\(\int \cos ^2(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [379]

Optimal result

Integrand size = 23, antiderivative size = 90 \[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f} \]

[Out]

AppellF1(1/2,-1/2,-p,3/2,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*(a+b*sin(f*x+e)^2)^p*(cos(f*x+e)^2)^(1/2)*tan(f*x+e)/
f/((1+b*sin(f*x+e)^2/a)^p)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3271, 441, 440} \[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\sqrt {\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{f} \]

[In]

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

[Out]

(AppellF1[1/2, -1/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt[Cos[e + f*x]^2]*(a + b*Sin[e + f*x
]^2)^p*Tan[e + f*x])/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 3271

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), 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/2] &&  !IntegerQ
[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \sqrt {1-x^2} \left (a+b x^2\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (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}\right ) \text {Subst}\left (\int \sqrt {1-x^2} \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(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} \tan (e+f x)}{f} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(195\) vs. \(2(90)=180\).

Time = 0.75 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \left (a+b \sin ^2(e+f x)\right )^p \sin (2 (e+f x))}{2 f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2},1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-a \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \]

[In]

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

[Out]

(3*a*AppellF1[1/2, -1/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*(a + b*Sin[e + f*x]^2)^p*Sin[2*(e +
f*x)])/(2*f*(3*a*AppellF1[1/2, -1/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + (2*b*p*AppellF1[3/2,
-1/2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] - a*AppellF1[3/2, 1/2, -p, 5/2, Sin[e + f*x]^2, -(
(b*Sin[e + f*x]^2)/a)])*Sin[e + f*x]^2))

Maple [F]

\[\int \left (\cos ^{2}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(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)^2, x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(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)^2, x)

Giac [F]

\[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(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)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]

[In]

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

[Out]

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