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]

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

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Rubi [A]  time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.20, 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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fricas [A]  time = 0.48, size = 87, normalized size = 0.88 \[ -\frac {35 \, a b \cos \left (d x + c\right )^{6} + {\left (15 \, b^{2} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 56 \, a^{2} - 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \]

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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giac [A]  time = 0.75, size = 136, normalized size = 1.37 \[ -\frac {a b \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {5 \, a b \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (4 \, a^{2} - 3 \, b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

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

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maple [A]  time = 0.20, size = 98, normalized size = 0.99 \[ \frac {b^{2} \left (-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {a b \left (\cos ^{6}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

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*cos(d*x+c)^6*sin(d*x+c)+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 = 0.32, size = 106, normalized size = 1.07 \[ \frac {15 \, b^{2} \sin \left (d x + c\right )^{7} + 35 \, a b \sin \left (d x + c\right )^{6} - 105 \, a b \sin \left (d x + c\right )^{4} + 21 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{5} + 105 \, a b \sin \left (d x + c\right )^{2} - 35 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \]

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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mupad [B]  time = 0.07, size = 104, normalized size = 1.05 \[ \frac {a^2\,\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}-\frac {b^2}{3}\right )+{\sin \left (c+d\,x\right )}^5\,\left (\frac {a^2}{5}-\frac {2\,b^2}{5}\right )+\frac {b^2\,{\sin \left (c+d\,x\right )}^7}{7}+a\,b\,{\sin \left (c+d\,x\right )}^2-a\,b\,{\sin \left (c+d\,x\right )}^4+\frac {a\,b\,{\sin \left (c+d\,x\right )}^6}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.45, size = 158, normalized size = 1.60 \[ \begin {cases} \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(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*cos(c + d*x)**6/(3*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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