3.1297 \(\int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=89 \[ \frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 89, 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 = {2837, 12, 772} \[ -\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^4*d) - ((a^2 - b^2)*Sin[c + d*x])/(b^3*d) + (a*Sin[c + d*x]^2)/(2*b
^2*d) - Sin[c + d*x]^3/(3*b*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 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 79, normalized size = 0.89 \[ \frac {6 b \left (b^2-a^2\right ) \sin (c+d x)+6 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+3 a b^2 \sin ^2(c+d x)-2 b^3 \sin ^3(c+d x)}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 78, normalized size = 0.88 \[ -\frac {3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 85, normalized size = 0.96 \[ -\frac {\frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right ) - 6 \, b^{2} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 106, normalized size = 1.19 \[ -\frac {\sin ^{3}\left (d x +c \right )}{3 b d}+\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{2 b^{2} d}-\frac {a^{2} \sin \left (d x +c \right )}{b^{3} d}+\frac {\sin \left (d x +c \right )}{b d}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4} d}-\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 79, normalized size = 0.89 \[ -\frac {\frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 78, normalized size = 0.88 \[ \frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2\,b^2}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a\,b^2-a^3\right )}{b^4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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