3.1.23 \(\int \sin ^3(a+b x) \sin ^5(2 a+2 b x) \, dx\) [23]

Optimal. Leaf size=46 \[ \frac {32 \sin ^9(a+b x)}{9 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^{13}(a+b x)}{13 b} \]

[Out]

32/9*sin(b*x+a)^9/b-64/11*sin(b*x+a)^11/b+32/13*sin(b*x+a)^13/b

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Rubi [A]
time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 2644, 276} \begin {gather*} \frac {32 \sin ^{13}(a+b x)}{13 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^9(a+b x)}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^9)/(9*b) - (64*Sin[a + b*x]^11)/(11*b) + (32*Sin[a + b*x]^13)/(13*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 4373

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

Rubi steps

\begin {align*} \int \sin ^3(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^8(a+b x) \, dx\\ &=\frac {32 \text {Subst}\left (\int x^8 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \text {Subst}\left (\int \left (x^8-2 x^{10}+x^{12}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \sin ^9(a+b x)}{9 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^{13}(a+b x)}{13 b}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 37, normalized size = 0.80 \begin {gather*} \frac {4 (505+540 \cos (2 (a+b x))+99 \cos (4 (a+b x))) \sin ^9(a+b x)}{1287 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(505 + 540*Cos[2*(a + b*x)] + 99*Cos[4*(a + b*x)])*Sin[a + b*x]^9)/(1287*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(40)=80\).
time = 0.17, size = 97, normalized size = 2.11

method	result	size
default	\(\frac {5 \sin \left (x b +a \right )}{32 b}-\frac {25 \sin \left (3 x b +3 a \right )}{384 b}-\frac {\sin \left (5 x b +5 a \right )}{128 b}+\frac {\sin \left (7 x b +7 a \right )}{64 b}-\frac {\sin \left (9 x b +9 a \right )}{576 b}-\frac {3 \sin \left (11 x b +11 a \right )}{1408 b}+\frac {\sin \left (13 x b +13 a \right )}{1664 b}\)	\(97\)
risch	\(\frac {5 \sin \left (x b +a \right )}{32 b}-\frac {25 \sin \left (3 x b +3 a \right )}{384 b}-\frac {\sin \left (5 x b +5 a \right )}{128 b}+\frac {\sin \left (7 x b +7 a \right )}{64 b}-\frac {\sin \left (9 x b +9 a \right )}{576 b}-\frac {3 \sin \left (11 x b +11 a \right )}{1408 b}+\frac {\sin \left (13 x b +13 a \right )}{1664 b}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x,method=_RETURNVERBOSE)

[Out]

5/32*sin(b*x+a)/b-25/384*sin(3*b*x+3*a)/b-1/128/b*sin(5*b*x+5*a)+1/64/b*sin(7*b*x+7*a)-1/576/b*sin(9*b*x+9*a)-
3/1408/b*sin(11*b*x+11*a)+1/1664/b*sin(13*b*x+13*a)

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Maxima [A]
time = 0.28, size = 80, normalized size = 1.74 \begin {gather*} \frac {99 \, \sin \left (13 \, b x + 13 \, a\right ) - 351 \, \sin \left (11 \, b x + 11 \, a\right ) - 286 \, \sin \left (9 \, b x + 9 \, a\right ) + 2574 \, \sin \left (7 \, b x + 7 \, a\right ) - 1287 \, \sin \left (5 \, b x + 5 \, a\right ) - 10725 \, \sin \left (3 \, b x + 3 \, a\right ) + 25740 \, \sin \left (b x + a\right )}{164736 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

1/164736*(99*sin(13*b*x + 13*a) - 351*sin(11*b*x + 11*a) - 286*sin(9*b*x + 9*a) + 2574*sin(7*b*x + 7*a) - 1287
*sin(5*b*x + 5*a) - 10725*sin(3*b*x + 3*a) + 25740*sin(b*x + a))/b

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Fricas [A]
time = 3.48, size = 73, normalized size = 1.59 \begin {gather*} \frac {32 \, {\left (99 \, \cos \left (b x + a\right )^{12} - 360 \, \cos \left (b x + a\right )^{10} + 458 \, \cos \left (b x + a\right )^{8} - 212 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{1287 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

32/1287*(99*cos(b*x + a)^12 - 360*cos(b*x + a)^10 + 458*cos(b*x + a)^8 - 212*cos(b*x + a)^6 + 3*cos(b*x + a)^4
+ 4*cos(b*x + a)^2 + 8)*sin(b*x + a)/b

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (39) = 78\).
time = 34.91, size = 447, normalized size = 9.72 \begin {gather*} \begin {cases} - \frac {1366 \sin ^{3}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{3003 b} - \frac {4960 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{9009 b} - \frac {256 \sin ^{3}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{1287 b} - \frac {271 \sin ^{2}{\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{3003 b} - \frac {48 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{143 b} - \frac {640 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {1388 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{3003 b} - \frac {2944 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {512 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{1001 b} + \frac {2234 \sin ^{5}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{9009 b} + \frac {4544 \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{9009 b} + \frac {256 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{1001 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{5}{\left (2 a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**5,x)

[Out]

Piecewise((-1366*sin(a + b*x)**3*sin(2*a + 2*b*x)**4*cos(2*a + 2*b*x)/(3003*b) - 4960*sin(a + b*x)**3*sin(2*a
+ 2*b*x)**2*cos(2*a + 2*b*x)**3/(9009*b) - 256*sin(a + b*x)**3*cos(2*a + 2*b*x)**5/(1287*b) - 271*sin(a + b*x)
**2*sin(2*a + 2*b*x)**5*cos(a + b*x)/(3003*b) - 48*sin(a + b*x)**2*sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a +
2*b*x)**2/(143*b) - 640*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(3003*b) - 1388*sin(
a + b*x)*sin(2*a + 2*b*x)**4*cos(a + b*x)**2*cos(2*a + 2*b*x)/(3003*b) - 2944*sin(a + b*x)*sin(2*a + 2*b*x)**2
*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(3003*b) - 512*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**5/(1001*b)
+ 2234*sin(2*a + 2*b*x)**5*cos(a + b*x)**3/(9009*b) + 4544*sin(2*a + 2*b*x)**3*cos(a + b*x)**3*cos(2*a + 2*b*x
)**2/(9009*b) + 256*sin(2*a + 2*b*x)*cos(a + b*x)**3*cos(2*a + 2*b*x)**4/(1001*b), Ne(b, 0)), (x*sin(a)**3*sin
(2*a)**5, True))

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Giac [A]
time = 0.43, size = 36, normalized size = 0.78 \begin {gather*} \frac {32 \, {\left (99 \, \sin \left (b x + a\right )^{13} - 234 \, \sin \left (b x + a\right )^{11} + 143 \, \sin \left (b x + a\right )^{9}\right )}}{1287 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

32/1287*(99*sin(b*x + a)^13 - 234*sin(b*x + a)^11 + 143*sin(b*x + a)^9)/b

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Mupad [B]
time = 0.07, size = 36, normalized size = 0.78 \begin {gather*} \frac {32\,\left (99\,{\sin \left (a+b\,x\right )}^{13}-234\,{\sin \left (a+b\,x\right )}^{11}+143\,{\sin \left (a+b\,x\right )}^9\right )}{1287\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^5,x)

[Out]

(32*(143*sin(a + b*x)^9 - 234*sin(a + b*x)^11 + 99*sin(a + b*x)^13))/(1287*b)

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