3.945 \(\int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=135 \[ \frac {4 \sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^{m+2} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {5}{2};-\frac {3}{2},-n;m+\frac {7}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a^2 f (2 m+5) \sqrt {1-\sin (e+f x)}} \]

[Out]

4*AppellF1(5/2+m,-n,-3/2,7/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^(2+m)*(
c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(5+2*m)/(((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 = 135, 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 = {2918, 140, 139, 138} \[ \frac {4 \sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^{m+2} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {5}{2};-\frac {3}{2},-n;m+\frac {7}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a^2 f (2 m+5) \sqrt {1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*AppellF1[5/2 + m, -3/2, -n, 7/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(2 + m)*(c + d*Sin[e + f*x])^n)/(a^2*f*(5 + 2*m)*Sqrt[1 - Sin[e + f*x]]*((c + d*S
in[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 140

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

Rule 2918

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

Rubi steps

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

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Mathematica [A]  time = 1.38, size = 160, normalized size = 1.19 \[ -\frac {4 \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \cos ^3(e+f x) \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{-m-\frac {3}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {5}{2};-m-\frac {3}{2},-n;\frac {7}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{5 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*AppellF1[5/2, -3/2 - m, -n, 7/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)]*C
os[e + f*x]^3*(a*(1 + Sin[e + f*x]))^m*(c + d*Sin[e + f*x])^n*Sin[(2*e - Pi + 2*f*x)/4]^2*(Sin[(2*e + Pi + 2*f
*x)/4]^2)^(-3/2 - m))/(5*f*((c + d*Sin[e + f*x])/(c + d))^n)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [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*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [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*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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