3.233 \(\int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=70 \[ \frac {\sin ^2(c+d x)}{2 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {3 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

3*ln(1+sin(d*x+c))/a^2/d-2*sin(d*x+c)/a^2/d+1/2*sin(d*x+c)^2/a^2/d+1/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ \frac {\sin ^2(c+d x)}{2 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {3 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(3*Log[1 + Sin[c + d*x]])/(a^2*d) - (2*Sin[c + d*x])/(a^2*d) + Sin[c + d*x]^2/(2*a^2*d) + 1/(d*(a^2 + a^2*Sin[
c + d*x]))

Rule 12

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

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 \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{a^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2 a+x-\frac {a^3}{(a+x)^2}+\frac {3 a^2}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {3 \log (1+\sin (c+d x))}{a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {\sin ^2(c+d x)}{2 a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 71, normalized size = 1.01 \[ \frac {\sin ^3(c+d x)-3 \sin ^2(c+d x)+\sin (c+d x) (6 \log (\sin (c+d x)+1)-4)+6 \log (\sin (c+d x)+1)+2}{2 a^2 d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2 + 6*Log[1 + Sin[c + d*x]] + (-4 + 6*Log[1 + Sin[c + d*x]])*Sin[c + d*x] - 3*Sin[c + d*x]^2 + Sin[c + d*x]^3
)/(2*a^2*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.55, size = 72, normalized size = 1.03 \[ \frac {6 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, \cos \left (d x + c\right )^{2} + 7\right )} \sin \left (d x + c\right ) - 3}{4 \, {\left (a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(6*cos(d*x + c)^2 + 12*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (2*cos(d*x + c)^2 + 7)*sin(d*x + c) - 3)
/(a^2*d*sin(d*x + c) + a^2*d)

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giac [A]  time = 0.17, size = 90, normalized size = 1.29 \[ -\frac {\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (\frac {6 \, a}{a \sin \left (d x + c\right ) + a} - 1\right )}}{a^{4}} + \frac {6 \, \log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a^{2}} - \frac {2}{{\left (a \sin \left (d x + c\right ) + a\right )} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*((a*sin(d*x + c) + a)^2*(6*a/(a*sin(d*x + c) + a) - 1)/a^4 + 6*log(abs(a*sin(d*x + c) + a)/((a*sin(d*x +
c) + a)^2*abs(a)))/a^2 - 2/((a*sin(d*x + c) + a)*a))/d

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maple [A]  time = 0.22, size = 66, normalized size = 0.94 \[ \frac {\sin ^{2}\left (d x +c \right )}{2 a^{2} d}-\frac {2 \sin \left (d x +c \right )}{a^{2} d}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{2} d}+\frac {1}{d \,a^{2} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

1/2*sin(d*x+c)^2/a^2/d-2*sin(d*x+c)/a^2/d+3*ln(1+sin(d*x+c))/a^2/d+1/d/a^2/(1+sin(d*x+c))

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maxima [A]  time = 0.32, size = 59, normalized size = 0.84 \[ \frac {\frac {2}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {\sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )}{a^{2}} + \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(2/(a^2*sin(d*x + c) + a^2) + (sin(d*x + c)^2 - 4*sin(d*x + c))/a^2 + 6*log(sin(d*x + c) + 1)/a^2)/d

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mupad [B]  time = 8.47, size = 59, normalized size = 0.84 \[ \frac {\frac {1}{a^2\,\sin \left (c+d\,x\right )+a^2}+\frac {3\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2}-\frac {2\,\sin \left (c+d\,x\right )}{a^2}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)

[Out]

(1/(a^2*sin(c + d*x) + a^2) + (3*log(sin(c + d*x) + 1))/a^2 - (2*sin(c + d*x))/a^2 + sin(c + d*x)^2/(2*a^2))/d

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sympy [A]  time = 2.81, size = 170, normalized size = 2.43 \[ \begin {cases} \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {\sin ^{3}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} - \frac {3 \sin ^{2}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((6*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 6*log(sin(c + d*x) + 1)/(
2*a**2*d*sin(c + d*x) + 2*a**2*d) + sin(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) - 3*sin(c + d*x)**2/(2*
a**2*d*sin(c + d*x) + 2*a**2*d) + 6/(2*a**2*d*sin(c + d*x) + 2*a**2*d), Ne(d, 0)), (x*sin(c)**3*cos(c)/(a*sin(
c) + a)**2, True))

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