\(\int \sec (e+f x) (a+b \sin ^4(e+f x))^p \, dx\) [423]

Optimal result

Integrand size = 21, antiderivative size = 158 \[ \int \sec (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f}+\frac {\operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \]

[Out]

AppellF1(1/4,1,-p,5/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x+e)^4/a)^
p)+1/3*AppellF1(3/4,1,-p,7/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)^3*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x
+e)^4/a)^p)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3302, 1254, 441, 440, 525, 524} \[ \int \sec (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{f}+\frac {\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{3 f} \]

[In]

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

[Out]

(AppellF1[1/4, 1, -p, 5/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(f*
(1 + (b*Sin[e + f*x]^4)/a)^p) + (AppellF1[3/4, 1, -p, 7/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*
x]^3*(a + b*Sin[e + f*x]^4)^p)/(3*f*(1 + (b*Sin[e + f*x]^4)/a)^p)

Rule 440

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

Rule 441

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

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 1254

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4)))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

Rule 3302

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^4\right )^p}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {\left (a+b x^4\right )^p}{1-x^4}-\frac {x^2 \left (a+b x^4\right )^p}{-1+x^4}\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {\left (a+b x^4\right )^p}{1-x^4} \, dx,x,\sin (e+f x)\right )}{f}-\frac {\text {Subst}\left (\int \frac {x^2 \left (a+b x^4\right )^p}{-1+x^4} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^4}{a}\right )^p}{1-x^4} \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left (\left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^4}{a}\right )^p}{-1+x^4} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f}+\frac {\operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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