3.1296 \(\int \frac{\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ -\frac{\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac{a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac{a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac{a \sin ^3(c+d x)}{3 b^2 d}-\frac{\sin ^4(c+d x)}{4 b d} \]

[Out]

-((a^2*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^5*d)) + (a*(a^2 - b^2)*Sin[c + d*x])/(b^4*d) - ((a^2 - b^2)*Sin
[c + d*x]^2)/(2*b^3*d) + (a*Sin[c + d*x]^3)/(3*b^2*d) - Sin[c + d*x]^4/(4*b*d)

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Rubi [A]  time = 0.172154, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ -\frac{\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac{a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac{a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac{a \sin ^3(c+d x)}{3 b^2 d}-\frac{\sin ^4(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^5*d)) + (a*(a^2 - b^2)*Sin[c + d*x])/(b^4*d) - ((a^2 - b^2)*Sin
[c + d*x]^2)/(2*b^3*d) + (a*Sin[c + d*x]^3)/(3*b^2*d) - Sin[c + d*x]^4/(4*b*d)

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]

Rule 12

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.347514, size = 104, normalized size = 0.87 \[ \frac{6 b^2 \left (b^2-a^2\right ) \sin ^2(c+d x)+12 a b \left (a^2-b^2\right ) \sin (c+d x)+12 a^2 \left (b^2-a^2\right ) \log (a+b \sin (c+d x))+4 a b^3 \sin ^3(c+d x)-3 b^4 \sin ^4(c+d x)}{12 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*(-a^2 + b^2)*Log[a + b*Sin[c + d*x]] + 12*a*b*(a^2 - b^2)*Sin[c + d*x] + 6*b^2*(-a^2 + b^2)*Sin[c + d*
x]^2 + 4*a*b^3*Sin[c + d*x]^3 - 3*b^4*Sin[c + d*x]^4)/(12*b^5*d)

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Maple [A]  time = 0.049, size = 144, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,bd}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,{b}^{2}d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d{b}^{3}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,bd}}+{\frac{{a}^{3}\sin \left ( dx+c \right ) }{d{b}^{4}}}-{\frac{a\sin \left ( dx+c \right ) }{{b}^{2}d}}-{\frac{{a}^{4}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{5}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.980881, size = 142, normalized size = 1.19 \begin{align*} -\frac{\frac{3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \,{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac{12 \,{\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*(a^2*b - b^3)*sin(d*x + c)^2 - 12*(a^3 - a*b^2)*sin(
d*x + c))/b^4 + 12*(a^4 - a^2*b^2)*log(b*sin(d*x + c) + a)/b^5)/d

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Fricas [A]  time = 1.58068, size = 230, normalized size = 1.93 \begin{align*} -\frac{3 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 12 \,{\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 2 \, a b^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^4*cos(d*x + c)^4 - 6*a^2*b^2*cos(d*x + c)^2 + 12*(a^4 - a^2*b^2)*log(b*sin(d*x + c) + a) + 4*(a*b^3
*cos(d*x + c)^2 - 3*a^3*b + 2*a*b^3)*sin(d*x + c))/(b^5*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.22803, size = 158, normalized size = 1.33 \begin{align*} -\frac{\frac{3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b \sin \left (d x + c\right )^{2} - 6 \, b^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) + 12 \, a b^{2} \sin \left (d x + c\right )}{b^{4}} + \frac{12 \,{\left (a^{4} - a^{2} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((3*b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*a^2*b*sin(d*x + c)^2 - 6*b^3*sin(d*x + c)^2 - 12*a^3
*sin(d*x + c) + 12*a*b^2*sin(d*x + c))/b^4 + 12*(a^4 - a^2*b^2)*log(abs(b*sin(d*x + c) + a))/b^5)/d