3.704 \(\int \frac{\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx\)

Optimal. Leaf size=160 \[ -\frac{4 (2 n+3) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{a^5 d (n+1)}+\frac{\sin ^{n+1}(c+d x) \left (a (2 n+7) \sin (c+d x)+a \left (8 n^2+30 n+27\right )\right )}{d \left (n^2+3 n+2\right ) \left (a^6 \sin (c+d x)+a^6\right )}-\frac{(a-a \sin (c+d x))^2 \sin ^{n+1}(c+d x)}{d (n+2) \left (a^7 \sin (c+d x)+a^7\right )} \]

[Out]

(-4*(3 + 2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(a^5*d*(1 + n)) - (Sin[c
+ d*x]^(1 + n)*(a - a*Sin[c + d*x])^2)/(d*(2 + n)*(a^7 + a^7*Sin[c + d*x])) + (Sin[c + d*x]^(1 + n)*(a*(27 +
30*n + 8*n^2) + a*(7 + 2*n)*Sin[c + d*x]))/(d*(2 + 3*n + n^2)*(a^6 + a^6*Sin[c + d*x]))

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Rubi [A]  time = 0.203876, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2836, 100, 146, 64} \[ -\frac{4 (2 n+3) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{a^5 d (n+1)}+\frac{\sin ^{n+1}(c+d x) \left (a (2 n+7) \sin (c+d x)+a \left (8 n^2+30 n+27\right )\right )}{d \left (n^2+3 n+2\right ) \left (a^6 \sin (c+d x)+a^6\right )}-\frac{(a-a \sin (c+d x))^2 \sin ^{n+1}(c+d x)}{d (n+2) \left (a^7 \sin (c+d x)+a^7\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^5,x]

[Out]

(-4*(3 + 2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(a^5*d*(1 + n)) - (Sin[c
+ d*x]^(1 + n)*(a - a*Sin[c + d*x])^2)/(d*(2 + n)*(a^7 + a^7*Sin[c + d*x])) + (Sin[c + d*x]^(1 + n)*(a*(27 +
30*n + 8*n^2) + a*(7 + 2*n)*Sin[c + d*x]))/(d*(2 + 3*n + n^2)*(a^6 + a^6*Sin[c + d*x]))

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]

Rule 100

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

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

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])))

Rubi steps

\begin{align*} \int \frac{\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^3 \left (\frac{x}{a}\right )^n}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=-\frac{\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{(a-x) \left (\frac{x}{a}\right )^n (a (3+2 n)+(-7-2 n) x)}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^6 d (2+n)}\\ &=-\frac{\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac{\sin ^{1+n}(c+d x) \left (a \left (27+30 n+8 n^2\right )+a (7+2 n) \sin (c+d x)\right )}{d (1+n) (2+n) \left (a^6+a^6 \sin (c+d x)\right )}-\frac{(4 (3+2 n)) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=-\frac{4 (3+2 n) \, _2F_1(1,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^5 d (1+n)}-\frac{\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac{\sin ^{1+n}(c+d x) \left (a \left (27+30 n+8 n^2\right )+a (7+2 n) \sin (c+d x)\right )}{d (1+n) (2+n) \left (a^6+a^6 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.206406, size = 108, normalized size = 0.68 \[ \frac{\sin ^{n+1}(c+d x) \left (-4 \left (2 n^2+7 n+6\right ) (\sin (c+d x)+1) \, _2F_1(1,n+1;n+2;-\sin (c+d x))-(n+1) \sin ^2(c+d x)+(4 n+9) \sin (c+d x)+8 n^2+29 n+26\right )}{a^5 d (n+1) (n+2) (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^5,x]

[Out]

(Sin[c + d*x]^(1 + n)*(26 + 29*n + 8*n^2 + (9 + 4*n)*Sin[c + d*x] - (1 + n)*Sin[c + d*x]^2 - 4*(6 + 7*n + 2*n^
2)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*(1 + Sin[c + d*x])))/(a^5*d*(1 + n)*(2 + n)*(1 + Sin[c +
d*x]))

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Maple [F]  time = 1.239, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{7} \left ( \sin \left ( dx+c \right ) \right ) ^{n}}{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^5,x)

[Out]

int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^7/(a*sin(d*x + c) + a)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{5 \, a^{5} \cos \left (d x + c\right )^{4} - 20 \, a^{5} \cos \left (d x + c\right )^{2} + 16 \, a^{5} +{\left (a^{5} \cos \left (d x + c\right )^{4} - 12 \, a^{5} \cos \left (d x + c\right )^{2} + 16 \, a^{5}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

integral(sin(d*x + c)^n*cos(d*x + c)^7/(5*a^5*cos(d*x + c)^4 - 20*a^5*cos(d*x + c)^2 + 16*a^5 + (a^5*cos(d*x +
c)^4 - 12*a^5*cos(d*x + c)^2 + 16*a^5)*sin(d*x + c)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**n/(a+a*sin(d*x+c))**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^5,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^7/(a*sin(d*x + c) + a)^5, x)