3.431 \(\int \cos ^5(e+f x) (a+b \sin ^n(e+f x))^p \, dx\)

Optimal. Leaf size=226 \[ \frac {\sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{f}+\frac {\sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{n},-p;\frac {n+5}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{5 f}-\frac {2 \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{n},-p;\frac {n+3}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 226, 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, 1893, 246, 245, 365, 364} \[ \frac {\sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{n},-p;\frac {n+5}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{5 f}-\frac {2 \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{n},-p;\frac {n+3}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{3 f}+\frac {\sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

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 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

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 \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \left (1-x^2\right )^2 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\left (a+b x^n\right )^p-2 x^2 \left (a+b x^n\right )^p+x^4 \left (a+b x^n\right )^p\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}+\frac {\operatorname {Subst}\left (\int x^4 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {2 \operatorname {Subst}\left (\int x^2 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (\left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (\left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int x^4 \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left (2 \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int x^2 \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{f}-\frac {2 \, _2F_1\left (\frac {3}{n},-p;\frac {3+n}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\, _2F_1\left (\frac {5}{n},-p;\frac {5+n}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 155, normalized size = 0.69 \[ \frac {\sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \left (15 \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right )+3 \sin ^4(e+f x) \, _2F_1\left (\frac {5}{n},-p;\frac {n+5}{n};-\frac {b \sin ^n(e+f x)}{a}\right )-10 \sin ^2(e+f x) \, _2F_1\left (\frac {3}{n},-p;\frac {n+3}{n};-\frac {b \sin ^n(e+f x)}{a}\right )\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^n)^p,x, algorithm="fricas")

[Out]

integral((b*sin(f*x + e)^n + a)^p*cos(f*x + e)^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^n)^p,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.69, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{n}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^n)^p,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (e+f\,x\right )}^5\,{\left (a+b\,{\sin \left (e+f\,x\right )}^n\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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