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

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.51 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}+2 \sqrt {1+\tan (e+f x)}}{f} \] Input:

Integrate[(1 + Tan[e + f*x])^(3/2),x]
 

Output:

((-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
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3963, 27, 3042, 4019, 3042, 4018, 216, 220}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3963

\(\displaystyle \int \frac {2 \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 4019

\(\displaystyle 2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 4018

\(\displaystyle 2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle 2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 220

\(\displaystyle 2 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

Input:

Int[(1 + Tan[e + f*x])^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

rule 220

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3963

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 4018

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[-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 4019

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> With[{q = Rt[a^2 + b^2, 2]}, Simp[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] - Simp[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] && N 
eQ[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] && NiceSqrtQ[a^2 + b^2]
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.42

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

Input:

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

Output:

1/f*(2*(tan(f*x+e)+1)^(1/2)-1/2*2^(1/2)*(1/2*(2*2^(1/2)+2)^(1/2)*ln(tan(f* 
x+e)+1+(tan(f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))+2*(1-2^(1/2))/(-2 
+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(tan(f*x+e)+1)^(1/2))/(-2+ 
2*2^(1/2))^(1/2)))-1/2*2^(1/2)*(-1/2*(2*2^(1/2)+2)^(1/2)*ln(tan(f*x+e)+1-( 
tan(f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))+2*(1-2^(1/2))/(-2+2*2^(1/ 
2))^(1/2)*arctan((2*(tan(f*x+e)+1)^(1/2)-(2*2^(1/2)+2)^(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 = 0.08 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.09 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {\sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, \sqrt {\tan \left (f x + e\right ) + 1}}{2 \, f} \] Input:

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

Output:

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)) - sq 
rt(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)*sqr 
t(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + sqrt(2)*f*sqr 
t(-(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)) + 4*sqrt(tan(f*x + 
e) + 1))/f
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[9]%%%}+%%%{4,[7]%%%}+%%%{6,[5]%%%}+%%%{4,[3]%%%}+%% 
%{1,[1]%%
 

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-2\,\mathrm {atanh}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}-2\,\mathrm {atanh}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}} \] Input:

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

Output:

(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)
 

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x) + 1),x) + int(sqrt(tan(e + f*x) + 1)*tan(e + f*x),x)