\(\int \tan ^3(x) (a+b \tan ^4(x))^{3/2} \, dx\) [393]

Optimal result

Integrand size = 17, antiderivative size = 171 \[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{16 \sqrt {b}}+\frac {1}{2} (a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )-\frac {1}{2} (a+b) \sqrt {a+b \tan ^4(x)}+\frac {1}{16} (3 a+4 b) \tan ^2(x) \sqrt {a+b \tan ^4(x)}-\frac {1}{6} \left (a+b \tan ^4(x)\right )^{3/2}+\frac {1}{8} \tan ^2(x) \left (a+b \tan ^4(x)\right )^{3/2} \] Output:

1/16*(3*a^2+12*a*b+8*b^2)*arctanh(b^(1/2)*tan(x)^2/(a+b*tan(x)^4)^(1/2))/b 
^(1/2)+1/2*(a+b)^(3/2)*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^( 
1/2))-1/2*(a+b)*(a+b*tan(x)^4)^(1/2)+1/16*(3*a+4*b)*tan(x)^2*(a+b*tan(x)^4 
)^(1/2)-1/6*(a+b*tan(x)^4)^(3/2)+1/8*tan(x)^2*(a+b*tan(x)^4)^(3/2)
 

Mathematica [A] (verified)

Time = 4.00 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.11 \[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {1}{48} \left (24 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+24 (a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )+\frac {3 \sqrt {a} (3 a+4 b) \text {arcsinh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a}}\right ) \sqrt {a+b \tan ^4(x)}}{\sqrt {b} \sqrt {1+\frac {b \tan ^4(x)}{a}}}+\sqrt {a+b \tan ^4(x)} \left (-8 (4 a+3 b)+3 (5 a+4 b) \tan ^2(x)-8 b \tan ^4(x)+6 b \tan ^6(x)\right )\right ) \] Input:

Integrate[Tan[x]^3*(a + b*Tan[x]^4)^(3/2),x]
 

Output:

(24*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]] + 24* 
(a + b)^(3/2)*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])] 
 + (3*Sqrt[a]*(3*a + 4*b)*ArcSinh[(Sqrt[b]*Tan[x]^2)/Sqrt[a]]*Sqrt[a + b*T 
an[x]^4])/(Sqrt[b]*Sqrt[1 + (b*Tan[x]^4)/a]) + Sqrt[a + b*Tan[x]^4]*(-8*(4 
*a + 3*b) + 3*(5*a + 4*b)*Tan[x]^2 - 8*b*Tan[x]^4 + 6*b*Tan[x]^6))/48
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {3042, 4153, 1579, 591, 25, 682, 27, 719, 224, 219, 488, 219}

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.

\(\displaystyle \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (x)^3 \left (a+b \tan (x)^4\right )^{3/2}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 1579

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

\(\Big \downarrow \) 591

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \int -\frac {\left (a-(3 a+4 b) \tan ^2(x)\right ) \sqrt {b \tan ^4(x)+a}}{\tan ^2(x)+1}d\tan ^2(x)-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int \frac {\left (a-(3 a+4 b) \tan ^2(x)\right ) \sqrt {b \tan ^4(x)+a}}{\tan ^2(x)+1}d\tan ^2(x)-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 682

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {\int \frac {b \left (a (5 a+4 b)-\left (3 a^2+12 b a+8 b^2\right ) \tan ^2(x)\right )}{\left (\tan ^2(x)+1\right ) \sqrt {b \tan ^4(x)+a}}d\tan ^2(x)}{2 b}-\frac {1}{2} \sqrt {a+b \tan ^4(x)} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right )\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {a (5 a+4 b)-\left (3 a^2+12 b a+8 b^2\right ) \tan ^2(x)}{\left (\tan ^2(x)+1\right ) \sqrt {b \tan ^4(x)+a}}d\tan ^2(x)-\frac {1}{2} \sqrt {a+b \tan ^4(x)} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right )\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{\sqrt {b \tan ^4(x)+a}}d\tan ^2(x)-8 (a+b)^2 \int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {b \tan ^4(x)+a}}d\tan ^2(x)\right )-\frac {1}{2} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{1-b \tan ^4(x)}d\frac {\tan ^2(x)}{\sqrt {b \tan ^4(x)+a}}-8 (a+b)^2 \int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {b \tan ^4(x)+a}}d\tan ^2(x)\right )-\frac {1}{2} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{\sqrt {b}}-8 (a+b)^2 \int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {b \tan ^4(x)+a}}d\tan ^2(x)\right )-\frac {1}{2} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (8 (a+b)^2 \int \frac {1}{-\tan ^4(x)+a+b}d\frac {a-b \tan ^2(x)}{\sqrt {b \tan ^4(x)+a}}+\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{\sqrt {b}}\right )-\frac {1}{2} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{\sqrt {b}}+8 (a+b)^{3/2} \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )\right )-\frac {1}{2} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}\right )-\frac {1}{12} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\right )\)

