\(\int (1+\tan (e+f x))^{3/2} \, dx\) [397]

Optimal result

Integrand size = 12, antiderivative size = 156 \[ \int (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f} \]

[Out]

-arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(2^(1/2)-1)^(1/2)/f-
arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(1+2^(1/2))^(1/2)/f+2
*(1+tan(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3563, 12, 3617, 3616, 209, 213} \[ \int (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {\sqrt {2}-1} \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f} \]

[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 12

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

Rule 209

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

Rule 213

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

Rule 3563

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 3616

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 3617

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])

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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}} \arctan \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}} \text {arctanh}\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] (verified)

Result contains complex when optimal does not.

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.51 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}+2 \sqrt {1+\tan (e+f x)}}{f} \]

[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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(221\)
default	\(\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(221\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (121) = 242\).

Time = 0.26 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.09 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {\sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, \sqrt {\tan \left (f x + e\right ) + 1}}{2 \, f} \]

[In]

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

[Out]

1/2*(sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(2)*(f^3*sqrt(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)
/f^2) + 2*sqrt(tan(f*x + e) + 1)) - sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f^4)
- f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2
)*log(sqrt(2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + sqrt(2)*f
*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*s
qrt(tan(f*x + e) + 1)) + 4*sqrt(tan(f*x + e) + 1))/f

Sympy [F]

\[ \int (1+\tan (e+f x))^{3/2} \, dx=\int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int (1+\tan (e+f x))^{3/2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  1which is not
of the expected type LIST

Giac [F]

\[ \int (1+\tan (e+f x))^{3/2} \, dx=\int { {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-2\,\mathrm {atanh}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}-2\,\mathrm {atanh}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}} \]

[In]

int((tan(e + f*x) + 1)^(3/2),x)

[Out]

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