3.374 \(\int \cos ^3(e+f x) (a+b \sin ^2(e+f x))^p \, dx\)

Optimal. Leaf size=124 \[ \frac {(a+b (2 p+3)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right )}{b f (2 p+3)}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b f (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3190, 388, 246, 245} \[ \frac {\left (\frac {a}{2 b p+3 b}+1\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right )}{f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

-((Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(1 + p))/(b*f*(3 + 2*p))) + ((1 + a/(3*b + 2*b*p))*Hypergeometric2F1[1/
2, -p, 3/2, -((b*Sin[e + f*x]^2)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 120, normalized size = 0.97 \[ -\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \left (\left (a+b \sin ^2(e+f x)\right ) \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^p-(a+b (2 p+3)) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sin ^2(e+f x)}{a}\right )\right )}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^3*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \cos \left (f x + e\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^p*cos(f*x + e)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e)^3, x)

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maple [F]  time = 7.10, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{3}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x)

[Out]

int(cos(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^p,x)

[Out]

int(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^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)**3*(a+b*sin(f*x+e)**2)**p,x)

[Out]

Timed out

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