3.1237 \(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=167 \[ \frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a b^4 d (n+1)}-\frac {a \left (a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-2 b^2\right ) \sin ^{n+2}(c+d x)}{b^3 d (n+2)}-\frac {a \sin ^{n+3}(c+d x)}{b^2 d (n+3)}+\frac {\sin ^{n+4}(c+d x)}{b d (n+4)} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 167, 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, 952, 1620, 64} \[ \frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a b^4 d (n+1)}-\frac {a \left (a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-2 b^2\right ) \sin ^{n+2}(c+d x)}{b^3 d (n+2)}-\frac {a \sin ^{n+3}(c+d x)}{b^2 d (n+3)}+\frac {\sin ^{n+4}(c+d x)}{b d (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(a^2 - 2*b^2)*Sin[c + d*x]^(1 + n))/(b^4*d*(1 + n))) + ((a^2 - b^2)^2*Hypergeometric2F1[1, 1 + n, 2 + n,
-((b*Sin[c + d*x])/a)]*Sin[c + d*x]^(1 + n))/(a*b^4*d*(1 + n)) + ((a^2 - 2*b^2)*Sin[c + d*x]^(2 + n))/(b^3*d*(
2 + n)) - (a*Sin[c + d*x]^(3 + n))/(b^2*d*(3 + n)) + Sin[c + d*x]^(4 + n)/(b*d*(4 + 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 952

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(c^p*(d
+ e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), 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] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

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

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Mathematica [A]  time = 0.58, size = 133, normalized size = 0.80 \[ \frac {\sin ^{n+1}(c+d x) \left (-\frac {a^3-2 a b^2}{n+1}+\frac {\left (a^2-b^2\right )^2 \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a (n+1)}+\frac {b \left (a^2-2 b^2\right ) \sin (c+d x)}{n+2}-\frac {a b^2 \sin ^2(c+d x)}{n+3}+\frac {b^3 \sin ^3(c+d x)}{n+4}\right )}{b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

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

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

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 3.27, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

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)),x)

[Out]

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

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

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)),x, algorithm="maxima")

[Out]

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

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

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)),x)

[Out]

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

[Out]

Timed out

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