3.424 \(\int \sec ^3(e+f x) (a+b \sin ^4(e+f x))^p \, dx\)

Optimal. Leaf size=239 \[ \frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {1}{4};2,-p;\frac {5}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{f}+\frac {\sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {5}{4};2,-p;\frac {9}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{5 f}+\frac {2 \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {3}{4};2,-p;\frac {7}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{3 f} \]

[Out]

AppellF1(1/4,2,-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)+2/3*AppellF1(3/4,2,-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)+1/5*AppellF1(5/4,2,-p,9/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)^5*(a+b*sin(f*x+e)^4)^p/f/((1+
b*sin(f*x+e)^4/a)^p)

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Rubi [A]  time = 0.22, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3223, 1240, 430, 429, 511, 510} \[ \frac {\sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {5}{4};2,-p;\frac {9}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{5 f}+\frac {2 \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {3}{4};2,-p;\frac {7}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{3 f}+\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {1}{4};2,-p;\frac {5}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p,x]

[Out]

(AppellF1[1/4, 2, -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) + (2*AppellF1[3/4, 2, -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) + (AppellF1[5/4, 2, -p, 9/4, Sin[e + f*x]^
4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p)/(5*f*(1 + (b*Sin[e + f*x]^4)/a)^p)

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 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 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])

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 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 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])

Rubi steps

\begin {align*} \int \sec ^3(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}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\left (a+b x^4\right )^p}{\left (-1+x^4\right )^2}+\frac {2 x^2 \left (a+b x^4\right )^p}{\left (-1+x^4\right )^2}+\frac {x^4 \left (a+b x^4\right )^p}{\left (-1+x^4\right )^2}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^4\right )^p}{\left (-1+x^4\right )^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^4\right )^p}{\left (-1+x^4\right )^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^4\right )^p}{\left (-1+x^4\right )^2} \, 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}{\left (-1+x^4\right )^2} \, 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^4 \left (1+\frac {b x^4}{a}\right )^p}{\left (-1+x^4\right )^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (2 \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}{\left (-1+x^4\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {F_1\left (\frac {1}{4};2,-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 {2 F_1\left (\frac {3}{4};2,-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}+\frac {F_1\left (\frac {5}{4};2,-p;\frac {9}{4};\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^5(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}}{5 f}\\ \end {align*}

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Mathematica [F]  time = 9.24, size = 0, normalized size = 0.00 \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p,x]

[Out]

Integrate[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p, x]

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (f x + e\right )^{4} - 2 \, b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \sec \left (f x + e\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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maple [F]  time = 2.92, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{4}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)

[Out]

int(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p}{{\cos \left (e+f\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^4)^p/cos(e + f*x)^3,x)

[Out]

int((a + b*sin(e + f*x)^4)^p/cos(e + f*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3*(a+b*sin(f*x+e)**4)**p,x)

[Out]

Timed out

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