Input:

Int[Tan[x]^3*(a + b*Tan[x]^4)^(3/2),x]
 

Output:

(-1/12*((4 - 3*Tan[x]^2)*(a + b*Tan[x]^4)^(3/2)) + ((((3*a^2 + 12*a*b + 8* 
b^2)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]])/Sqrt[b] + 8*(a + b) 
^(3/2)*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/2 - ( 
(8*(a + b) - (3*a + 4*b)*Tan[x]^2)*Sqrt[a + b*Tan[x]^4])/2)/4)/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 488

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

rule 591

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^p*((c*(2*p + 1) - d*(n + 2*p + 1)*x)/ 
(d^2*(n + 2*p + 1)*(n + 2*p + 2))), x] + Simp[2*(p/(d^2*(n + 2*p + 1)*(n + 
2*p + 2)))   Int[(c + d*x)^n*(a + b*x^2)^(p - 1)*Simp[a*c*d*n + (b*c^2*(2*p 
 + 1) + a*d^2*(n + 2*p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && 
GtQ[p, 0] && LeQ[-1, n, 0] &&  !ILtQ[n + 2*p, 0]
 

rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p 
+ 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p 
+ 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)))   Int[(d + e*x) 
^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* 
d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x 
], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  ! 
RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1579

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], 
 x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
 

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(139)=278\).

Time = 0.20 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.80

method	result	size
derivativedivides	\(\frac {3 a^{2} \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{16 \sqrt {b}}+\frac {b \tan \left (x \right )^{6} \sqrt {a +b \tan \left (x \right )^{4}}}{8}+\frac {5 a \tan \left (x \right )^{2} \sqrt {a +b \tan \left (x \right )^{4}}}{16}-\frac {b \sqrt {a +b \tan \left (x \right )^{4}}}{2}-\frac {b^{2} \left (\frac {\tan \left (x \right )^{4} \sqrt {a +b \tan \left (x \right )^{4}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (x \right )^{4}}}{3 b^{2}}\right )}{2}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2}+a \sqrt {b}\, \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )+\frac {b^{2} \left (\frac {\tan \left (x \right )^{2} \sqrt {a +b \tan \left (x \right )^{4}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2 b^{\frac {3}{2}}}\right )}{2}-a \sqrt {a +b \tan \left (x \right )^{4}}\)	\(307\)
default	\(\frac {3 a^{2} \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{16 \sqrt {b}}+\frac {b \tan \left (x \right )^{6} \sqrt {a +b \tan \left (x \right )^{4}}}{8}+\frac {5 a \tan \left (x \right )^{2} \sqrt {a +b \tan \left (x \right )^{4}}}{16}-\frac {b \sqrt {a +b \tan \left (x \right )^{4}}}{2}-\frac {b^{2} \left (\frac {\tan \left (x \right )^{4} \sqrt {a +b \tan \left (x \right )^{4}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (x \right )^{4}}}{3 b^{2}}\right )}{2}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2}+a \sqrt {b}\, \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )+\frac {b^{2} \left (\frac {\tan \left (x \right )^{2} \sqrt {a +b \tan \left (x \right )^{4}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2 b^{\frac {3}{2}}}\right )}{2}-a \sqrt {a +b \tan \left (x \right )^{4}}\)	\(307\)

Input:

int(tan(x)^3*(a+b*tan(x)^4)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

3/16*a^2*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))/b^(1/2)+1/8*b*tan(x)^6* 
(a+b*tan(x)^4)^(1/2)+5/16*a*tan(x)^2*(a+b*tan(x)^4)^(1/2)-1/2*b*(a+b*tan(x 
)^4)^(1/2)-1/2*b^2*(1/3*tan(x)^4/b*(a+b*tan(x)^4)^(1/2)-2/3*a/b^2*(a+b*tan 
(x)^4)^(1/2))+1/2*(a^2+2*a*b+b^2)/(a+b)^(1/2)*ln((2*a+2*b-2*b*(1+tan(x)^2) 
+2*(a+b)^(1/2)*(b*(1+tan(x)^2)^2-2*b*(1+tan(x)^2)+a+b)^(1/2))/(1+tan(x)^2) 
)+1/2*b^(3/2)*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))+a*b^(1/2)*ln(b^(1/ 
2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))+1/2*b^2*(1/2*tan(x)^2/b*(a+b*tan(x)^4)^( 
1/2)-1/2*a/b^(3/2)*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2)))-a*(a+b*tan(x 
)^4)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 758, normalized size of antiderivative = 4.43 \[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="fricas")
 

Output:

