3.492 \(\int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=134 \[ \frac {\cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{a d (n+1) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{a d (n+2) \sqrt {\cos ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2839, 2577} \[ \frac {\cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{a d (n+1) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{a d (n+2) \sqrt {\cos ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^n)/(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2839

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

Rubi steps

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

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Mathematica [B]  time = 11.48, size = 441, normalized size = 3.29 \[ \frac {2^{n+1} \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^n \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^n \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\frac {\, _2F_1\left (\frac {n+1}{2},n+4;\frac {n+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+1}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\tan ^2\left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac {n+5}{2};\frac {n+7}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+5}+\frac {4 \tan \left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (\frac {n+4}{2},n+4;\frac {n+6}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+4}-\frac {\, _2F_1\left (\frac {n+3}{2},n+4;\frac {n+5}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+3}+\frac {\tan ^4\left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac {n+7}{2};\frac {n+9}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+7}-\frac {2 \tan ^3\left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac {n+6}{2};\frac {n+8}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+6}\right )-\frac {2 \, _2F_1\left (\frac {n+2}{2},n+4;\frac {n+4}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n+2}\right )\right )}{d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^n)/(a + a*Sin[c + d*x]),x]

[Out]

(2^(1 + n)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*Tan[(c + d*x)/2]*(Tan[(c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2)
)^n*(1 + Tan[(c + d*x)/2]^2)^n*(Hypergeometric2F1[(1 + n)/2, 4 + n, (3 + n)/2, -Tan[(c + d*x)/2]^2]/(1 + n) +
Tan[(c + d*x)/2]*((-2*Hypergeometric2F1[(2 + n)/2, 4 + n, (4 + n)/2, -Tan[(c + d*x)/2]^2])/(2 + n) + Tan[(c +
d*x)/2]*(-(Hypergeometric2F1[(3 + n)/2, 4 + n, (5 + n)/2, -Tan[(c + d*x)/2]^2]/(3 + n)) + (4*Hypergeometric2F1
[(4 + n)/2, 4 + n, (6 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2])/(4 + n) - (Hypergeometric2F1[4 + n, (5 +
n)/2, (7 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(5 + n) - (2*Hypergeometric2F1[4 + n, (6 + n)/2, (8
+ n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^3)/(6 + n) + (Hypergeometric2F1[4 + n, (7 + n)/2, (9 + n)/2, -Ta
n[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4)/(7 + n)))))/(d*(a + a*Sin[c + d*x]))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^4/(a*sin(d*x + c) + a), x)

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maple [F]  time = 4.78, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +a \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c)),x)

[Out]

int(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^4/(a*sin(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n}{a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^n)/(a + a*sin(c + d*x)),x)

[Out]

int((cos(c + d*x)^4*sin(c + d*x)^n)/(a + a*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)**4*sin(d*x+c)**n/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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