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

Optimal. Leaf size=99 \[ \frac{\left (a^2-2 b^2\right ) \sin ^5(c+d x)}{5 d}-\frac{\left (2 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{a b \cos ^6(c+d x)}{3 d}+\frac{b^2 \sin ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.089251, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2668, 696, 1810} \[ \frac{\left (a^2-2 b^2\right ) \sin ^5(c+d x)}{5 d}-\frac{\left (2 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{a b \cos ^6(c+d x)}{3 d}+\frac{b^2 \sin ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 696

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*m*d^(m - 1)*(a + c*x^2)^(p + 1))
/(2*c*(p + 1)), x] + Int[((d + e*x)^m - e*m*d^(m - 1)*x)*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*
d^2 + a*e^2, 0] && IGtQ[p, 1] && IGtQ[m, 0] && LeQ[m, p]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.204956, size = 104, normalized size = 1.05 \[ \frac{\sin (c+d x) \left (21 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+35 \left (b^2-2 a^2\right ) \sin ^2(c+d x)+105 a^2+35 a b \sin ^5(c+d x)-105 a b \sin ^3(c+d x)+105 a b \sin (c+d x)+15 b^2 \sin ^6(c+d x)\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(105*a^2 + 105*a*b*Sin[c + d*x] + 35*(-2*a^2 + b^2)*Sin[c + d*x]^2 - 105*a*b*Sin[c + d*x]^3 + 21
*(a^2 - 2*b^2)*Sin[c + d*x]^4 + 35*a*b*Sin[c + d*x]^5 + 15*b^2*Sin[c + d*x]^6))/(105*d)

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Maple [A]  time = 0.043, size = 98, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) -{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^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))-1/3*a*b*cos(d*x+c)
^6+1/5*a^2*(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 = 0.960065, size = 143, normalized size = 1.44 \begin{align*} \frac{15 \, b^{2} \sin \left (d x + c\right )^{7} + 35 \, a b \sin \left (d x + c\right )^{6} - 105 \, a b \sin \left (d x + c\right )^{4} + 21 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{5} + 105 \, a b \sin \left (d x + c\right )^{2} - 35 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*b^2*sin(d*x + c)^7 + 35*a*b*sin(d*x + c)^6 - 105*a*b*sin(d*x + c)^4 + 21*(a^2 - 2*b^2)*sin(d*x + c)^
5 + 105*a*b*sin(d*x + c)^2 - 35*(2*a^2 - b^2)*sin(d*x + c)^3 + 105*a^2*sin(d*x + c))/d

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Fricas [A]  time = 2.23771, size = 211, normalized size = 2.13 \begin{align*} -\frac{35 \, a b \cos \left (d x + c\right )^{6} +{\left (15 \, b^{2} \cos \left (d x + c\right )^{6} - 3 \,{\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 56 \, a^{2} - 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(35*a*b*cos(d*x + c)^6 + (15*b^2*cos(d*x + c)^6 - 3*(7*a^2 + b^2)*cos(d*x + c)^4 - 4*(7*a^2 + b^2)*cos(
d*x + c)^2 - 56*a^2 - 8*b^2)*sin(d*x + c))/d

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Sympy [A]  time = 8.25853, size = 202, normalized size = 2.04 \begin{align*} \begin{cases} \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{a b \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac{a b \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a b \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \cos ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08869, size = 184, normalized size = 1.86 \begin{align*} -\frac{a b \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac{5 \, a b \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac{b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{{\left (4 \, a^{2} - 3 \, b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (20 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \,{\left (8 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*a*b*cos(6*d*x + 6*c)/d - 1/16*a*b*cos(4*d*x + 4*c)/d - 5/32*a*b*cos(2*d*x + 2*c)/d - 1/448*b^2*sin(7*d*x
+ 7*c)/d + 1/320*(4*a^2 - 3*b^2)*sin(5*d*x + 5*c)/d + 1/192*(20*a^2 - b^2)*sin(3*d*x + 3*c)/d + 5/64*(8*a^2 +
b^2)*sin(d*x + c)/d