Integrand size = 23, antiderivative size = 88 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},1,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{5 f} \]
1/5*AppellF1(5/2,1,-p,7/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*tan(f*x+e)^ 5*(a+b+b*tan(f*x+e)^2)^p/f/((1+b*tan(f*x+e)^2/(a+b))^p)
Leaf count is larger than twice the leaf count of optimal. \(2777\) vs. \(2(88)=176\).
Time = 18.57 (sec) , antiderivative size = 2777, normalized size of antiderivative = 31.56 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\text {Result too large to show} \]
((a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*(a + b*Sec[e + f*x]^2 )^p*Tan[e + f*x]^5*((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x] ^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x]^2)/(3*(a + b)*AppellF1[1/2, -p , 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF 1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f* x]^2])*Tan[e + f*x]^2) + (-3*Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))] + Hypergeometric2F1[3/2, -p, 5/2, -((b*Tan[e + f*x]^2)/( a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/(3*f*(((a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(1 + p)*((9*(a + b)*AppellF1[ 1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f *x]^2)/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^ 2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan [e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (-3*Hypergeomet ric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))] + Hypergeometric2F1[3/ 2, -p, 5/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/3 - (2*a*p*(a + 2*b + a*Cos[2*(e + f*x)])^(-1 + p)*( Sec[e + f*x]^2)^p*Sin[2*(e + f*x)]*Tan[e + f*x]*((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Cos[e + f*x...
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 395, 394}
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.
\(\displaystyle \int \tan ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \frac {\tan ^4(e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},1,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{5 f}\) |
(AppellF1[5/2, 1, -p, 7/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] *Tan[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^p)/(5*f*(1 + (b*Tan[e + f*x]^2) /(a + b))^p)
3.5.47.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
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])
\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \tan \left (f x +e \right )^{4}d x\]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\text {Timed out} \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^4(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]