3.1204 \(\int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=83 \[ \frac{a \sin ^3(c+d x)}{3 d}-\frac{2 a \sin (c+d x)}{d}-\frac{a \csc (c+d x)}{d}+\frac{b \sin ^4(c+d x)}{4 d}-\frac{b \sin ^2(c+d x)}{d}+\frac{b \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0926954, antiderivative size = 83, normalized size of antiderivative = 1., 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, 766} \[ \frac{a \sin ^3(c+d x)}{3 d}-\frac{2 a \sin (c+d x)}{d}-\frac{a \csc (c+d x)}{d}+\frac{b \sin ^4(c+d x)}{4 d}-\frac{b \sin ^2(c+d x)}{d}+\frac{b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x])/d) + (b*Log[Sin[c + d*x]])/d - (2*a*Sin[c + d*x])/d - (b*Sin[c + d*x]^2)/d + (a*Sin[c + d*x
]^3)/(3*d) + (b*Sin[c + d*x]^4)/(4*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 766

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

Rubi steps

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

Mathematica [A]  time = 0.0301926, size = 83, normalized size = 1. \[ \frac{a \sin ^3(c+d x)}{3 d}-\frac{2 a \sin (c+d x)}{d}-\frac{a \csc (c+d x)}{d}+\frac{b \sin ^4(c+d x)}{4 d}-\frac{b \sin ^2(c+d x)}{d}+\frac{b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 116, normalized size = 1.4 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{8\,a\sin \left ( dx+c \right ) }{3\,d}}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}a}{d}}-{\frac{4\,a\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c)),x)

[Out]

-1/d*a/sin(d*x+c)*cos(d*x+c)^6-8/3*a*sin(d*x+c)/d-1/d*cos(d*x+c)^4*sin(d*x+c)*a-4/3/d*a*sin(d*x+c)*cos(d*x+c)^
2+1/4/d*b*cos(d*x+c)^4+1/2*b*cos(d*x+c)^2/d+b*ln(sin(d*x+c))/d

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Maxima [A]  time = 0.971597, size = 93, normalized size = 1.12 \begin{align*} \frac{3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, b \sin \left (d x + c\right )^{2} + 12 \, b \log \left (\sin \left (d x + c\right )\right ) - 24 \, a \sin \left (d x + c\right ) - \frac{12 \, a}{\sin \left (d x + c\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*b*sin(d*x + c)^2 + 12*b*log(sin(d*x + c)) - 24*a*sin(d*x +
c) - 12*a/sin(d*x + c))/d

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Fricas [A]  time = 1.75939, size = 250, normalized size = 3.01 \begin{align*} \frac{32 \, a \cos \left (d x + c\right )^{4} + 128 \, a \cos \left (d x + c\right )^{2} + 96 \, b \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \,{\left (8 \, b \cos \left (d x + c\right )^{4} + 16 \, b \cos \left (d x + c\right )^{2} - 11 \, b\right )} \sin \left (d x + c\right ) - 256 \, a}{96 \, d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/96*(32*a*cos(d*x + c)^4 + 128*a*cos(d*x + c)^2 + 96*b*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*b*cos(d*x +
c)^4 + 16*b*cos(d*x + c)^2 - 11*b)*sin(d*x + c) - 256*a)/(d*sin(d*x + c))

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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)**5*csc(d*x+c)**2*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.29756, size = 107, normalized size = 1.29 \begin{align*} \frac{3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, b \sin \left (d x + c\right )^{2} + 12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 24 \, a \sin \left (d x + c\right ) - \frac{12 \,{\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*b*sin(d*x + c)^2 + 12*b*log(abs(sin(d*x + c))) - 24*a*sin(d
*x + c) - 12*(b*sin(d*x + c) + a)/sin(d*x + c))/d