f*x]])/(3*f)

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Rubi [A]  time = 0.20787, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3566, 3630, 12, 3536, 3535, 203, 207} \[ \frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f}+\frac{2 \tan (e+f x) \sqrt{\tan (e+f x)+1}}{3 f}-\frac{4 \sqrt{\tan (e+f x)+1}}{3 f}+\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^3/Sqrt[1 + Tan[e + f*x]],x]

[Out]

(Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan
[e + f*x]])])/(2*f) + (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*S
qrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*f) - (4*Sqrt[1 + Tan[e + f*x]])/(3*f) + (2*Tan[e + f*x]*Sqrt[1 + Tan[e +
f*x]])/(3*f)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3536

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> With[{q =
Rt[a^2 + b^2, 2]}, Dist[1/(2*q), Int[(a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*
x]], x], x] - Dist[1/(2*q), Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]], x
], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && NeQ[2
*a*c*d - b*(c^2 - d^2), 0] && (PerfectSquareQ[a^2 + b^2] || RationalQ[a, b, c, d])

Rule 3535

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*
d^2)/f, Subst[Int[1/(2*b*c*d - 4*a*d^2 + x^2), x], x, (b*c - 2*a*d - b*d*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]
]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[2
*a*c*d - b*(c^2 - d^2), 0]

Rule 203

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

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\tan ^3(e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx &=\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2}{3} \int \frac{-1-\frac{3}{2} \tan (e+f x)-\tan ^2(e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{4 \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2}{3} \int -\frac{3 \tan (e+f x)}{2 \sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{4 \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}-\int \frac{\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{4 \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}+\frac{\int \frac{1+\left (-1-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}-\frac{\int \frac{1+\left (-1+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}\\ &=-\frac{4 \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}-\frac{\left (4-3 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \left (-1+\sqrt{2}\right )-4 \left (-1+\sqrt{2}\right )^2+x^2} \, dx,x,\frac{1-2 \left (-1+\sqrt{2}\right )-\left (-1+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{2 f}-\frac{\left (4+3 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \left (-1-\sqrt{2}\right )-4 \left (-1-\sqrt{2}\right )^2+x^2} \, dx,x,\frac{1-2 \left (-1-\sqrt{2}\right )-\left (-1-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{2 f}\\ &=\frac{\sqrt{-1+\sqrt{2}} \tan ^{-1}\left (\frac{3-2 \sqrt{2}+\left (1-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{2 \left (-7+5 \sqrt{2}\right )} \sqrt{1+\tan (e+f x)}}\right )}{2 f}+\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{3+2 \sqrt{2}+\left (1+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{1+\tan (e+f x)}}\right )}{2 f}-\frac{4 \sqrt{1+\tan (e+f x)}}{3 f}+\frac{2 \tan (e+f x) \sqrt{1+\tan (e+f x)}}{3 f}\\ \end{align*}

Mathematica [C]  time = 0.196591, size = 90, normalized size = 0.48 \[ \frac{4 (\tan (e+f x)-2) \sqrt{\tan (e+f x)+1}+\frac{6 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )}{\sqrt{1-i}}+\frac{6 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{\sqrt{1+i}}}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^3/Sqrt[1 + Tan[e + f*x]],x]

[Out]

((6*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] + (6*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])
/Sqrt[1 + I] + 4*(-2 + Tan[e + f*x])*Sqrt[1 + Tan[e + f*x]])/(6*f)

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Maple [B]  time = 0.041, size = 329, normalized size = 1.8 \begin{align*}{\frac{2}{3\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,f}\ln \left ( 1+\sqrt{2}-\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{2\,f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }-{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }+{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,f}\ln \left ( 1+\sqrt{2}+\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{2\,f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) } \right ) } \right ) }-{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(1+tan(f*x+e))^(3/2)/f-2*(1+tan(f*x+e))^(1/2)/f-1/8/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(1+2^(1/2)-(2*2^(1/2)+
2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+1/2/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e))^(1/2)-(2*2^(1/2)
+2)^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)-1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e))^(1/2)-(2*2^(1/2)+2)
^(1/2))/(-2+2*2^(1/2))^(1/2))+1/8/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(1+2^(1/2)+(2*2^(1/2)+2)^(1/2)*(1+tan(f*x+e)
)^(1/2)+tan(f*x+e))+1/2/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/
2))^(1/2))*2^(1/2)-1/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))
^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{3}}{\sqrt{\tan \left (f x + e\right ) + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(f*x + e)^3/sqrt(tan(f*x + e) + 1), x)

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Fricas [B]  time = 2.06189, size = 3008, normalized size = 16.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(1/2)^(3/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*sqrt(f
^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e
) + (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)
) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x
+ e))*(f^(-4))^(3/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4
)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - 2*sqrt(1/2))*cos(f*x +
e) + 12*(1/2)^(3/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*sqrt(f^
(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e)
- (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4))
+ 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x +
e))*(f^(-4))^(3/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)
*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + 2*sqrt(1/2))*cos(f*x + e
) + 3*(1/2)^(1/4)*(sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)
) + 4)*(f^(-4))^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4
))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/c
os(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 3*(1/2)^(1/4)*(sqrt(1/2)*f^3*sqrt(f
^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log((2*sqrt(1/2)*
f^2*sqrt(f^(-4))*cos(f*x + e) - (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(
-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x +
e) + sin(f*x + e))/cos(f*x + e)) - 8*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(2*cos(f*x + e) - sin(f
*x + e)))/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (e + f x \right )}}{\sqrt{\tan{\left (e + f x \right )} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)**3/sqrt(tan(e + f*x) + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{3}}{\sqrt{\tan \left (f x + e\right ) + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(f*x + e)^3/sqrt(tan(f*x + e) + 1), x)