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

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Rubi [A]  time = 0.21, antiderivative size = 170, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.76, 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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fricas [B]  time = 1.00, size = 572, normalized size = 3.36 \[ -\frac {{\left (2 \, {\left (a b n^{6} + 22 \, a b n^{5} + 190 \, a b n^{4} + 820 \, a b n^{3} + 1849 \, a b n^{2} + 2038 \, a b n + 840 \, a b\right )} \cos \left (d x + c\right )^{6} - 16 \, a b n^{4} - 256 \, a b n^{3} - 2 \, {\left (a b n^{6} + 18 \, a b n^{5} + 118 \, a b n^{4} + 348 \, a b n^{3} + 457 \, a b n^{2} + 210 \, a b n\right )} \cos \left (d x + c\right )^{4} - 1376 \, a b n^{2} - 2816 \, a b n - 8 \, {\left (a b n^{5} + 16 \, a b n^{4} + 86 \, a b n^{3} + 176 \, a b n^{2} + 105 \, a b n\right )} \cos \left (d x + c\right )^{2} - 1680 \, a b + {\left ({\left (b^{2} n^{6} + 21 \, b^{2} n^{5} + 175 \, b^{2} n^{4} + 735 \, b^{2} n^{3} + 1624 \, b^{2} n^{2} + 1764 \, b^{2} n + 720 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 8 \, {\left (a^{2} + b^{2}\right )} n^{4} - {\left ({\left (a^{2} + b^{2}\right )} n^{6} + {\left (23 \, a^{2} + 17 \, b^{2}\right )} n^{5} + 3 \, {\left (69 \, a^{2} + 37 \, b^{2}\right )} n^{4} + 5 \, {\left (185 \, a^{2} + 71 \, b^{2}\right )} n^{3} + 8 \, {\left (268 \, a^{2} + 73 \, b^{2}\right )} n^{2} + 1008 \, a^{2} + 144 \, b^{2} + 36 \, {\left (67 \, a^{2} + 13 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (19 \, a^{2} + 13 \, b^{2}\right )} n^{3} - 64 \, {\left (16 \, a^{2} + 7 \, b^{2}\right )} n^{2} - 4 \, {\left ({\left (a^{2} + b^{2}\right )} n^{5} + 2 \, {\left (10 \, a^{2} + 7 \, b^{2}\right )} n^{4} + 3 \, {\left (49 \, a^{2} + 23 \, b^{2}\right )} n^{3} + 4 \, {\left (121 \, a^{2} + 37 \, b^{2}\right )} n^{2} + 336 \, a^{2} + 48 \, b^{2} + 4 \, {\left (173 \, a^{2} + 35 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{2} - 2688 \, a^{2} - 384 \, b^{2} - 32 \, {\left (89 \, a^{2} + 23 \, b^{2}\right )} n\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{7} + 28 \, d n^{6} + 322 \, d n^{5} + 1960 \, d n^{4} + 6769 \, d n^{3} + 13132 \, d n^{2} + 13068 \, d n + 5040 \, d} \]

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

[Out]

Timed out

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maple [F]  time = 26.34, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{2}\, dx \]

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 [A]  time = 0.37, size = 178, normalized size = 1.05 \[ \frac {\frac {b^{2} \sin \left (d x + c\right )^{n + 7}}{n + 7} + \frac {2 \, a b \sin \left (d x + c\right )^{n + 6}}{n + 6} + \frac {a^{2} \sin \left (d x + c\right )^{n + 5}}{n + 5} - \frac {2 \, b^{2} \sin \left (d x + c\right )^{n + 5}}{n + 5} - \frac {4 \, a b \sin \left (d x + c\right )^{n + 4}}{n + 4} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {b^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {2 \, a b \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{2} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

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]

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

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mupad [B]  time = 18.92, size = 887, normalized size = 5.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)*sin(c + d*x)^n*(44*n + 12*n^2 + n^3 + 48)*(1272*a^2*n + 581*b^2*n + 4200*a^2 + 525*b^2 + 152*a^2
*n^2 + 8*a^2*n^3 + 59*b^2*n^2 + 3*b^2*n^3)*1i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i
+ n^6*28i + n^7*1i + 5040i)) + (sin(c + d*x)^n*sin(5*c + 5*d*x)*(4*a^2*n - b^2*n + 28*a^2 - 21*b^2)*(324*n +
260*n^2 + 95*n^3 + 16*n^4 + n^5 + 144)*1i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n
^6*28i + n^7*1i + 5040i)) - (b^2*sin(c + d*x)^n*sin(7*c + 7*d*x)*(1764*n + 1624*n^2 + 735*n^3 + 175*n^4 + 21*n
^5 + n^6 + 720)*1i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i
)) + (sin(c + d*x)^n*sin(3*c + 3*d*x)*(92*n + 56*n^2 + 13*n^3 + n^4 + 48)*(184*a^2*n + 40*b^2*n + 700*a^2 - 35
*b^2 + 12*a^2*n^2 + 3*b^2*n^2)*1i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i +
n^7*1i + 5040i)) + (a*b*sin(c + d*x)^n*(n*16958i + n^2*10137i + n^3*2788i + n^4*398i + n^5*30i + n^6*1i + 924
0i))/(8*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i)) - (a*b*sin(c
+ d*x)^n*cos(6*c + 6*d*x)*(n*2038i + n^2*1849i + n^3*820i + n^4*190i + n^5*22i + n^6*1i + 840i))/(16*d*(n*1306
8i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i)) - (a*b*sin(c + d*x)^n*cos(4*c
+ 4*d*x)*(n*5694i + n^2*4633i + n^3*1764i + n^4*334i + n^5*30i + n^6*1i + 2520i))/(8*d*(n*13068i + n^2*13132i
+ n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i)) - (a*b*sin(c + d*x)^n*cos(2*c + 2*d*x)*(n*2049
0i + n^2*9159i + n^3*1228i - n^4*62i - n^5*22i - n^6*1i + 12600i))/(16*d*(n*13068i + n^2*13132i + n^3*6769i +
n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i))

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

[Out]

Timed out

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