3.5.20 \(\int \cos ^5(e+f x) (a+b \sin ^4(e+f x))^p \, dx\) [420]

Optimal. Leaf size=197 \[ \frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}-\frac {(a-b (5+4 p)) \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{b f (5+4 p)}-\frac {2 \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \]

[Out]

sin(f*x+e)*(a+b*sin(f*x+e)^4)^(1+p)/b/f/(5+4*p)-(a-b*(5+4*p))*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/b/f/(5+4*p)/((1+b*sin(f*x+e)^4/a)^p)-2/3*hypergeom([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)

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Rubi [A]
time = 0.15, antiderivative size = 191, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3302, 1221, 1218, 252, 251, 372, 371} \begin {gather*} \frac {\left (1-\frac {a}{4 b p+5 b}\right ) \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 {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} \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\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+1}}{b f (4 p+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p,x]

[Out]

(Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^(1 + p))/(b*f*(5 + 4*p)) + ((1 - a/(5*b + 4*b*p))*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*Sin[e + f*x]^4)/a)^p) - (2
*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 1221

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e^q*x^(2*q - 3)*((a + c*x^4)^(p +
1)/(c*(4*p + 2*q + 1))), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d
+ e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, c, d, e, p},
x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]

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*} \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \left (1-x^2\right )^2 \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}+\frac {\text {Subst}\left (\int \left (-a+b (5+4 p)-2 b (5+4 p) x^2\right ) \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{b f (5+4 p)}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}+\frac {\text {Subst}\left (\int \left (-a \left (1-\frac {b (5+4 p)}{a}\right ) \left (a+b x^4\right )^p-2 b (5+4 p) x^2 \left (a+b x^4\right )^p\right ) \, dx,x,\sin (e+f x)\right )}{b f (5+4 p)}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}-\frac {2 \text {Subst}\left (\int x^2 \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {(a-5 b-4 b p) \text {Subst}\left (\int \left (a+b x^4\right )^p \, dx,x,\sin (e+f x)\right )}{b f (5+4 p)}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}-\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 ) \text {Subst}\left (\int x^2 \left (1+\frac {b x^4}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left ((a-5 b-4 b p) \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 )}{b f (5+4 p)}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}-\frac {(a-b (5+4 p)) \, _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}}{b f (5+4 p)}-\frac {2 \, _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*}

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Mathematica [A]
time = 0.11, size = 141, normalized size = 0.72 \begin {gather*} \frac {\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} \left (15 \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b \sin ^4(e+f x)}{a}\right )-10 \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^2(e+f x)+3 \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^4(e+f x)\right )}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p,x]

[Out]

(Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p*(15*Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)] - 10*Hyper
geometric2F1[3/4, -p, 7/4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^2 + 3*Hypergeometric2F1[5/4, -p, 9/4, -((b*Si
n[e + f*x]^4)/a)]*Sin[e + f*x]^4))/(15*f*(1 + (b*Sin[e + f*x]^4)/a)^p)

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Maple [F]
time = 0.88, size = 0, normalized size = 0.00 \[\int \left (\cos ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{4}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x)

[Out]

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(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)^5, x)

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Fricas [F]
time = 0.42, size = 37, normalized size = 0.19 \begin {gather*} {\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 )^{5}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(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)^5, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**5*(a+b*sin(f*x+e)**4)**p,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(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)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^4)^p,x)

[Out]

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^4)^p, x)

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