\(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^5(e+f x) \, dx\) [430]

Optimal result

Integrand size = 25, antiderivative size = 220 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\frac {\left (8 a^2+40 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 \sqrt {a+b} f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac {(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f} \] Output:

1/8*(8*a^2+40*a*b+35*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a 
+b)^(1/2)/f-1/8*(8*a^2+40*a*b+35*b^2)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)/f-1/2 
4*(8*a^2+40*a*b+35*b^2)*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/f-1/8*(8*a+9*b)*s 
ec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(5/2)/(a+b)^2/f+1/4*sec(f*x+e)^4*(a+b*sin(f 
*x+e)^2)^(5/2)/(a+b)/f
 

Mathematica [A] (verified)

Time = 2.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.73 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=-\frac {3 (8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}-6 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+\left (8 a^2+40 a b+35 b^2\right ) \left (-3 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+3 b+b \sin ^2(e+f x)\right )\right )}{24 (a+b)^2 f} \] Input:

Integrate[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]
 

Output:

-1/24*(3*(8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) - 6*(a + 
b)*Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(5/2) + (8*a^2 + 40*a*b + 35*b^2) 
*(-3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[ 
a + b*Sin[e + f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/((a + b)^2*f)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3673, 100, 27, 87, 60, 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.

\(\displaystyle \int \tan ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^5 \left (a+b \sin (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2} \left (4 (a+b) \sin ^2(e+f x)+4 a+5 b\right )}{2 \left (1-\sin ^2(e+f x)\right )^2}d\sin ^2(e+f x)}{2 (a+b)}}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2} \left (4 (a+b) \sin ^2(e+f x)+4 a+5 b\right )}{\left (1-\sin ^2(e+f x)\right )^2}d\sin ^2(e+f x)}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+9 b) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \int \frac {\left (b \sin ^2(e+f x)+a\right )^{3/2}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+9 b) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left ((a+b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+9 b) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left ((a+b) \left ((a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+9 b) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left ((a+b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2}-\frac {\frac {(8 a+9 b) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{(a+b) \left (1-\sin ^2(e+f x)\right )}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \left ((a+b) \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \sin ^2(e+f x)}\right )-\frac {2}{3} \left (a+b \sin ^2(e+f x)\right )^{3/2}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

Input:

Int[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]
 

Output:

((a + b*Sin[e + f*x]^2)^(5/2)/(2*(a + b)*(1 - Sin[e + f*x]^2)^2) - (((8*a 
+ 9*b)*(a + b*Sin[e + f*x]^2)^(5/2))/((a + b)*(1 - Sin[e + f*x]^2)) - ((8* 
a^2 + 40*a*b + 35*b^2)*((-2*(a + b*Sin[e + f*x]^2)^(3/2))/3 + (a + b)*(2*S 
qrt[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)))/(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. \(710\) vs. \(2(196)=392\).

Time = 0.57 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.23

method	result	size
default	\(\frac {16 \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, b \cos \left (f x +e \right )^{6}+\left (-64 a \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {5}{2}}-160 b \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {5}{2}}+24 \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}+168 \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 +369 \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}+330 \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}+105 \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}+24 \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}+168 \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 +369 \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}+330 \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}+105 \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}\right ) \cos \left (f x +e \right )^{4}-6 \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (8 a +13 b \right ) \cos \left (f x +e \right )^{2}+12 a \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {5}{2}}+12 b \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {5}{2}}}{48 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} f}\)	\(711\)

Input:

int((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
 

Output:

1/48*(16*(a+b)^(5/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b*cos(f*x+e)^6+(-64*a*(a+b 
-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)-160*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^ 
(5/2)+24*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+168*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+369*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c 
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2+330*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+105*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+24*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+168*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^3*b+369*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+330*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+105*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)*cos(f*x+e)^4-6*(a 
+b)^(5/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(8*a+13*b)*cos(f*x+e)^2+12*a*(a+b-b*c 
os(f*x+e)^2)^(1/2)*(a+b)^(5/2)+12*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2) 
)/(a+b)^(5/2)/cos(f*x+e)^4/f
 

Fricas [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.81 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\left [\frac {3 \, {\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \, {\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{48 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}, \frac {3 \, {\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) \cos \left (f x + e\right )^{4} + {\left (8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \, {\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{24 \, {\left (a + b\right )} f \cos \left (f x + e\right )^{4}}\right ] \] Input:

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="fricas")
 

Output:

[1/48*(3*(8*a^2 + 40*a*b + 35*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*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5*b^2 
)*cos(f*x + e)^4 - 3*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12 
*a*b + 6*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a + b)*f*cos(f*x + e)^4), 
 1/24*(3*(8*a^2 + 40*a*b + 35*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*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5*b^2)*cos(f*x + e)^4 - 3 
*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12*a*b + 6*b^2)*sqrt(- 
b*cos(f*x + e)^2 + a + b))/((a + b)*f*cos(f*x + e)^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+b*sin(f*x+e)**2)**(3/2)*tan(f*x+e)**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.08 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=-\frac {16 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{3} + 48 \, {\left (a b^{3} + 3 \, b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (8 \, a^{2} b^{3} + 40 \, a b^{4} + 35 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}} - \frac {6 \, {\left ({\left (8 \, a b^{4} + 13 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 19 \, a b^{5} + 11 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a + b\right )} + a^{2} + 2 \, a b + b^{2}}}{48 \, b^{3} f} \] Input:

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="maxima")
 

Output:

-1/48*(16*(b*sin(f*x + e)^2 + a)^(3/2)*b^3 + 48*(a*b^3 + 3*b^4)*sqrt(b*sin 
(f*x + e)^2 + a) + 3*(8*a^2*b^3 + 40*a*b^4 + 35*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)))/sqrt 
(a + b) - 6*((8*a*b^4 + 13*b^5)*(b*sin(f*x + e)^2 + a)^(3/2) - (8*a^2*b^4 
+ 19*a*b^5 + 11*b^6)*sqrt(b*sin(f*x + e)^2 + a))/((b*sin(f*x + e)^2 + a)^2 
 - 2*(b*sin(f*x + e)^2 + a)*(a + b) + a^2 + 2*a*b + b^2))/(b^3*f)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3585 vs. \(2 (196) = 392\).

Time = 2.13 (sec) , antiderivative size = 3585, normalized size of antiderivative = 16.30 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x, algorithm="giac")
 

Output:

-1/12*(3*(8*a^2 + 40*a*b + 35*b^2)*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*ta 
n(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/sqrt(-a - b) - 16*(6*(s 
qrt(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))^5*a*b + 9*(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))^5*b^2 + 18*(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))^4*a^(3/2)*b + 39*(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))^4*sqrt(a)*b^2 + 12*(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^2*b + 66*(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*b^2 + 88*(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^3 - 12*(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^(5/2)*b + 6*(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 +...
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \] Input:

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^5*(a + b*sin(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^5(e+f x) \, dx=\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{5}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{5}d x \right ) a \] Input:

int((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^5,x)
 

Output:

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