\(\int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [442]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{3/2} f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a+b}}{2 f} \] Input:

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

Output:

-1/2*(((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b)^ 
(3/2) - (Sec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(a + b))/f
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3673, 87, 73, 221}

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.

\(\displaystyle \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^3}{\sqrt {a+b \sin (e+f x)^2}}dx\)

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\sqrt {a+b \sin ^2(e+f x)}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-(((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b)^(3/ 
2)) + Sqrt[a + b*Sin[e + f*x]^2]/((a + b)*(1 - Sin[e + f*x]^2)))/(2*f)
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

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

rule 3042

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

rule 3673

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ 
(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m 
 + 1)/2)/(2*f)   Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 
)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ 
erQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(352\) vs. \(2(69)=138\).

Time = 0.54 (sec) , antiderivative size = 353, normalized size of antiderivative = 4.36

method	result	size
default	\(\frac {-\left (2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {3}{2}}}{4 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2} f}\)	\(353\)

Input:

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

Output:

1/4*(-(2*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin 
(f*x+e)+a))*a^2+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1 
/2)+b*sin(f*x+e)+a))*a*b+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e 
)^2)^(1/2)+b*sin(f*x+e)+a))*b^2+2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b* 
cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2+3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2 
)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b+ln(2/(1+sin(f*x+e))*((a+ 
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^2)*cos(f*x+e)^2+2*( 
a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2))/(a+b)^(5/2)/cos(f*x+e)^2/f
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.90 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (2 \, a + b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}, -\frac {{\left (2 \, a + b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) \cos \left (f x + e\right )^{2} - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}\right ] \] Input:

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

Output:

[1/4*((2*a + b)*sqrt(a + b)*cos(f*x + e)^2*log((b*cos(f*x + e)^2 + 2*sqrt( 
-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*sq 
rt(-b*cos(f*x + e)^2 + a + b)*(a + b))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e) 
^2), -1/2*((2*a + b)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*s 
qrt(-a - b)/(b*cos(f*x + e)^2 - a - b))*cos(f*x + e)^2 - sqrt(-b*cos(f*x + 
 e)^2 + a + b)*(a + b))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.53 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a + b\right )} - a^{2} - 2 \, a b - b^{2}} - \frac {{\left (2 \, a b^{2} + b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}}}{4 \, b^{2} f} \] Input:

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

Output:

-1/4*(2*sqrt(b*sin(f*x + e)^2 + a)*b^3/((b*sin(f*x + e)^2 + a)*(a + b) - a 
^2 - 2*a*b - b^2) - (2*a*b^2 + b^3)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt 
(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/(a + b)^(3/2))/(b^2*f 
)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (69) = 138\).

Time = 0.45 (sec) , antiderivative size = 753, normalized size of antiderivative = 9.30 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*a + b)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f* 
x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a 
) - sqrt(a))/sqrt(-a - b))/((a + b)*sqrt(-a - b)) - 2*(2*(sqrt(a)*tan(1/2* 
f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^ 
2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a + (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 
 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/ 
2*f*x + 1/2*e)^2 + a))^3*b + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2* 
e)^2 + a))^2*a^(3/2) + 5*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2* 
f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + 
 a))^2*sqrt(a)*b - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x 
+ 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)) 
*a^2 - (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2 
*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b + 4*(sqrt 
(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f 
*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b^2 - 2*a^(5/2) - 5*a^(3/ 
2)*b - 4*sqrt(a)*b^2)/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f 
*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + 
a))^2 - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 
+ 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a)...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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