3.1197 \(\int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=129 \[ \frac {a \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {b \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2838, 2577} \[ \frac {a \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {b \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[c + d*x]*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)
*Sqrt[Cos[c + d*x]^2]) + (b*Cos[c + d*x]*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c +
d*x]^(2 + n))/(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 2838

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 111, normalized size = 0.86 \[ \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{n+1}(c+d x) \left (a (n+2) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )+b (n+1) \sin (c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )\right )}{d (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 7.33, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)^4*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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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