\(\int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) [387]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a b \cos ^6(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {\left (2 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^7(c+d x)}{7 d} \]

[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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2747, 710, 1824} \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {\left (2 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {a b \cos ^6(c+d x)}{3 d}+\frac {b^2 \sin ^7(c+d x)}{7 d} \]

[In]

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

[Out]

-1/3*(a*b*Cos[c + d*x]^6)/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 710

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 1824

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 2747

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*} \text {integral}& = \frac {\text {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 {\text {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 {\text {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] (verified)

Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin (c+d x) \left (105 a^2+105 a b \sin (c+d x)+35 \left (-2 a^2+b^2\right ) \sin ^2(c+d x)-105 a b \sin ^3(c+d x)+21 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+35 a b \sin ^5(c+d x)+15 b^2 \sin ^6(c+d x)\right )}{105 d} \]

[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)

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {a b \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}-a b \left (\sin ^{4}\left (d x +c \right )\right )+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\left (\sin ^{2}\left (d x +c \right )\right ) a b +a^{2} \sin \left (d x +c \right )}{d}\)	\(102\)
default	\(\frac {\frac {b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {a b \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}-a b \left (\sin ^{4}\left (d x +c \right )\right )+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\left (\sin ^{2}\left (d x +c \right )\right ) a b +a^{2} \sin \left (d x +c \right )}{d}\)	\(102\)
parallelrisch	\(\frac {-420 a b \cos \left (4 d x +4 c \right )-15 b^{2} \sin \left (7 d x +7 c \right )-70 a b \cos \left (6 d x +6 c \right )+84 \sin \left (5 d x +5 c \right ) a^{2}-63 \sin \left (5 d x +5 c \right ) b^{2}+700 \sin \left (3 d x +3 c \right ) a^{2}-35 \sin \left (3 d x +3 c \right ) b^{2}-1050 a b \cos \left (2 d x +2 c \right )+4200 a^{2} \sin \left (d x +c \right )+525 \sin \left (d x +c \right ) b^{2}+1540 a b}{6720 d}\)	\(142\)
risch	\(\frac {5 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {5 b^{2} \sin \left (d x +c \right )}{64 d}-\frac {b^{2} \sin \left (7 d x +7 c \right )}{448 d}-\frac {a b \cos \left (6 d x +6 c \right )}{96 d}+\frac {\sin \left (5 d x +5 c \right ) a^{2}}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{2}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{192 d}-\frac {5 a b \cos \left (2 d x +2 c \right )}{32 d}\)	\(163\)
norman	\(\frac {\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (91 a^{2}+38 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {2 \left (113 a^{2}-16 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 \left (113 a^{2}-16 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {40 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {40 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\)	\(297\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {35 \, a b \cos \left (d x + c\right )^{6} + {\left (15 \, b^{2} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 56 \, a^{2} - 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \]

[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

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.60 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {15 \, b^{2} \sin \left (d x + c\right )^{7} + 35 \, a b \sin \left (d x + c\right )^{6} - 105 \, a b \sin \left (d x + c\right )^{4} + 21 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{5} + 105 \, a b \sin \left (d x + c\right )^{2} - 35 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.37 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {5 \, a b \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (4 \, a^{2} - 3 \, b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 4.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}-\frac {b^2}{3}\right )+{\sin \left (c+d\,x\right )}^5\,\left (\frac {a^2}{5}-\frac {2\,b^2}{5}\right )+\frac {b^2\,{\sin \left (c+d\,x\right )}^7}{7}+a\,b\,{\sin \left (c+d\,x\right )}^2-a\,b\,{\sin \left (c+d\,x\right )}^4+\frac {a\,b\,{\sin \left (c+d\,x\right )}^6}{3}}{d} \]

[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