3.224 \(\int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=85 \[ \frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x)+1)}{a d} \]

[Out]

ln(1+sin(d*x+c))/a/d-sin(d*x+c)/a/d+1/2*sin(d*x+c)^2/a/d-1/3*sin(d*x+c)^3/a/d+1/4*sin(d*x+c)^4/a/d

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Rubi [A]  time = 0.09, antiderivative size = 85, 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} \[ \frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

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

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 ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{a^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a^3+a^2 x-a x^2+x^3+\frac {a^4}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\log (1+\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 60, normalized size = 0.71 \[ \frac {3 \sin ^4(c+d x)-4 \sin ^3(c+d x)+6 \sin ^2(c+d x)-12 \sin (c+d x)+12 \log (\sin (c+d x)+1)}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

(12*Log[1 + Sin[c + d*x]] - 12*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(12*a*d)

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fricas [A]  time = 0.64, size = 58, normalized size = 0.68 \[ \frac {3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{12 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 76, normalized size = 0.89 \[ \frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right )}{a^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 80, normalized size = 0.94 \[ \frac {\ln \left (1+\sin \left (d x +c \right )\right )}{a d}-\frac {\sin \left (d x +c \right )}{a d}+\frac {\sin ^{2}\left (d x +c \right )}{2 a d}-\frac {\sin ^{3}\left (d x +c \right )}{3 d a}+\frac {\sin ^{4}\left (d x +c \right )}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

ln(1+sin(d*x+c))/a/d-sin(d*x+c)/a/d+1/2*sin(d*x+c)^2/a/d-1/3*sin(d*x+c)^3/d/a+1/4*sin(d*x+c)^4/d/a

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maxima [A]  time = 0.41, size = 63, normalized size = 0.74 \[ \frac {\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.46, size = 68, normalized size = 0.80 \[ \frac {\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a}-\frac {\sin \left (c+d\,x\right )}{a}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*sin(c + d*x)^4)/(a + a*sin(c + d*x)),x)

[Out]

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

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sympy [A]  time = 3.20, size = 80, normalized size = 0.94 \[ \begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} - \frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} + \frac {\sin ^{2}{\left (c + d x \right )}}{2 a d} - \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\relax (c )} \cos {\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((log(sin(c + d*x) + 1)/(a*d) + sin(c + d*x)**4/(4*a*d) - sin(c + d*x)**3/(3*a*d) + sin(c + d*x)**2/(
2*a*d) - sin(c + d*x)/(a*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a), True))

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