3.1238 \(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=191 \[ \frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (n+4)\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a^2 b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x)}{a b^4 d (a+b \sin (c+d x))}-\frac {2 a \sin ^{n+2}(c+d x)}{b^3 d (n+2)}+\frac {\sin ^{n+3}(c+d x)}{b^2 d (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 950, 1620, 64} \[ \frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (n+4)\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a^2 b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x)}{a b^4 d (a+b \sin (c+d x))}-\frac {2 a \sin ^{n+2}(c+d x)}{b^3 d (n+2)}+\frac {\sin ^{n+3}(c+d x)}{b^2 d (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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 950

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = P
olynomialQuotient[(a + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + c*x^2)^p, d + e*x, x]}, Simp[(R*(d
+ e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f - d*g)), Int[(d + e*x)^(m +
1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 143, normalized size = 0.75 \[ \frac {\sin ^{n+1}(c+d x) \left (\frac {\left (a^2-b^2\right )^2 \, _2F_1\left (2,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a^2 (n+1)}-\frac {4 \left (a^2-b^2\right ) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{n+1}+\frac {3 a^2-2 b^2}{n+1}-\frac {2 a b \sin (c+d x)}{n+2}+\frac {b^2 \sin ^2(c+d x)}{n+3}\right )}{b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(b*sin(d*x + c) + a)^2, x)

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maple [F]  time = 8.67, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +b \sin \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(b*sin(d*x + c) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + b*sin(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + b*sin(c + d*x))^2, 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+b*sin(d*x+c))**2,x)

[Out]

Timed out

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