\(\int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx\) [420]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 \sqrt {a+b} f}+\frac {(2 a+3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) f} \] Output:

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+(2+\cos (2 (e+f x))) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3673, 87, 60, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {(2 a+3 b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )}{2 (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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. \(402\) vs. \(2(102)=204\).

Time = 0.53 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.42

method	result	size
default	\(\frac {-\left (-4 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, a -6 b \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \sqrt {a +b}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+5 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \cos \left (f x +e \right )^{2}+2 \left (a +b -b \cos \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \sqrt {a +b}}{4 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2} f}\)	\(403\)

Input:

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

Output:

1/4*(-(-4*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a-6*b*(a+b-b*cos(f*x+e)^2 
)^(1/2)*(a+b)^(1/2)+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+5*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+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))*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+5*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+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))*b^2) 
*cos(f*x+e)^2+2*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(1/2))/(a+b)^(3/2)/cos(f* 
x+e)^2/f
 

Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.11 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\left [\frac {{\left (2 \, a + 3 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a + b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}, -\frac {{\left (2 \, a + 3 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) \cos \left (f x + e\right )^{2} - {\left (2 \, {\left (a + b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.08 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {4 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2} - \frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3}}{b \sin \left (f x + e\right )^{2} - b} + \frac {{\left (2 \, a b^{2} + 3 \, b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}}}{4 \, b^{2} f} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 959 vs. \(2 (102) = 204\).

Time = 0.64 (sec) , antiderivative size = 959, normalized size of antiderivative = 8.13 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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