Integrand size = 25, antiderivative size = 214 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {(a-b) \left (a^2+10 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b f}+\frac {(7 a+b) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \]
a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f-1/16*(a-b) *(a^2+10*a*b+b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b ^(3/2)/f+1/16*(a^2-8*a*b-b^2)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f+1/ 24*(7*a+b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f+1/6*b*(a+b+b*tan(f*x+ e)^2)^(1/2)*tan(f*x+e)^5/f
Time = 5.29 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.21 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\frac {\left (16 a^{3/2} b \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )-\frac {(a-b) \left (a^2+10 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 \sqrt {2} b f (a+2 b+a \cos (2 e+2 f x))^{3/2}}+\frac {\left (9 a^2-58 a b+17 b^2+4 \left (3 a^2-24 a b-11 b^2\right ) \cos (2 (e+f x))+\left (3 a^2-38 a b+3 b^2\right ) \cos (4 (e+f x))\right ) \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x)}{384 b f} \]
((16*a^(3/2)*b*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2 ]] - ((a - b)*(a^2 + 10*a*b + b^2)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3 /2))/(4*Sqrt[2]*b*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)) + ((9*a^2 - 58*a *b + 17*b^2 + 4*(3*a^2 - 24*a*b - 11*b^2)*Cos[2*(e + f*x)] + (3*a^2 - 38*a *b + 3*b^2)*Cos[4*(e + f*x)])*Sec[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2]*Ta n[e + f*x])/(384*b*f)
Time = 0.55 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4629, 2075, 379, 444, 27, 444, 25, 398, 224, 219, 291, 216}
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.
| \(\displaystyle \int \tan ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 379 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\tan ^4(e+f x) \left (b (7 a+b) \tan ^2(e+f x)+(a+b) (6 a+b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\int \frac {3 b \tan ^2(e+f x) \left ((a+b) (7 a+b)-\left (a^2-8 b a-b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 b}\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \int \frac {\tan ^2(e+f x) \left ((a+b) (7 a+b)-\left (a^2-8 b a-b^2\right ) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (-\frac {\int -\frac {(a-b) \left (a^2+10 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (a^2-8 b a-b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {\int \frac {(a-b) \left (a^2+10 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (a^2-8 b a-b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {(a-b) \left (a^2+10 a b+b^2\right ) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-16 a^2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {(a-b) \left (a^2+10 a b+b^2\right ) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-16 a^2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {\frac {(a-b) \left (a^2+10 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 a^2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {\frac {(a-b) \left (a^2+10 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 a^2 b \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} (7 a+b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {3}{4} \left (\frac {\frac {(a-b) \left (a^2+10 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 a^{3/2} b \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b}-\frac {\left (a^2-8 a b-b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}\right )\right )+\frac {1}{6} b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
((b*Tan[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/6 + (((7*a + b)*Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/4 - (3*((-16*a^(3/2)*b*ArcTan[(Sqr t[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] + ((a - b)*(a^2 + 10*a* b + b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/S qrt[b])/(2*b) - ((a^2 - 8*a*b - b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f *x]^2])/(2*b)))/4)/6)/f
3.4.96.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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])
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]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[d*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*e*(m + 2*(p + q) + 1))), x] + Simp[1/(b*(m + 2*(p + q) + 1)) Int[(e *x)^m*(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*((b*c - a*d)*(m + 1) + b*c*2 *(p + q)) + (d*(b*c - a*d)*(m + 1) + d*2*(q - 1)*(b*c - a*d) + b*c*d*2*(p + q))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d, 0 ] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && GtQ[m, 1]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(1618\) vs. \(2(188)=376\).
Time = 21.72 (sec) , antiderivative size = 1619, normalized size of antiderivative = 7.57
-1/96/f/(-a)^(1/2)/b^(9/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1 +cos(f*x+e))^2)^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))*(-6*cos(f*x+e)^2*s in(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(11/2)- 96*cos(f*x+e)^3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*si n(f*x+e)*a)*b^(9/2)*a^2-6*cos(f*x+e)*sin(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e )^2)/(1+cos(f*x+e))^2)^(1/2)*b^(11/2)+76*cos(f*x+e)^2*sin(f*x+e)*(-a)^(1/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(9/2)*a+28*(-a)^(1/2)*b^(1 1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)+76*cos(f*x+e)* sin(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(9/2)* a-6*cos(f*x+e)^2*sin(f*x+e)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*b^(7/2)*a^2+28*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 1/2)*b^(11/2)*tan(f*x+e)-28*(-a)^(1/2)*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*a*sin(f*x+e)-6*cos(f*x+e)*sin(f*x+e)*(-a)^(1/2)*((b+a*cos (f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)*a^2-16*(-a)^(1/2)*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(11/2)*tan(f*x+e)*sec(f*x+e)-28*(-a)^(1/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(9/2)*a*tan(f*x+e)+3*cos(f *x+e)^3*(-a)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/ 2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+ e)*a-a-b)/(sin(f*x+e)+1))*a^3*b^3+27*cos(f*x+e)^3*(-a)^(1/2)*ln(4*(((b+...
Time = 4.15 (sec) , antiderivative size = 1777, normalized size of antiderivative = 8.30 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\text {Too large to display} \]
[1/192*(24*sqrt(-a)*a*b^2*cos(f*x + e)^5*log(128*a^4*cos(f*x + e)^8 - 256* (a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a ^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^ 2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 3*(a^3 + 9*a^2*b - 9*a*b^2 - b^3)*s qrt(b)*cos(f*x + e)^5*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b ^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b) *sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*((3*a^2*b - 38*a*b^2 + 3*b^3)*cos(f*x + e)^4 + 8*b^3 + 14*(a* b^2 - b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin (f*x + e))/(b^2*f*cos(f*x + e)^5), 1/96*(12*sqrt(-a)*a*b^2*cos(f*x + e)^5* log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28* a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(1 6*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2 *b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e ))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 3* (a^3 + 9*a^2*b - 9*a*b^2 - b^3)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x +...
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{4}{\left (e + f x \right )}\, dx \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]