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

Optimal. Leaf size=140 \[ \frac{\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b \sin ^4(e+f x)}{a}\right )}{f}-\frac{\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b \sin ^4(e+f x)}{a}\right )}{3 f} \]

[Out]

(Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(f*(1 + (b*Si
n[e + f*x]^4)/a)^p) - (Hypergeometric2F1[3/4, -p, 7/4, -((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 = 0.0980214, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3223, 1204, 246, 245, 365, 364} \[ \frac{\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b \sin ^4(e+f x)}{a}\right )}{f}-\frac{\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b \sin ^4(e+f x)}{a}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(f*(1 + (b*Si
n[e + f*x]^4)/a)^p) - (Hypergeometric2F1[3/4, -p, 7/4, -((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 1204

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (a+b x^4\right )^p-x^2 \left (a+b x^4\right )^p\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac{\operatorname{Subst}\left (\int x^2 \left (a+b x^4\right )^p \, 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 \left (1+\frac{b x^4}{a}\right )^p \, 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 x^2 \left (1+\frac{b x^4}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\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{\, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\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 [A]  time = 0.0486709, size = 106, normalized size = 0.76 \[ -\frac{\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac{b \sin ^4(e+f x)}{a}+1\right )^{-p} \left (\sin ^2(e+f x) \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b \sin ^4(e+f x)}{a}\right )-3 \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b \sin ^4(e+f x)}{a}\right )\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sin[e + f*x]*(-3*Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)] + Hypergeometric2F1[3/4, -p, 7/4,
-((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^2)*(a + b*Sin[e + f*x]^4)^p)/(3*f*(1 + (b*Sin[e + f*x]^4)/a)^p)

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Maple [F]  time = 3.415, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{3} \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(cos(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)

[Out]

int(cos(f*x+e)^3*(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} \cos \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(f*x + e)^3, 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} \cos \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(f*x + e)^3, 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(cos(f*x+e)**3*(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} \cos \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(f*x + e)^3, x)