3.989 \(\int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=51 \[ \frac {B (a \sin (c+d x)+a)^5}{5 a^2 d}+\frac {(A-B) (a \sin (c+d x)+a)^4}{4 a d} \]

[Out]

1/4*(A-B)*(a+a*sin(d*x+c))^4/a/d+1/5*B*(a+a*sin(d*x+c))^5/a^2/d

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Rubi [A]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2833, 43} \[ \frac {B (a \sin (c+d x)+a)^5}{5 a^2 d}+\frac {(A-B) (a \sin (c+d x)+a)^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

((A - B)*(a + a*Sin[c + d*x])^4)/(4*a*d) + (B*(a + a*Sin[c + d*x])^5)/(5*a^2*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 36, normalized size = 0.71 \[ \frac {a^3 (\sin (c+d x)+1)^4 (5 A+4 B \sin (c+d x)-B)}{20 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(1 + Sin[c + d*x])^4*(5*A - B + 4*B*Sin[c + d*x]))/(20*d)

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fricas [A]  time = 0.56, size = 94, normalized size = 1.84 \[ \frac {5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 4 \, {\left (B a^{3} \cos \left (d x + c\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/20*(5*(A + 3*B)*a^3*cos(d*x + c)^4 - 40*(A + B)*a^3*cos(d*x + c)^2 + 4*(B*a^3*cos(d*x + c)^4 - (5*A + 7*B)*a
^3*cos(d*x + c)^2 + 2*(5*A + 3*B)*a^3)*sin(d*x + c))/d

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giac [B]  time = 0.23, size = 116, normalized size = 2.27 \[ \frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, A a^{3} \sin \left (d x + c\right )^{4} + 15 \, B a^{3} \sin \left (d x + c\right )^{4} + 20 \, A a^{3} \sin \left (d x + c\right )^{3} + 20 \, B a^{3} \sin \left (d x + c\right )^{3} + 30 \, A a^{3} \sin \left (d x + c\right )^{2} + 10 \, B a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/20*(4*B*a^3*sin(d*x + c)^5 + 5*A*a^3*sin(d*x + c)^4 + 15*B*a^3*sin(d*x + c)^4 + 20*A*a^3*sin(d*x + c)^3 + 20
*B*a^3*sin(d*x + c)^3 + 30*A*a^3*sin(d*x + c)^2 + 10*B*a^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d

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maple [B]  time = 0.23, size = 98, normalized size = 1.92 \[ \frac {\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (a^{3} A +3 B \,a^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (3 a^{3} A +3 B \,a^{3}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 a^{3} A +B \,a^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{3} A \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/5*B*a^3*sin(d*x+c)^5+1/4*(A*a^3+3*B*a^3)*sin(d*x+c)^4+1/3*(3*A*a^3+3*B*a^3)*sin(d*x+c)^3+1/2*(3*A*a^3+B
*a^3)*sin(d*x+c)^2+a^3*A*sin(d*x+c))

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maxima [A]  time = 0.31, size = 84, normalized size = 1.65 \[ \frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{3} + 10 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/20*(4*B*a^3*sin(d*x + c)^5 + 5*(A + 3*B)*a^3*sin(d*x + c)^4 + 20*(A + B)*a^3*sin(d*x + c)^3 + 10*(3*A + B)*a
^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d

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mupad [B]  time = 9.07, size = 81, normalized size = 1.59 \[ \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A+3\,B\right )}{4}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^5}{5}+A\,a^3\,\sin \left (c+d\,x\right )+a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)

[Out]

((a^3*sin(c + d*x)^2*(3*A + B))/2 + (a^3*sin(c + d*x)^4*(A + 3*B))/4 + (B*a^3*sin(c + d*x)^5)/5 + A*a^3*sin(c
+ d*x) + a^3*sin(c + d*x)^3*(A + B))/d

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sympy [A]  time = 2.02, size = 151, normalized size = 2.96 \[ \begin {cases} \frac {A a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{d} - \frac {3 A a^{3} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} - \frac {B a^{3} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{3} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((A*a**3*sin(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)**3/d + A*a**3*sin(c + d*x)/d - 3*A*a**3*cos(c +
d*x)**2/(2*d) + B*a**3*sin(c + d*x)**5/(5*d) + 3*B*a**3*sin(c + d*x)**4/(4*d) + B*a**3*sin(c + d*x)**3/d - B*a
**3*cos(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c), True))

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