3.942 \(\int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\sqrt {2} (1-\sin (e+f x)) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {3}{2};\frac {1}{2},-n;\frac {5}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{3 a f \sqrt {\sin (e+f x)+1}} \]

[Out]

-1/3*AppellF1(3/2,-n,1/2,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(1-sin(f*x+e))*(c+d*sin(f*x
+e))^n*2^(1/2)/a/f/(((c+d*sin(f*x+e))/(c+d))^n)/(1+sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2914, 2755, 139, 138} \[ -\frac {\sqrt {2} (1-\sin (e+f x)) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {3}{2};\frac {1}{2},-n;\frac {5}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{3 a f \sqrt {\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]),x]

[Out]

-(Sqrt[2]*AppellF1[3/2, 1/2, -n, 5/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(1 -
Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(3*a*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2755

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*C
os[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]]), Subst[Int[((a + b*x)^m*Sqrt[1 + (d*x)/c])/Sqrt
[1 - (d*x)/c], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b
^2, 0] &&  !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]

Rule 2914

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(
x_)])^(n_), x_Symbol] :> Dist[a^(2*m), Int[(c + d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a,
b, c, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx &=\frac {\int (a-a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx}{a^2}\\ &=\frac {\cos (e+f x) \operatorname {Subst}\left (\int \frac {\sqrt {1-x} (c+d x)^n}{\sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=\frac {\left (\cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac {c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-x} \left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^n}{\sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {\sqrt {2} F_1\left (\frac {3}{2};\frac {1}{2},-n;\frac {5}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{3 a f \sqrt {1+\sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 229, normalized size = 1.92 \[ -\frac {\sec (e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {-\frac {d (\sin (e+f x)-1)}{c+d}} (c+d \sin (e+f x))^{n+1} \left ((n+1) (c+d \sin (e+f x)) F_1\left (n+2;\frac {1}{2},\frac {1}{2};n+3;\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )-(n+2) (c+d) F_1\left (n+1;\frac {1}{2},\frac {1}{2};n+2;\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )\right )}{a d f (n+1) (n+2) (d-c) \sqrt {\frac {d (\sin (e+f x)+1)}{d-c}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[e + f*x]^2*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]),x]

[Out]

-((Sec[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*(c + d*Sin[e
+ f*x])^(1 + n)*(-((c + d)*(2 + n)*AppellF1[1 + n, 1/2, 1/2, 2 + n, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e
+ f*x])/(c + d)]) + (1 + n)*AppellF1[2 + n, 1/2, 1/2, 3 + n, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + f*x
])/(c + d)]*(c + d*Sin[e + f*x])))/(a*d*(-c + d)*f*(1 + n)*(2 + n)*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d)]))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}}{a \sin \left (f x + e\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((d*sin(f*x + e) + c)^n*cos(f*x + e)^2/(a*sin(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^2/(a*sin(f*x + e) + a), x)

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maple [F]  time = 1.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{a +a \sin \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x)

[Out]

int(cos(f*x+e)^2*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^2/(a*sin(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^2*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x)),x)

[Out]

int((cos(e + f*x)^2*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x)), 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)**2*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e)),x)

[Out]

Timed out

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