3.490 \(\int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=200 \[ \frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {2 a^2 \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}}+\frac {a^2 \cos (c+d x) \sin ^{n+3}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+3}{2};\frac {n+5}{2};\sin ^2(c+d x)\right )}{d (n+3) \sqrt {\cos ^2(c+d x)}} \]

[Out]

a^2*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)+2*a^2*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)+a^2*cos(d*x+c)*hypergeom([-3/2, 3/2+1/2*n],[5/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(3+n)/d/(3+n)/(cos(d*x
+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2873, 2577} \[ \frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {2 a^2 \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}}+\frac {a^2 \cos (c+d x) \sin ^{n+3}(c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+3}{2};\frac {n+5}{2};\sin ^2(c+d x)\right )}{d (n+3) \sqrt {\cos ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*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]) + (2*a^2*Cos[c + d*x]*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*S
in[c + d*x]^(2 + n))/(d*(2 + n)*Sqrt[Cos[c + d*x]^2]) + (a^2*Cos[c + d*x]*Hypergeometric2F1[-3/2, (3 + n)/2, (
5 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(3 + n))/(d*(3 + 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 2873

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 164, normalized size = 0.82 \[ \frac {a^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{n+1}(c+d x) \left (\left (n^2+5 n+6\right ) \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )+(n+1) \sin (c+d x) \left (2 (n+3) \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )+(n+2) \sin (c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+3}{2};\frac {n+5}{2};\sin ^2(c+d x)\right )\right )\right )}{d (n+1) (n+2) (n+3)} \]

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]

(a^2*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*((6 + 5*n + n^2)*Hypergeometric2F1[-3/2, (1 + n)/2
, (3 + n)/2, Sin[c + d*x]^2] + (1 + n)*Sin[c + d*x]*(2*(3 + n)*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, S
in[c + d*x]^2] + (2 + n)*Hypergeometric2F1[-3/2, (3 + n)/2, (5 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x])))/(d*(1 +
n)*(2 + n)*(3 + n))

________________________________________________________________________________________

fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} \cos \left (d x + c\right )^{6} - 2 \, a^{2} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-(a^2*cos(d*x + c)^6 - 2*a^2*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*cos(d*x + c)^4)*sin(d*x + c)^n, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 11.09, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{2}\, dx \]

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))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]

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))^2,x)

[Out]

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

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________