3.951 \(\int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=121 \[ -\frac {(1-\sin (e+f x))^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {5}{2};\frac {5}{2},-n;\frac {7}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{10 \sqrt {2} a^4 f \sqrt {\sin (e+f x)+1}} \]

[Out]

-1/20*AppellF1(5/2,-n,5/2,7/2,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))^2*(c+d*sin(
f*x+e))^n/a^4/f/(((c+d*sin(f*x+e))/(c+d))^n)*2^(1/2)/(1+sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2917, 139, 138} \[ -\frac {(1-\sin (e+f x))^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {5}{2};\frac {5}{2},-n;\frac {7}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{10 \sqrt {2} a^4 f \sqrt {\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^4*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^4,x]

[Out]

-(AppellF1[5/2, 5/2, -n, 7/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(1 - Sin[e +
f*x])^2*(c + d*Sin[e + f*x])^n)/(10*Sqrt[2]*a^4*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 2917

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx &=\frac {\cos (e+f x) \operatorname {Subst}\left (\int \frac {(1-x)^{3/2} (c+d x)^n}{(1+x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{a^4 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 {(1-x)^{3/2} \left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^n}{(1+x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{a^4 f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {F_1\left (\frac {5}{2};\frac {5}{2},-n;\frac {7}{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))^2 (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{10 \sqrt {2} a^4 f \sqrt {1+\sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 25.13, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^4} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(Cos[e + f*x]^4*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^4,x]

[Out]

Integrate[(Cos[e + f*x]^4*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^4, x]

________________________________________________________________________________________

fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{a^{4} \cos \left (f x + e\right )^{4} - 8 \, a^{4} \cos \left (f x + e\right )^{2} + 8 \, a^{4} - 4 \, {\left (a^{4} \cos \left (f x + e\right )^{2} - 2 \, a^{4}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

integral((d*sin(f*x + e) + c)^n*cos(f*x + e)^4/(a^4*cos(f*x + e)^4 - 8*a^4*cos(f*x + e)^2 + 8*a^4 - 4*(a^4*cos
(f*x + e)^2 - 2*a^4)*sin(f*x + e)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^4/(a*sin(f*x + e) + a)^4, x)

________________________________________________________________________________________

maple [F]  time = 6.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{4}\left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x)

[Out]

int(cos(f*x+e)^4*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^4/(a*sin(f*x + e) + a)^4, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^4\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^4*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^4,x)

[Out]

int((cos(e + f*x)^4*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^4, x)

________________________________________________________________________________________

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)**4*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**4,x)

[Out]

Timed out

________________________________________________________________________________________