\(\int (a+b \sec ^2(e+f x))^p \sin ^4(e+f x) \, dx\) [138]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},3,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{5 f} \]

[Out]

1/5*AppellF1(5/2,3,-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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4217, 525, 524} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},3,-p,\frac {7}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{5 f} \]

[In]

Int[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^4,x]

[Out]

(AppellF1[5/2, 3, -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)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 4217

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1
+ ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b+b x^2\right )^p}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\left (\left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \text {Subst}\left (\int \frac {x^4 \left (1+\frac {b x^2}{a+b}\right )^p}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\operatorname {AppellF1}\left (\frac {5}{2},3,-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} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5878\) vs. \(2(88)=176\).

Time = 25.84 (sec) , antiderivative size = 5878, normalized size of antiderivative = 66.80 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\text {Result too large to show} \]

[In]

Integrate[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^4,x]

[Out]

Result too large to show

Maple [F]

\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{4}d x\]

[In]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^4,x)

[Out]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^4,x)

Fricas [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e)^2 + a)^p, x)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(f*x+e)**2)**p*sin(f*x+e)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^4, x)

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]

[In]

int(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^p,x)

[Out]

int(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^p, x)