3.1524 \(\int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=375 \[ -\frac {\sqrt {2} (c+d) \cos (e+f x) \left (a c d (n+3)-b \left (2 c^2-d^2 (n+2)\right )\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^3 f (n+2) (n+3) \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} \left (c^2-d^2\right ) \cos (e+f x) (2 b c-a d (n+3)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^3 f (n+2) (n+3) \sqrt {\sin (e+f x)+1}}-\frac {\cos (e+f x) (2 b c-a d (n+3)) (c+d \sin (e+f x))^{n+1}}{d^2 f (n+2) (n+3)}+\frac {b \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.64, antiderivative size = 373, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2922, 3034, 3023, 2756, 2665, 139, 138} \[ \frac {\sqrt {2} (c+d) \cos (e+f x) \left (-a c d (n+3)+2 b c^2-b d^2 (n+2)\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^3 f (n+2) (n+3) \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} \left (c^2-d^2\right ) \cos (e+f x) (2 b c-a d (n+3)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac {1}{2};\frac {1}{2},-n;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^3 f (n+2) (n+3) \sqrt {\sin (e+f x)+1}}-\frac {\cos (e+f x) (2 b c-a d (n+3)) (c+d \sin (e+f x))^{n+1}}{d^2 f (n+2) (n+3)}+\frac {b \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*b*c - a*d*(3 + n))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d^2*f*(2 + n)*(3 + n))) + (b*Cos[e + f*x]
*Sin[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(3 + n)) + (Sqrt[2]*(c + d)*(2*b*c^2 - b*d^2*(2 + n) - a*c*d*
(3 + n))*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c
+ d*Sin[e + f*x])^n)/(d^3*f*(2 + n)*(3 + n)*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n) - (Sqrt[
2]*(c^2 - d^2)*(2*b*c - a*d*(3 + n))*AppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/
(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(d^3*f*(2 + n)*(3 + n)*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 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 2756

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

Rule 2922

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a,
b, c, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3034

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m
+ 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*d*(C*(m + 2) + A*(m
+ 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [F]  time = 3.57, size = 0, normalized size = 0.00 \[ \int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.19, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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