3.1215 \(\int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=138 \[ \frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^8(c+d x)}{4 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \]

[Out]

1/3*a^2*sin(d*x+c)^3/d+1/2*a*b*sin(d*x+c)^4/d-1/5*(2*a^2-b^2)*sin(d*x+c)^5/d-2/3*a*b*sin(d*x+c)^6/d+1/7*(a^2-2
*b^2)*sin(d*x+c)^7/d+1/4*a*b*sin(d*x+c)^8/d+1/9*b^2*sin(d*x+c)^9/d

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Rubi [A]  time = 0.19, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 948} \[ \frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^8(c+d x)}{4 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sin[c + d*x]^3)/(3*d) + (a*b*Sin[c + d*x]^4)/(2*d) - ((2*a^2 - b^2)*Sin[c + d*x]^5)/(5*d) - (2*a*b*Sin[c
+ d*x]^6)/(3*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^7)/(7*d) + (a*b*Sin[c + d*x]^8)/(4*d) + (b^2*Sin[c + d*x]^9)/(9*
d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+x)^2 \left (b^2-x^2\right )^2}{b^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int x^2 (a+x)^2 \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 b^4 x^2+2 a b^4 x^3+b^2 \left (-2 a^2+b^2\right ) x^4-4 a b^2 x^5+\left (a^2-2 b^2\right ) x^6+2 a x^7+x^8\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}+\frac {a b \sin ^8(c+d x)}{4 d}+\frac {b^2 \sin ^9(c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 169, normalized size = 1.22 \[ \frac {12600 a^2 \sin (c+d x)-840 a^2 \sin (3 (c+d x))-1512 a^2 \sin (5 (c+d x))-360 a^2 \sin (7 (c+d x))-7560 a b \cos (2 (c+d x))-1260 a b \cos (4 (c+d x))+840 a b \cos (6 (c+d x))+315 a b \cos (8 (c+d x))+3780 b^2 \sin (c+d x)-840 b^2 \sin (3 (c+d x))-504 b^2 \sin (5 (c+d x))+90 b^2 \sin (7 (c+d x))+70 b^2 \sin (9 (c+d x))}{161280 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7560*a*b*Cos[2*(c + d*x)] - 1260*a*b*Cos[4*(c + d*x)] + 840*a*b*Cos[6*(c + d*x)] + 315*a*b*Cos[8*(c + d*x)]
+ 12600*a^2*Sin[c + d*x] + 3780*b^2*Sin[c + d*x] - 840*a^2*Sin[3*(c + d*x)] - 840*b^2*Sin[3*(c + d*x)] - 1512*
a^2*Sin[5*(c + d*x)] - 504*b^2*Sin[5*(c + d*x)] - 360*a^2*Sin[7*(c + d*x)] + 90*b^2*Sin[7*(c + d*x)] + 70*b^2*
Sin[9*(c + d*x)])/(161280*d)

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fricas [A]  time = 0.71, size = 121, normalized size = 0.88 \[ \frac {315 \, a b \cos \left (d x + c\right )^{8} - 420 \, a b \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, b^{2} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1260*(315*a*b*cos(d*x + c)^8 - 420*a*b*cos(d*x + c)^6 + 4*(35*b^2*cos(d*x + c)^8 - 5*(9*a^2 + 10*b^2)*cos(d*
x + c)^6 + 3*(3*a^2 + b^2)*cos(d*x + c)^4 + 4*(3*a^2 + b^2)*cos(d*x + c)^2 + 24*a^2 + 8*b^2)*sin(d*x + c))/d

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giac [A]  time = 0.33, size = 173, normalized size = 1.25 \[ \frac {a b \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} + \frac {a b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {3 \, a b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {b^{2} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {{\left (4 \, a^{2} - b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (3 \, a^{2} + b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (a^{2} + b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {{\left (10 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*a*b*cos(8*d*x + 8*c)/d + 1/192*a*b*cos(6*d*x + 6*c)/d - 1/128*a*b*cos(4*d*x + 4*c)/d - 3/64*a*b*cos(2*d*
x + 2*c)/d + 1/2304*b^2*sin(9*d*x + 9*c)/d - 1/1792*(4*a^2 - b^2)*sin(7*d*x + 7*c)/d - 1/320*(3*a^2 + b^2)*sin
(5*d*x + 5*c)/d - 1/192*(a^2 + b^2)*sin(3*d*x + 3*c)/d + 1/128*(10*a^2 + 3*b^2)*sin(d*x + c)/d

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maple [A]  time = 0.31, size = 155, normalized size = 1.12 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+2 a b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )+b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{9}-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{21}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{105}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+2*a*b*(-1/8*sin(d*
x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)+b^2*(-1/9*sin(d*x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*
(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)))

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maxima [A]  time = 0.33, size = 108, normalized size = 0.78 \[ \frac {140 \, b^{2} \sin \left (d x + c\right )^{9} + 315 \, a b \sin \left (d x + c\right )^{8} - 840 \, a b \sin \left (d x + c\right )^{6} + 180 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{7} + 630 \, a b \sin \left (d x + c\right )^{4} - 252 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{5} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1260*(140*b^2*sin(d*x + c)^9 + 315*a*b*sin(d*x + c)^8 - 840*a*b*sin(d*x + c)^6 + 180*(a^2 - 2*b^2)*sin(d*x +
c)^7 + 630*a*b*sin(d*x + c)^4 - 252*(2*a^2 - b^2)*sin(d*x + c)^5 + 420*a^2*sin(d*x + c)^3)/d

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mupad [B]  time = 11.44, size = 108, normalized size = 0.78 \[ \frac {{\sin \left (c+d\,x\right )}^7\,\left (\frac {a^2}{7}-\frac {2\,b^2}{7}\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {2\,a^2}{5}-\frac {b^2}{5}\right )+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,b\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^8}{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)^7*(a^2/7 - (2*b^2)/7) - sin(c + d*x)^5*((2*a^2)/5 - b^2/5) + (a^2*sin(c + d*x)^3)/3 + (b^2*sin(c
+ d*x)^9)/9 + (a*b*sin(c + d*x)^4)/2 - (2*a*b*sin(c + d*x)^6)/3 + (a*b*sin(c + d*x)^8)/4)/d

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sympy [A]  time = 14.27, size = 214, normalized size = 1.55 \[ \begin {cases} \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {a b \sin ^{8}{\left (c + d x \right )}}{12 d} + \frac {a b \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a b \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {8 b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin ^{2}{\relax (c )} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a**2*sin(c + d*x)**7/(105*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a**2*sin(c + d*x)*
*3*cos(c + d*x)**4/(3*d) + a*b*sin(c + d*x)**8/(12*d) + a*b*sin(c + d*x)**6*cos(c + d*x)**2/(3*d) + a*b*sin(c
+ d*x)**4*cos(c + d*x)**4/(2*d) + 8*b**2*sin(c + d*x)**9/(315*d) + 4*b**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*
d) + b**2*sin(c + d*x)**5*cos(c + d*x)**4/(5*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**2*cos(c)**5, True))

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