[1/96*(3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan 
(x)^4 + a)*sqrt(b)*tan(x)^2 - a) + 24*(a*b + b^2)*sqrt(a + b)*log(((a*b + 
2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a) 
*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + 2*(6*b^2*tan(x) 
^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b 
*tan(x)^4 + a))/b, -1/48*(3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt( 
b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2)) - 12*(a*b + b^2)*sqrt(a + b)*log((( 
a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^ 
2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) - (6*b^2*ta 
n(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sq 
rt(b*tan(x)^4 + a))/b, 1/96*(48*(a*b + b^2)*sqrt(-a - b)*arctan(sqrt(b*tan 
(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b 
)) + 3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x 
)^4 + a)*sqrt(b)*tan(x)^2 - a) + 2*(6*b^2*tan(x)^6 - 8*b^2*tan(x)^4 + 3*(5 
*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b*tan(x)^4 + a))/b, 1/48*(2 
4*(a*b + b^2)*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sq 
rt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b)) - 3*(3*a^2 + 12*a*b + 8*b^2 
)*sqrt(-b)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2)) + (6*b^2*tan 
(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqr 
t(b*tan(x)^4 + a))/b]
 

Sympy [F]

\[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int \left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (x \right )}\, dx \] Input:

integrate(tan(x)**3*(a+b*tan(x)**4)**(3/2),x)
 

Output:

Integral((a + b*tan(x)**4)**(3/2)*tan(x)**3, x)
 

Maxima [F]

\[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int { {\left (b \tan \left (x\right )^{4} + a\right )}^{\frac {3}{2}} \tan \left (x\right )^{3} \,d x } \] Input:

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(x)^4 + a)^(3/2)*tan(x)^3, x)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.41 \[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {1}{48} \, \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (2 \, {\left (3 \, b \tan \left (x\right )^{2} - 4 \, b\right )} \tan \left (x\right )^{2} + \frac {3 \, {\left (5 \, a b^{2} + 4 \, b^{3}\right )}}{b^{2}}\right )} \tan \left (x\right )^{2} - \frac {8 \, {\left (4 \, a b^{2} + 3 \, b^{3}\right )}}{b^{2}}\right )} \] Input:

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="giac")
 

Output:

1/48*sqrt(b*tan(x)^4 + a)*((2*(3*b*tan(x)^2 - 4*b)*tan(x)^2 + 3*(5*a*b^2 + 
 4*b^3)/b^2)*tan(x)^2 - 8*(4*a*b^2 + 3*b^3)/b^2)
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\int {\mathrm {tan}\left (x\right )}^3\,{\left (b\,{\mathrm {tan}\left (x\right )}^4+a\right )}^{3/2} \,d x \] Input:

int(tan(x)^3*(a + b*tan(x)^4)^(3/2),x)
 

Output:

int(tan(x)^3*(a + b*tan(x)^4)^(3/2), x)
 

Reduce [F]

\[ \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx=\frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{6} b}{8}-\frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{4} b}{6}+\frac {5 \sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{2} a}{16}+\frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{2} b}{4}-\frac {2 \sqrt {\tan \left (x \right )^{4} b +a}\, a}{3}-\frac {\sqrt {\tan \left (x \right )^{4} b +a}\, b}{2}+\frac {3 \left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{3}}{\tan \left (x \right )^{4} b +a}d x \right ) a^{2}}{8}+\frac {3 \left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{3}}{\tan \left (x \right )^{4} b +a}d x \right ) a b}{2}+\left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{3}}{\tan \left (x \right )^{4} b +a}d x \right ) b^{2}-\frac {5 \left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )}{\tan \left (x \right )^{4} b +a}d x \right ) a^{2}}{8}-\frac {\left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )}{\tan \left (x \right )^{4} b +a}d x \right ) a b}{2} \] Input:

int(tan(x)^3*(a+b*tan(x)^4)^(3/2),x)
 

Output:

(6*sqrt(tan(x)**4*b + a)*tan(x)**6*b - 8*sqrt(tan(x)**4*b + a)*tan(x)**4*b 
 + 15*sqrt(tan(x)**4*b + a)*tan(x)**2*a + 12*sqrt(tan(x)**4*b + a)*tan(x)* 
*2*b - 32*sqrt(tan(x)**4*b + a)*a - 24*sqrt(tan(x)**4*b + a)*b + 18*int((s 
qrt(tan(x)**4*b + a)*tan(x)**3)/(tan(x)**4*b + a),x)*a**2 + 72*int((sqrt(t 
an(x)**4*b + a)*tan(x)**3)/(tan(x)**4*b + a),x)*a*b + 48*int((sqrt(tan(x)* 
*4*b + a)*tan(x)**3)/(tan(x)**4*b + a),x)*b**2 - 30*int((sqrt(tan(x)**4*b 
+ a)*tan(x))/(tan(x)**4*b + a),x)*a**2 - 24*int((sqrt(tan(x)**4*b + a)*tan 
(x))/(tan(x)**4*b + a),x)*a*b)/48