3.493 \(\int \frac{\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=173 \[ \frac{(2 n+3) \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{a^2 d (n+1) (n+2) \sqrt{\cos ^2(c+d x)}}-\frac{2 \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{a^2 d (n+2) \sqrt{\cos ^2(c+d x)}}-\frac{\cos (c+d x) \sin ^{n+1}(c+d x)}{a^2 d (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.242226, antiderivative size = 173, normalized size of antiderivative = 1., 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 = {2869, 2763, 2748, 2643} \[ \frac{(2 n+3) \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{a^2 d (n+1) (n+2) \sqrt{\cos ^2(c+d x)}}-\frac{2 \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{a^2 d (n+2) \sqrt{\cos ^2(c+d x)}}-\frac{\cos (c+d x) \sin ^{n+1}(c+d x)}{a^2 d (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2869

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

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

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

Mathematica [A]  time = 4.6507, size = 312, normalized size = 1.8 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) \sin ^n(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )^n \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (\frac{\, _2F_1\left (\frac{n+1}{2},n+3;\frac{n+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+1}+\tan \left (\frac{1}{2} (c+d x)\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan ^2\left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (n+3,\frac{n+5}{2};\frac{n+7}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+5}-\frac{4 \tan \left (\frac{1}{2} (c+d x)\right ) \, _2F_1\left (n+3,\frac{n+4}{2};\frac{n+6}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+4}+\frac{6 \, _2F_1\left (\frac{n+3}{2},n+3;\frac{n+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+3}\right )-\frac{4 \, _2F_1\left (\frac{n+2}{2},n+3;\frac{n+4}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n+2}\right )\right )}{d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*(Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Sin[c + d*x]^n*Tan[(c + d*x)/2]*(Hypergeomet
ric2F1[(1 + n)/2, 3 + n, (3 + n)/2, -Tan[(c + d*x)/2]^2]/(1 + n) + Tan[(c + d*x)/2]*((-4*Hypergeometric2F1[(2
+ n)/2, 3 + n, (4 + n)/2, -Tan[(c + d*x)/2]^2])/(2 + n) + Tan[(c + d*x)/2]*((6*Hypergeometric2F1[(3 + n)/2, 3
+ n, (5 + n)/2, -Tan[(c + d*x)/2]^2])/(3 + n) - (4*Hypergeometric2F1[3 + n, (4 + n)/2, (6 + n)/2, -Tan[(c + d*
x)/2]^2]*Tan[(c + d*x)/2])/(4 + n) + (Hypergeometric2F1[3 + n, (5 + n)/2, (7 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[
(c + d*x)/2]^2)/(5 + n)))))/(d*(a + a*Sin[c + d*x])^2)

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Maple [F]  time = 1.661, 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}}{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int(cos(d*x+c)^4*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,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 )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^4/(a*sin(d*x + c) + a)^2, 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 )^{4}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(-sin(d*x + c)^n*cos(d*x + c)^4/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), 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)**4*sin(d*x+c)**n/(a+a*sin(d*x+c))**2,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 )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^4/(a*sin(d*x + c) + a)^2, x)