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

Optimal. Leaf size=208 \[ \frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}-\frac{4 (\tan (e+f x)+1)^{3/2}}{15 f}-\frac{2 \sqrt{\tan (e+f x)+1}}{f}+\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

[Out]

(Sqrt[(-1 + Sqrt[2])/2]*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 + T
an[e + f*x]])])/f + (Sqrt[(1 + Sqrt[2])/2]*ArcTanh[(4 + 3*Sqrt[2] + (2 + Sqrt[2])*Tan[e + f*x])/(2*Sqrt[7 + 5*
Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/f - (2*Sqrt[1 + Tan[e + f*x]])/f - (4*(1 + Tan[e + f*x])^(3/2))/(15*f) + (2
*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(5*f)

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Rubi [A]  time = 0.264328, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3566, 3630, 12, 3528, 3536, 3535, 203, 207} \[ \frac{\sqrt{\frac{1}{2} \left (\sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\left (2-\sqrt{2}\right ) \tan (e+f x)-3 \sqrt{2}+4}{2 \sqrt{5 \sqrt{2}-7} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}-\frac{4 (\tan (e+f x)+1)^{3/2}}{15 f}-\frac{2 \sqrt{\tan (e+f x)+1}}{f}+\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan (e+f x)+3 \sqrt{2}+4}{2 \sqrt{7+5 \sqrt{2}} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(-1 + Sqrt[2])/2]*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 + T
an[e + f*x]])])/f + (Sqrt[(1 + Sqrt[2])/2]*ArcTanh[(4 + 3*Sqrt[2] + (2 + Sqrt[2])*Tan[e + f*x])/(2*Sqrt[7 + 5*
Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/f - (2*Sqrt[1 + Tan[e + f*x]])/f - (4*(1 + Tan[e + f*x])^(3/2))/(15*f) + (2
*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(5*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 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

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 \tan ^3(e+f x) \sqrt{1+\tan (e+f x)} \, dx &=\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}+\frac{2}{5} \int \sqrt{1+\tan (e+f x)} \left (-1-\frac{5}{2} \tan (e+f x)-\tan ^2(e+f x)\right ) \, dx\\ &=-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}+\frac{2}{5} \int -\frac{5}{2} \tan (e+f x) \sqrt{1+\tan (e+f x)} \, dx\\ &=-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}-\int \tan (e+f x) \sqrt{1+\tan (e+f x)} \, dx\\ &=-\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}-\int \frac{-1+\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=-\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}+\frac{\int \frac{\sqrt{2}+\left (-2-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}-\frac{\int \frac{-\sqrt{2}+\left (-2+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{2 \sqrt{2}}\\ &=-\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}-\frac{\left (4-3 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \sqrt{2} \left (-2+\sqrt{2}\right )-4 \left (-2+\sqrt{2}\right )^2+x^2} \, dx,x,\frac{-\sqrt{2}-2 \left (-2+\sqrt{2}\right )-\left (-2+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{f}-\frac{\left (4+3 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \sqrt{2} \left (-2-\sqrt{2}\right )-4 \left (-2-\sqrt{2}\right )^2+x^2} \, dx,x,\frac{\sqrt{2}-2 \left (-2-\sqrt{2}\right )-\left (-2-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{4-3 \sqrt{2}+\left (2-\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{-7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}+\frac{\sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tanh ^{-1}\left (\frac{4+3 \sqrt{2}+\left (2+\sqrt{2}\right ) \tan (e+f x)}{2 \sqrt{7+5 \sqrt{2}} \sqrt{1+\tan (e+f x)}}\right )}{f}-\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac{2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f}\\ \end{align*}

Mathematica [C]  time = 0.32758, size = 100, normalized size = 0.48 \[ \frac{2 \sqrt{\tan (e+f x)+1} \left (3 \tan ^2(e+f x)+\tan (e+f x)-17\right )+15 \sqrt{1-i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+15 \sqrt{1+i} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 15*Sqrt[1 + I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sq
rt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(-17 + Tan[e + f*x] + 3*Tan[e + f*x]^2))/(15*f)

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Maple [B]  time = 0.046, size = 336, normalized size = 1.6 \begin{align*}{\frac{2}{5\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\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}}{4\,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{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}}{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}}{4\,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}}{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((1+tan(f*x+e))^(1/2)*tan(f*x+e)^3,x)

[Out]

2/5*(1+tan(f*x+e))^(5/2)/f-2/3*(1+tan(f*x+e))^(3/2)/f-2*(1+tan(f*x+e))^(1/2)/f-1/4/f*(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/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/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)+1/4/f*(2*2^(1/2)+2)^(1/2)*ln(1+2^(1/2)+(2*2^(1/2)+2)^(1/2)*(1+t
an(f*x+e))^(1/2)+tan(f*x+e))+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))*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 \sqrt{\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.04108, size = 2990, normalized size = 14.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(60*2^(3/4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*(f^5*sq
rt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e)
+ 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqr
t((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*
(f^(-4))^(3/4) - 1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(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)) - sqrt(2))*cos(f*x + e)^2 + 60*2^(3/4
)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*(f^5*sqrt(f^(-4)) + sqrt
(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2^(1/4)*(sqrt(2
)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) +
sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) -
1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(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)) + sqrt(2))*cos(f*x + e)^2 - 15*2^(1/4)*(sqrt(2)*f^3*sq
rt(f^(-4))*cos(f*x + e)^2 + 2*f*cos(f*x + e)^2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(1/2*(
2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*s
qrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f
*x + e) + 2*sin(f*x + e))/cos(f*x + e)) + 15*2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^2 + 2*f*cos(f*x +
e)^2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(1/2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) -
2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt(
(cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) +
16*(20*cos(f*x + e)^2 - cos(f*x + e)*sin(f*x + e) - 3)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*co
s(f*x + e)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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