Integrand size = 25, antiderivative size = 120 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}} \]
arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f+1/3*(a-3*b )*tan(f*x+e)/a^2/b/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*(a+b)*tan(f*x+e)/a/b/f /(a+b+b*tan(f*x+e)^2)^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(409\) vs. \(2(120)=240\).
Time = 6.55 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.41 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^{5/2} \sec ^4(e+f x) \left (\frac {\sqrt {2} \csc (e+f x) \sec (e+f x) \left (\frac {\sin ^2(e+f x)}{a+b}+\frac {(a+2 b+a \cos (2 (e+f x))) \sin ^2(e+f x)}{(a+b)^2}-\frac {12 \sin ^4(e+f x)}{a+b}+\frac {16 \left (a+b-a \sin ^2(e+f x)\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-\frac {6 a (a+b) \sin ^2(e+f x)}{a+2 b+a \cos (2 (e+f x))}+\frac {a^2 (a+b) \sin ^4(e+f x)}{\left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sqrt {a} \sqrt {a+b} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) \sin (e+f x)}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{a^3}\right )}{\left (a+b-a \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (2 a+3 b+a \cos (2 (e+f x))) \tan (e+f x)}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}-\frac {12 (b+(3 a+2 b) \cos (2 (e+f x))) \tan (e+f x)}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}\right )}{384 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
((a + 2*b + a*Cos[2*(e + f*x)])^(5/2)*Sec[e + f*x]^4*((Sqrt[2]*Csc[e + f*x ]*Sec[e + f*x]*(Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*S in[e + f*x]^2)/(a + b)^2 - (12*Sin[e + f*x]^4)/(a + b) + (16*(a + b - a*Si n[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x] ^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x] )/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^ 3))/(a + b - a*Sin[e + f*x]^2)^(3/2) + (8*(2*a + 3*b + a*Cos[2*(e + f*x)]) *Tan[e + f*x])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)) - (12*(b + (3*a + 2*b)*Cos[2*(e + f*x)])*Tan[e + f*x])/((a + b)^2*(a + 2*b + a*Cos[2 *(e + f*x)])^(3/2))))/(384*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4629, 2075, 372, 402, 27, 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.
| \(\displaystyle \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {(a-2 b) \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a b}-\frac {(a+b) \tan (e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 b (a+b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}+\frac {(a-3 b) \tan (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a b}-\frac {(a+b) \tan (e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}+\frac {(a-3 b) \tan (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a b}-\frac {(a+b) \tan (e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {3 b \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}+\frac {(a-3 b) \tan (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a b}-\frac {(a+b) \tan (e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {3 b \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{a \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a b}-\frac {(a+b) \tan (e+f x)}{3 a b \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
(-1/3*((a + b)*Tan[e + f*x])/(a*b*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3* b*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^(3/2) + ((a - 3*b)*Tan[e + f*x])/(a*Sqrt[a + b + b*Tan[e + f*x]^2]))/(3*a*b))/f
3.5.35.3.1 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[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[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -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. \(656\) vs. \(2(106)=212\).
Time = 4.59 (sec) , antiderivative size = 657, normalized size of antiderivative = 5.48
| method | result | size |
| default | \(-\frac {\left (a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b \right ) \left (-6 \sqrt {-a}\, a \left (1-\cos \left (f x +e \right )\right )^{5} \csc \left (f x +e \right )^{5}-6 \sqrt {-a}\, b \left (1-\cos \left (f x +e \right )\right )^{5} \csc \left (f x +e \right )^{5}+20 \left (1-\cos \left (f x +e \right )\right )^{3} a \sqrt {-a}\, \csc \left (f x +e \right )^{3}-12 \sqrt {-a}\, b \left (1-\cos \left (f x +e \right )\right )^{3} \csc \left (f x +e \right )^{3}+3 \ln \left (\frac {4 \sqrt {-a}\, \sqrt {a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b}-8 a \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1}\right ) \left (a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b \right )^{\frac {3}{2}}-6 \sqrt {-a}\, a \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )-6 \sqrt {-a}\, b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{3 f \,a^{2} \sqrt {-a}\, \left (\frac {a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b}{\left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2}}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{5}}\) | \(657\) |
-1/3/f/a^2/(-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4* csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f* x+e)^2+a+b)*(-6*(-a)^(1/2)*a*(1-cos(f*x+e))^5*csc(f*x+e)^5-6*(-a)^(1/2)*b* (1-cos(f*x+e))^5*csc(f*x+e)^5+20*(1-cos(f*x+e))^3*a*(-a)^(1/2)*csc(f*x+e)^ 3-12*(-a)^(1/2)*b*(1-cos(f*x+e))^3*csc(f*x+e)^3+3*ln(4*((-a)^(1/2)*(a*(1-c os(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e ))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f* x+e)-cot(f*x+e)))/((1-cos(f*x+e))^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*c sc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e )^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(3/2)-6*(-a)^(1/2)*a*(csc(f*x+e )-cot(f*x+e))-6*(-a)^(1/2)*b*(csc(f*x+e)-cot(f*x+e)))/((a*(1-cos(f*x+e))^4 *csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x +e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)/((1-cos(f*x+e))^2*csc(f*x+e)^ 2-1)^2)^(5/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^5
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (106) = 212\).
Time = 0.95 (sec) , antiderivative size = 661, normalized size of antiderivative = 5.51 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{24 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{12 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \]
[-1/24*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*log(1 28*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))*sq rt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(4*a^ 2*cos(f*x + e)^3 - (a^2 - 3*a*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b) /cos(f*x + e)^2)*sin(f*x + e))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), -1/12*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e) ^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b) /cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)* cos(f*x + e)^2)*sin(f*x + e))) + 4*(4*a^2*cos(f*x + e)^3 - (a^2 - 3*a*b)*c os(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^ 5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)]
\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]