3.397 \(\int (1+\tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=156 \[ -\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

[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 + T
an[e + f*x]])])/f) - (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sq
rt[2])]*Sqrt[1 + Tan[e + f*x]])])/f + (2*Sqrt[1 + Tan[e + f*x]])/f

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Rubi [A]  time = 0.157521, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3482, 12, 3536, 3535, 203, 207} \[ -\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[e + f*x])^(3/2),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 + T
an[e + f*x]])])/f) - (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sq
rt[2])]*Sqrt[1 + Tan[e + f*x]])])/f + (2*Sqrt[1 + Tan[e + f*x]])/f

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 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 (1+\tan (e+f x))^{3/2} \, dx &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\int \frac{2 \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+2 \int \frac{\tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}-\frac{\int \frac{1+\left (-1-\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{\sqrt{2}}+\frac{\int \frac{1+\left (-1+\sqrt{2}\right ) \tan (e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx}{\sqrt{2}}\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{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 )}{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 )}{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 )}{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 )}{f}+\frac{2 \sqrt{1+\tan (e+f x)}}{f}\\ \end{align*}

Mathematica [C]  time = 0.0647281, size = 79, normalized size = 0.51 \[ \frac{2 \sqrt{\tan (e+f x)+1}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )}{\sqrt{1-i}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{\sqrt{1+i}}}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[e + f*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.02, size = 314, normalized size = 2. \begin{align*} 2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}+{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{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 ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }+2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{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 ) }+2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{\sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) }}{\sqrt{-2+2\,\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1+tan(f*x+e))^(1/2)/f+1/4/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/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)+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))
-1/4/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/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)+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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.73673, size = 2776, normalized size = 17.79 \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))^(3/2),x, algorithm="fricas")

[Out]

-1/8*(4*8^(1/4)*sqrt(2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/16*8^(3/4)*sqrt(2)*(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) + 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 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) - 1/8*8^(3/4)*(2*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)) + 4*8^(1/4)*sqrt(2)*
sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/16*8^(3/4)*sqrt(2)*(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) - 8^(1/4)*(sqrt(2)
*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + si
n(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) - 1/8
*8^(3/4)*(2*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)) + 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4)) + 2*f)*s
qrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 8^(1/4)*(
sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 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)) - 8^(1/4)*(sq
rt(2)*f^3*sqrt(f^(-4)) + 2*f)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f
^(-4))*cos(f*x + e) - 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqr
t(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*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/f

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tan{\left (e + f x \right )} + 1\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\tan \left (f x + e\right ) + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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