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[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])

________________________________________________________________________________________

Rubi [A]  time = 0.176057, antiderivative size = 134, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [B]  time = 11.1801, size = 441, normalized size = 3.29 \[ \frac{2^{n+1} \tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^n \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^n \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (\frac{\, _2F_1\left (\frac{n+1}{2},n+4;\frac{n+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+1}+\tan \left (\frac{1}{2} (c+d x)\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan ^4\left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac{n+7}{2};\frac{n+9}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+7}-\frac{2 \tan ^3\left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac{n+6}{2};\frac{n+8}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+6}-\frac{\tan ^2\left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (n+4,\frac{n+5}{2};\frac{n+7}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+5}+\frac{4 \tan \left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (\frac{n+4}{2},n+4;\frac{n+6}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+4}-\frac{\, _2F_1\left (\frac{n+3}{2},n+4;\frac{n+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+3}\right )-\frac{2 \, _2F_1\left (\frac{n+2}{2},n+4;\frac{n+4}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+2}\right )\right )}{d (a \sin (c+d x)+a)} \]

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]))

________________________________________________________________________________________

Maple [F]  time = 1.718, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{n}}{a+a\sin \left ( dx+c \right ) }}\, dx \end{align*}

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)

________________________________________________________________________________________

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 )^{4}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

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)

________________________________________________________________________________________

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 )^{4}}{a \sin \left (d x + c\right ) + a}, x\right ) \end{align*}

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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{4}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

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)