3.1235 \(\int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=170 \[ -\frac{\left (2 a^2-b^2\right ) \sin ^{n+3}(c+d x)}{d (n+3)}+\frac{\left (a^2-2 b^2\right ) \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{a^2 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac{2 a b \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{4 a b \sin ^{n+4}(c+d x)}{d (n+4)}+\frac{2 a b \sin ^{n+6}(c+d x)}{d (n+6)}+\frac{b^2 \sin ^{n+7}(c+d x)}{d (n+7)} \]

[Out]

(a^2*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (2*a*b*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - ((2*a^2 - b^2)*Sin[c + d*x
]^(3 + n))/(d*(3 + n)) - (4*a*b*Sin[c + d*x]^(4 + n))/(d*(4 + n)) + ((a^2 - 2*b^2)*Sin[c + d*x]^(5 + n))/(d*(5
+ n)) + (2*a*b*Sin[c + d*x]^(6 + n))/(d*(6 + n)) + (b^2*Sin[c + d*x]^(7 + n))/(d*(7 + n))

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Rubi [A]  time = 0.213198, antiderivative size = 170, 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 = {2837, 948} \[ -\frac{\left (2 a^2-b^2\right ) \sin ^{n+3}(c+d x)}{d (n+3)}+\frac{\left (a^2-2 b^2\right ) \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{a^2 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac{2 a b \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{4 a b \sin ^{n+4}(c+d x)}{d (n+4)}+\frac{2 a b \sin ^{n+6}(c+d x)}{d (n+6)}+\frac{b^2 \sin ^{n+7}(c+d x)}{d (n+7)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]

[Out]

(a^2*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (2*a*b*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - ((2*a^2 - b^2)*Sin[c + d*x
]^(3 + n))/(d*(3 + n)) - (4*a*b*Sin[c + d*x]^(4 + n))/(d*(4 + n)) + ((a^2 - 2*b^2)*Sin[c + d*x]^(5 + n))/(d*(5
+ n)) + (2*a*b*Sin[c + d*x]^(6 + n))/(d*(6 + n)) + (b^2*Sin[c + d*x]^(7 + n))/(d*(7 + n))

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

\begin{align*} \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{x}{b}\right )^n (a+x)^2 \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2 b^4 \left (\frac{x}{b}\right )^n+2 a b^5 \left (\frac{x}{b}\right )^{1+n}-b^4 \left (2 a^2-b^2\right ) \left (\frac{x}{b}\right )^{2+n}-4 a b^5 \left (\frac{x}{b}\right )^{3+n}+b^4 \left (a^2-2 b^2\right ) \left (\frac{x}{b}\right )^{4+n}+2 a b^5 \left (\frac{x}{b}\right )^{5+n}+b^6 \left (\frac{x}{b}\right )^{6+n}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{a^2 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac{2 a b \sin ^{2+n}(c+d x)}{d (2+n)}-\frac{\left (2 a^2-b^2\right ) \sin ^{3+n}(c+d x)}{d (3+n)}-\frac{4 a b \sin ^{4+n}(c+d x)}{d (4+n)}+\frac{\left (a^2-2 b^2\right ) \sin ^{5+n}(c+d x)}{d (5+n)}+\frac{2 a b \sin ^{6+n}(c+d x)}{d (6+n)}+\frac{b^2 \sin ^{7+n}(c+d x)}{d (7+n)}\\ \end{align*}

Mathematica [A]  time = 0.661141, size = 139, normalized size = 0.82 \[ \frac{\sin ^{n+1}(c+d x) \left (\frac{\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{n+5}-\frac{\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{n+3}+\frac{a^2}{n+1}+\frac{2 a b \sin ^5(c+d x)}{n+6}-\frac{4 a b \sin ^3(c+d x)}{n+4}+\frac{2 a b \sin (c+d x)}{n+2}+\frac{b^2 \sin ^6(c+d x)}{n+7}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]

[Out]

(Sin[c + d*x]^(1 + n)*(a^2/(1 + n) + (2*a*b*Sin[c + d*x])/(2 + n) - ((2*a^2 - b^2)*Sin[c + d*x]^2)/(3 + n) - (
4*a*b*Sin[c + d*x]^3)/(4 + n) + ((a^2 - 2*b^2)*Sin[c + d*x]^4)/(5 + n) + (2*a*b*Sin[c + d*x]^5)/(6 + n) + (b^2
*Sin[c + d*x]^6)/(7 + n)))/d

