\(\int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx\) [419]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{3/2} f}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac {(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f} \] Output:

1/8*(8*a^2+24*a*b+15*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a 
+b)^(3/2)/f-1/8*(8*a^2+24*a*b+15*b^2)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^2/f-1 
/8*(8*a+7*b)*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/f+1/4*sec(f*x+e 
)^4*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)/f
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {-\left ((8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )+2 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}+\left (8 a^2+24 a b+15 b^2\right ) \left (\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-\sqrt {a+b \sin ^2(e+f x)}\right )}{8 (a+b)^2 f} \] Input:

Integrate[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]
 

Output:

(-((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2)) + 2*(a + b)*Se 
c[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2) + (8*a^2 + 24*a*b + 15*b^2)*(Sqr 
t[a + b]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - Sqrt[a + b*Sin[ 
e + f*x]^2]))/(8*(a + b)^2*f)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3673, 100, 27, 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.

\(\displaystyle \int \tan ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^5 \sqrt {a+b \sin (e+f x)^2}dx\)

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+7 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+7 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \left ((a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)-2 \sqrt {a+b \sin ^2(e+f x)}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+7 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+7 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \sin ^2(e+f x)}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

Input:

Int[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]
 

Output:

((a + b*Sin[e + f*x]^2)^(3/2)/(2*(a + b)*(1 - Sin[e + f*x]^2)^2) - (((8*a 
+ 7*b)*(a + b*Sin[e + f*x]^2)^(3/2))/((a + b)*(1 - Sin[e + f*x]^2)) - ((8* 
a^2 + 24*a*b + 15*b^2)*(2*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/S 
qrt[a + b]] - 2*Sqrt[a + b*Sin[e + f*x]^2]))/(2*(a + b)))/(4*(a + b)))/(2* 
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 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 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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. \(720\) vs. \(2(157)=314\).

Time = 0.62 (sec) , antiderivative size = 721, normalized size of antiderivative = 4.07

method	result	size
default	\(\frac {\left (-16 \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {3}{2}} a^{2}-48 \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {3}{2}} a b -30 b^{2} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {3}{2}}+8 \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^{4}+40 \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^{3} b +71 \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} b^{2}+54 \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}+15 \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^{4}+8 \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^{4}+40 \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^{3} b +71 \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} b^{2}+54 \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}+15 \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^{4}\right ) \cos \left (f x +e \right )^{4}-2 \left (a +b -b \cos \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}} \left (8 a +7 b \right ) \cos \left (f x +e \right )^{2}+4 \left (a +b -b \cos \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}} a +4 b \left (a +b -b \cos \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}}}{16 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\)	\(721\)

Input:

int((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
 

Output:

1/16*((-16*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a^2-48*(a+b-b*cos(f*x+e) 
^2)^(1/2)*(a+b)^(3/2)*a*b-30*b^2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)+8* 
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^4+40*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*si 
n(f*x+e)+a))*a^3*b+71*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*b^2+54*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+15*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^4+8*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^4+40*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^3* 
b+71*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*b^2+54*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+15*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^4)*cos(f*x+e)^4-2*(a+b-b*cos(f*x+e)^2) 
^(3/2)*(a+b)^(3/2)*(8*a+7*b)*cos(f*x+e)^2+4*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+ 
b)^(3/2)*a+4*b*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2))/(a+b)^(3/2)/cos(f*x 
+e)^4/(a^2+2*a*b+b^2)/f
 

Fricas [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.08 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\left [\frac {{\left (8 \, a^{2} + 24 \, a b + 15 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (8 \, a^{2} + 17 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, \frac {{\left (8 \, a^{2} + 24 \, a b + 15 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) \cos \left (f x + e\right )^{4} - {\left (8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (8 \, a^{2} + 17 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}\right ] \] Input:

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="fricas")
 

Output:

[1/16*((8*a^2 + 24*a*b + 15*b^2)*sqrt(a + b)*cos(f*x + e)^4*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*(8*(a^2 + 2*a*b + b^2)*cos(f*x + e)^4 + (8*a^2 + 17*a*b + 9 
*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + 
 b))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4), 1/8*((8*a^2 + 24*a*b + 15*b^2 
)*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)^4 - (8*(a^2 + 2*a*b + b^2)*cos(f*x + e)^ 
4 + (8*a^2 + 17*a*b + 9*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt( 
-b*cos(f*x + e)^2 + a + b))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4)]
 

Sympy [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx \] Input:

integrate((a+b*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**5,x)
 

Output:

Integral(sqrt(a + b*sin(e + f*x)**2)*tan(e + f*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.30 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=-\frac {16 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3} + \frac {{\left (8 \, a^{2} b^{3} + 24 \, a b^{4} + 15 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} - \frac {2 \, {\left ({\left (8 \, a b^{4} + 9 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 15 \, a b^{5} + 7 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} {\left (a + b\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{16 \, b^{3} f} \] Input:

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="maxima")
 

Output:

-1/16*(16*sqrt(b*sin(f*x + e)^2 + a)*b^3 + (8*a^2*b^3 + 24*a*b^4 + 15*b^5) 
*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) - 2*((8*a*b^4 + 9*b^5)*(b*sin(f*x + e)^2 + 
 a)^(3/2) - (8*a^2*b^4 + 15*a*b^5 + 7*b^6)*sqrt(b*sin(f*x + e)^2 + a))/((b 
*sin(f*x + e)^2 + a)^2*(a + b) + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(b*sin( 
f*x + e)^2 + a)*(a^2 + 2*a*b + b^2)))/(b^3*f)
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \] Input:

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(1/2),x)
 

Output:

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{5}d x \] Input:

int((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x)
 

Output:

int(sqrt(sin(e + f*x)**2*b + a)*tan(e + f*x)**5,x)