Integrand size = 23, antiderivative size = 83 \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{3 f} \]
1/3*AppellF1(3/2,2,-p,5/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*tan(f*x+e)^3*(a +b*tan(f*x+e)^2)^p/f/((1+b*tan(f*x+e)^2/a)^p)
Result contains complex when optimal does not.
Time = 17.39 (sec) , antiderivative size = 3698, normalized size of antiderivative = 44.55 \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \]
(3*a*Cos[e + f*x]^3*Sin[e + f*x]*(a + b*Tan[e + f*x]^2)^p*(AppellF1[1/2, 2 , -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]/(-3*a*AppellF1[1/2, 2 , -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*(b*p*AppellF1[3/2 , 2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*a*AppellF1[ 3/2, 3, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2 ) + (AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*S ec[e + f*x]^2)/(3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Ta n[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))*(-1/4*(Cos[2*(e + f*x)]^3*(a + b*Tan[e + f*x]^2)^p) + (I/4)*Sin[2*(e + f*x)]*(a + b*Tan[e + f*x]^2)^p + (Sin[2*( e + f*x)]^2*(a + b*Tan[e + f*x]^2)^p)/2 - (I/4)*Sin[2*(e + f*x)]^3*(a + b* Tan[e + f*x]^2)^p + Cos[2*(e + f*x)]^2*((a + b*Tan[e + f*x]^2)^p/2 - (I/4) *Sin[2*(e + f*x)]*(a + b*Tan[e + f*x]^2)^p) + Cos[2*(e + f*x)]*(-1/4*(a + b*Tan[e + f*x]^2)^p - (Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2)^p)/4)))/( f*(6*a*b*p*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p)*(AppellF1[1/2, 2 , -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]/(-3*a*AppellF1[1/2, 2 , -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*(b*p*AppellF1[3/2 , 2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*a*AppellF1[ 3/2, 3, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x...
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, 4146, 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.
\(\displaystyle \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^2 \left (a+b \tan (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \left (b \tan ^2(e+f x)+a\right )^p}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\tan ^2(e+f x) \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^p}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {\tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{3 f}\) |
(AppellF1[3/2, 2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p)/(3*f*(1 + (b*Tan[e + f*x]^2)/a)^p)
3.2.59.3.1 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[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
\[\int \sin \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]
\[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]