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Maple [F]  time = 10.642, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+b\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)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.34373, size = 1408, normalized size = 8.28 \begin{align*} -\frac{{\left (2 \,{\left (a b n^{6} + 22 \, a b n^{5} + 190 \, a b n^{4} + 820 \, a b n^{3} + 1849 \, a b n^{2} + 2038 \, a b n + 840 \, a b\right )} \cos \left (d x + c\right )^{6} - 16 \, a b n^{4} - 256 \, a b n^{3} - 2 \,{\left (a b n^{6} + 18 \, a b n^{5} + 118 \, a b n^{4} + 348 \, a b n^{3} + 457 \, a b n^{2} + 210 \, a b n\right )} \cos \left (d x + c\right )^{4} - 1376 \, a b n^{2} - 2816 \, a b n - 8 \,{\left (a b n^{5} + 16 \, a b n^{4} + 86 \, a b n^{3} + 176 \, a b n^{2} + 105 \, a b n\right )} \cos \left (d x + c\right )^{2} - 1680 \, a b +{\left ({\left (b^{2} n^{6} + 21 \, b^{2} n^{5} + 175 \, b^{2} n^{4} + 735 \, b^{2} n^{3} + 1624 \, b^{2} n^{2} + 1764 \, b^{2} n + 720 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 8 \,{\left (a^{2} + b^{2}\right )} n^{4} -{\left ({\left (a^{2} + b^{2}\right )} n^{6} +{\left (23 \, a^{2} + 17 \, b^{2}\right )} n^{5} + 3 \,{\left (69 \, a^{2} + 37 \, b^{2}\right )} n^{4} + 5 \,{\left (185 \, a^{2} + 71 \, b^{2}\right )} n^{3} + 8 \,{\left (268 \, a^{2} + 73 \, b^{2}\right )} n^{2} + 1008 \, a^{2} + 144 \, b^{2} + 36 \,{\left (67 \, a^{2} + 13 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{4} - 8 \,{\left (19 \, a^{2} + 13 \, b^{2}\right )} n^{3} - 64 \,{\left (16 \, a^{2} + 7 \, b^{2}\right )} n^{2} - 4 \,{\left ({\left (a^{2} + b^{2}\right )} n^{5} + 2 \,{\left (10 \, a^{2} + 7 \, b^{2}\right )} n^{4} + 3 \,{\left (49 \, a^{2} + 23 \, b^{2}\right )} n^{3} + 4 \,{\left (121 \, a^{2} + 37 \, b^{2}\right )} n^{2} + 336 \, a^{2} + 48 \, b^{2} + 4 \,{\left (173 \, a^{2} + 35 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{2} - 2688 \, a^{2} - 384 \, b^{2} - 32 \,{\left (89 \, a^{2} + 23 \, b^{2}\right )} n\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{7} + 28 \, d n^{6} + 322 \, d n^{5} + 1960 \, d n^{4} + 6769 \, d n^{3} + 13132 \, d n^{2} + 13068 \, d n + 5040 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-(2*(a*b*n^6 + 22*a*b*n^5 + 190*a*b*n^4 + 820*a*b*n^3 + 1849*a*b*n^2 + 2038*a*b*n + 840*a*b)*cos(d*x + c)^6 -
16*a*b*n^4 - 256*a*b*n^3 - 2*(a*b*n^6 + 18*a*b*n^5 + 118*a*b*n^4 + 348*a*b*n^3 + 457*a*b*n^2 + 210*a*b*n)*cos(
d*x + c)^4 - 1376*a*b*n^2 - 2816*a*b*n - 8*(a*b*n^5 + 16*a*b*n^4 + 86*a*b*n^3 + 176*a*b*n^2 + 105*a*b*n)*cos(d
*x + c)^2 - 1680*a*b + ((b^2*n^6 + 21*b^2*n^5 + 175*b^2*n^4 + 735*b^2*n^3 + 1624*b^2*n^2 + 1764*b^2*n + 720*b^
2)*cos(d*x + c)^6 - 8*(a^2 + b^2)*n^4 - ((a^2 + b^2)*n^6 + (23*a^2 + 17*b^2)*n^5 + 3*(69*a^2 + 37*b^2)*n^4 + 5
*(185*a^2 + 71*b^2)*n^3 + 8*(268*a^2 + 73*b^2)*n^2 + 1008*a^2 + 144*b^2 + 36*(67*a^2 + 13*b^2)*n)*cos(d*x + c)
^4 - 8*(19*a^2 + 13*b^2)*n^3 - 64*(16*a^2 + 7*b^2)*n^2 - 4*((a^2 + b^2)*n^5 + 2*(10*a^2 + 7*b^2)*n^4 + 3*(49*a
^2 + 23*b^2)*n^3 + 4*(121*a^2 + 37*b^2)*n^2 + 336*a^2 + 48*b^2 + 4*(173*a^2 + 35*b^2)*n)*cos(d*x + c)^2 - 2688
*a^2 - 384*b^2 - 32*(89*a^2 + 23*b^2)*n)*sin(d*x + c))*sin(d*x + c)^n/(d*n^7 + 28*d*n^6 + 322*d*n^5 + 1960*d*n
^4 + 6769*d*n^3 + 13132*d*n^2 + 13068*d*n + 5040*d)

