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

Optimal. Leaf size=158 \[ \frac{\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{3}{4};1,-p;\frac{7}{4};\sin ^4(e+f x),-\frac{b \sin ^4(e+f x)}{a}\right )}{3 f}+\frac{\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{4};1,-p;\frac{5}{4};\sin ^4(e+f x),-\frac{b \sin ^4(e+f x)}{a}\right )}{f} \]

[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)

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Rubi [A]  time = 0.150238, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3223, 1240, 430, 429, 511, 510} \[ \frac{\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{3}{4};1,-p;\frac{7}{4};\sin ^4(e+f x),-\frac{b \sin ^4(e+f x)}{a}\right )}{3 f}+\frac{\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{4};1,-p;\frac{5}{4};\sin ^4(e+f x),-\frac{b \sin ^4(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[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 3223

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])

Rule 1240

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 430

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 429

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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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])

Rubi steps

\begin{align*} \int \sec (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^4\right )^p}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{\left (a+b x^4\right )^p}{1-x^4} \, dx,x,\sin (e+f x)\right )}{f}-\frac{\operatorname{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 ) \operatorname{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 ) \operatorname{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{F_1\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{F_1\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]  time = 5.95045, size = 0, normalized size = 0. \[ \int \sec (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx \]

Verification is Not applicable to the result.

[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]

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Maple [F]  time = 1.202, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( fx+e \right ) \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{4} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \cos \left (f x + e\right )^{4} - 2 \, b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \sec \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)