\(\int (a+b \tan ^2(e+f x))^{3/2} \, dx\) [112]

Optimal result

Integrand size = 16, antiderivative size = 125 \[ \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {(a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {(3 a-2 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.14 \[ \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {-2 (a-b)^{3/2} \arctan \left (\frac {\sqrt {b}+\sqrt {b} \tan ^2(e+f x)-\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\sqrt {b} (-3 a+2 b) \log \left (-\sqrt {b} \tan (e+f x)+\sqrt {a+b \tan ^2(e+f x)}\right )+b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \] Input:

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

Output:

(-2*(a - b)^(3/2)*ArcTan[(Sqrt[b] + Sqrt[b]*Tan[e + f*x]^2 - Tan[e + f*x]* 
Sqrt[a + b*Tan[e + f*x]^2])/Sqrt[a - b]] + Sqrt[b]*(-3*a + 2*b)*Log[-(Sqrt 
[b]*Tan[e + f*x]) + Sqrt[a + b*Tan[e + f*x]^2]] + b*Tan[e + f*x]*Sqrt[a + 
b*Tan[e + f*x]^2])/(2*f)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4144, 318, 398, 224, 219, 291, 216}

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.

\(\displaystyle \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \tan (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4144

\(\displaystyle \frac {\int \frac {\left (b \tan ^2(e+f x)+a\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (2 (a-b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+\sqrt {b} (3 a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )\right )+\frac {1}{2} b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{2} \left (2 (a-b)^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}+\sqrt {b} (3 a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )\right )+\frac {1}{2} b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{2} \left (2 (a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )+\sqrt {b} (3 a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )\right )+\frac {1}{2} b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

Input:

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

Output:

((2*(a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x 
]^2]] + (3*a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[ 
e + f*x]^2]])/2 + (b*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/2)/f
 

Defintions of rubi rules used

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 219

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

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

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 3042

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

rule 4144

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(296\) vs. \(2(107)=214\).

Time = 0.34 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.38

method	result	size
derivativedivides	\(\frac {b \tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 f}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{f}-\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f b \left (a -b \right )}+\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \,b^{2} \left (a -b \right )}\)	\(297\)
default	\(\frac {b \tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 f}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \left (a -b \right )}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{f}-\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f b \left (a -b \right )}+\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{f \,b^{2} \left (a -b \right )}\)	\(297\)

Input:

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

Output:

1/2*b*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/f+3/2/f*b^(1/2)*a*ln(b^(1/2)*tan 
(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*(b^4*(a-b))^(1/2)/(a-b)*arctan(b^2*( 
a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))-1/f*b^(3/2)*ln 
(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))-2/f*a/b*(b^4*(a-b))^(1/2)/(a 
-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e) 
)+1/f*a^2*(b^4*(a-b))^(1/2)/b^2/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/( 
a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 521, normalized size of antiderivative = 4.17 \[ \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {{\left (3 \, a - 2 \, b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, {\left (a - b\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, -\frac {{\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) - {\left (-a + b\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, f}, \frac {4 \, {\left (a - b\right )}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a - b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) - {\left (3 \, a - 2 \, b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, \frac {2 \, {\left (a - b\right )}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a - b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) - {\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) + \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, f}\right ] \] Input:

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

Output:

[-1/4*((3*a - 2*b)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^ 
2 + a)*sqrt(b)*tan(f*x + e) + a) + 2*(a - b)*sqrt(-a + b)*log(-((a - 2*b)* 
tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - 
a)/(tan(f*x + e)^2 + 1)) - 2*sqrt(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/f, 
 -1/2*((3*a - 2*b)*sqrt(-b)*arctan(sqrt(-b)*tan(f*x + e)/sqrt(b*tan(f*x + 
e)^2 + a)) - (-a + b)^(3/2)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan( 
f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - sqr 
t(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/f, 1/4*(4*(a - b)^(3/2)*arctan(sqr 
t(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) - (3*a - 2*b)*sqrt(b)*lo 
g(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + 
 a) + 2*sqrt(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/f, 1/2*(2*(a - b)^(3/2) 
*arctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) - (3*a - 2*b) 
*sqrt(-b)*arctan(sqrt(-b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) + sqrt( 
b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/f]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x)**2*b + a),x)*a + int(sqrt(tan(e + f*x)**2*b + a)*tan 
(e + f*x)**2,x)*b