3.570 \(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=85 \[ \frac {4 \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{a^3 d (n+1)}-\frac {3 \sin ^{n+1}(c+d x)}{a^3 d (n+1)}+\frac {\sin ^{n+2}(c+d x)}{a^3 d (n+2)} \]

[Out]

-3*sin(d*x+c)^(1+n)/a^3/d/(1+n)+4*hypergeom([1, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/a^3/d/(1+n)+sin(d*x+c
)^(2+n)/a^3/d/(2+n)

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Rubi [A]  time = 0.14, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 88, 64} \[ \frac {4 \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{a^3 d (n+1)}-\frac {3 \sin ^{n+1}(c+d x)}{a^3 d (n+1)}+\frac {\sin ^{n+2}(c+d x)}{a^3 d (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*Sin[c + d*x]^(1 + n))/(a^3*d*(1 + n)) + (4*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^
(1 + n))/(a^3*d*(1 + n)) + Sin[c + d*x]^(2 + n)/(a^3*d*(2 + n))

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2 \left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a \left (\frac {x}{a}\right )^n+a \left (\frac {x}{a}\right )^{1+n}+\frac {4 a^2 \left (\frac {x}{a}\right )^n}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=-\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)}+\frac {4 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=-\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {4 \, _2F_1(1,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 64, normalized size = 0.75 \[ \frac {\sin ^{n+1}(c+d x) (4 (n+2) \, _2F_1(1,n+1;n+2;-\sin (c+d x))+(n+1) \sin (c+d x)-3 (n+2))}{a^3 d (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

(Sin[c + d*x]^(1 + n)*(-3*(2 + n) + 4*(2 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]] + (1 + n)*Sin[
c + d*x]))/(a^3*d*(1 + n)*(2 + n))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-sin(d*x + c)^n*cos(d*x + c)^5/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x +
c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)

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maple [F]  time = 6.21, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3, 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(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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