Integrand size = 23, antiderivative size = 83 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},1,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{5 f} \]
1/5*AppellF1(5/2,1,-p,7/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*tan(f*x+e)^5*(a +b*tan(f*x+e)^2)^p/f/((1+b*tan(f*x+e)^2/a)^p)
Leaf count is larger than twice the leaf count of optimal. \(1896\) vs. \(2(83)=166\).
Time = 13.89 (sec) , antiderivative size = 1896, normalized size of antiderivative = 22.84 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx =\text {Too large to display} \]
(-2*Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]* (a + b*Tan[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/a)^p) + (Tan[e + f*x] *(a + b*Tan[e + f*x]^2)^p*((-a + b*(3 + 2*p))*Hypergeometric2F1[1/2, -p, 3 /2, -((b*Tan[e + f*x]^2)/a)] + (a + b*Tan[e + f*x]^2)*(1 + (b*Tan[e + f*x] ^2)/a)^p))/(b*f*(3 + 2*p)*(1 + (b*Tan[e + f*x]^2)/a)^p) + (3*a*AppellF1[1/ 2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]*Sin[ e + f*x]*(a + b*Tan[e + f*x]^2)^(2*p))/(f*(3*a*AppellF1[1/2, -p, 1, 3/2, - ((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] - a*AppellF1[3/2, -p, 2, 5/ 2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)*((6*a*b*p*Ap pellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p))/(3*a*AppellF1[1/2, -p, 1, 3/2, -(( b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/ 2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] - a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (3*a*AppellF 1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]^ 2*(a + b*Tan[e + f*x]^2)^p)/(3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f* x]^2)/a), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[ e + f*x]^2)/a), -Tan[e + f*x]^2] - a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (3*a*AppellF1[1/2, -p...
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4153, 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.
\(\displaystyle \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^4 \left (a+b \tan (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \left (b \tan ^2(e+f x)+a\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\tan ^4(e+f x) \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},1,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{5 f}\) |
(AppellF1[5/2, 1, -p, 7/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p)/(5*f*(1 + (b*Tan[e + f*x]^2)/a)^p)
3.4.68.3.1 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[((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]))
\[\int \tan \left (f x +e \right )^{4} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
\[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p} \tan ^{4}{\left (e + f x \right )}\, dx \]
\[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
\[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]