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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)**5*sin(d*x+c)**n*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.18212, size = 776, normalized size = 4.56 \begin{align*} \frac{\frac{{\left (n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} + 4 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} - 2 \, n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + 3 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} - 12 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right ) - 10 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + 8 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right ) + 15 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )\right )} a^{2}}{n^{3} + 9 \, n^{2} + 23 \, n + 15} + \frac{2 \,{\left (n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} + 6 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} - 2 \, n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + 8 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{6} - 16 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} - 24 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4} + 10 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}\right )} a b}{n^{3} + 12 \, n^{2} + 44 \, n + 48} + \frac{{\left (n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{7} + 8 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{7} - 2 \, n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{7} - 20 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} + n^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} - 42 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5} + 12 \, n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3} + 35 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}\right )} b^{2}}{n^{3} + 15 \, n^{2} + 71 \, n + 105}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

((n^2*sin(d*x + c)^n*sin(d*x + c)^5 + 4*n*sin(d*x + c)^n*sin(d*x + c)^5 - 2*n^2*sin(d*x + c)^n*sin(d*x + c)^3
+ 3*sin(d*x + c)^n*sin(d*x + c)^5 - 12*n*sin(d*x + c)^n*sin(d*x + c)^3 + n^2*sin(d*x + c)^n*sin(d*x + c) - 10*
sin(d*x + c)^n*sin(d*x + c)^3 + 8*n*sin(d*x + c)^n*sin(d*x + c) + 15*sin(d*x + c)^n*sin(d*x + c))*a^2/(n^3 + 9
*n^2 + 23*n + 15) + 2*(n^2*sin(d*x + c)^n*sin(d*x + c)^6 + 6*n*sin(d*x + c)^n*sin(d*x + c)^6 - 2*n^2*sin(d*x +
c)^n*sin(d*x + c)^4 + 8*sin(d*x + c)^n*sin(d*x + c)^6 - 16*n*sin(d*x + c)^n*sin(d*x + c)^4 + n^2*sin(d*x + c)
^n*sin(d*x + c)^2 - 24*sin(d*x + c)^n*sin(d*x + c)^4 + 10*n*sin(d*x + c)^n*sin(d*x + c)^2 + 24*sin(d*x + c)^n*
sin(d*x + c)^2)*a*b/(n^3 + 12*n^2 + 44*n + 48) + (n^2*sin(d*x + c)^n*sin(d*x + c)^7 + 8*n*sin(d*x + c)^n*sin(d
*x + c)^7 - 2*n^2*sin(d*x + c)^n*sin(d*x + c)^5 + 15*sin(d*x + c)^n*sin(d*x + c)^7 - 20*n*sin(d*x + c)^n*sin(d
*x + c)^5 + n^2*sin(d*x + c)^n*sin(d*x + c)^3 - 42*sin(d*x + c)^n*sin(d*x + c)^5 + 12*n*sin(d*x + c)^n*sin(d*x
+ c)^3 + 35*sin(d*x + c)^n*sin(d*x + c)^3)*b^2/(n^3 + 15*n^2 + 71*n + 105))/d