\(\int \cos ^3(e+f x) (a+b \sin ^4(e+f x))^p \, dx\) [421]

Optimal result

Integrand size = 23, antiderivative size = 140 \[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f}-\frac {\operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \]

[Out]

hypergeom([1/4, -p],[5/4],-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*hy
pergeom([3/4, -p],[7/4],-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 = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3302, 1218, 252, 251, 372, 371} \[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right )}{f}-\frac {\sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right )}{3 f} \]

[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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 1218

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 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 \left (1-x^2\right ) \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {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 {\text {Subst}\left (\int \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {\text {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 ) \text {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 ) \text {Subst}\left (\int x^2 \left (1+\frac {b x^4}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\operatorname {Hypergeometric2F1}\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 {\operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.76 \[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=-\frac {\sin (e+f x) \left (-3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^2(e+f x)\right ) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\cos ^{3}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{4}\left (f x +e \right )\right )\right )}^{p}d x\]

[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)

Fricas [F]

\[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{3} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{3} \,d x } \]

[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)

Giac [F]

\[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{3} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p \,d x \]

[In]

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

[Out]

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