3.5.32 \(\int \cos ^3(e+f x) (a+b \sin ^n(e+f x))^p \, dx\) [432]

Optimal. Leaf size=148 \[ \frac {\, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{f}-\frac {\, _2F_1\left (\frac {3}{n},-p;\frac {3+n}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f} \]

[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)-1/3*
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)

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Rubi [A]
time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 = {3302, 1907, 252, 251, 372, 371} \begin {gather*} \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 ^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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3*(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) - (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)

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 1907

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 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 ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \left (1-x^2\right ) \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (\left (a+b x^n\right )^p-x^2 \left (a+b x^n\right )^p\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {\text {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 ) \text {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 ) \text {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 {\, _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}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 114, normalized size = 0.77 \begin {gather*} -\frac {\sin (e+f x) \left (-3 \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b \sin ^n(e+f x)}{a}\right )+\, _2F_1\left (\frac {3}{n},-p;\frac {3+n}{n};-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^2(e+f x)\right ) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.41, size = 0, normalized size = 0.00 \[\int \left (\cos ^{3}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{n}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.39, size = 25, normalized size = 0.17 \begin {gather*} {\rm integral}\left ({\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{3}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(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)^3, 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)**3*(a+b*sin(f*x+e)**n)**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)^3*(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)^3, 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 )}^3\,{\left (a+b\,{\sin \left (e+f\,x\right )}^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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