3.1218 \(\int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ \frac{\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 d}-\frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{a b \sin ^4(c+d x)}{2 d}-\frac{2 a b \sin ^2(c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.159245, antiderivative size = 125, 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, 948} \[ \frac{\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 d}-\frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{a b \sin ^4(c+d x)}{2 d}-\frac{2 a b \sin ^2(c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Csc[c + d*x])/d) + (2*a*b*Log[Sin[c + d*x]])/d - ((2*a^2 - b^2)*Sin[c + d*x])/d - (2*a*b*Sin[c + d*x]^2
)/d + ((a^2 - 2*b^2)*Sin[c + d*x]^3)/(3*d) + (a*b*Sin[c + d*x]^4)/(2*d) + (b^2*Sin[c + d*x]^5)/(5*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 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 0.0497468, size = 142, normalized size = 1.14 \[ \frac{a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin (c+d x)}{d}-\frac{a^2 \csc (c+d x)}{d}+\frac{a b \sin ^4(c+d x)}{2 d}-\frac{2 a b \sin ^2(c+d x)}{d}+\frac{2 a b \log (\sin (c+d x))}{d}+\frac{b^2 \sin ^5(c+d x)}{5 d}-\frac{2 b^2 \sin ^3(c+d x)}{3 d}+\frac{b^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 185, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{8\,{a}^{2}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{4\,{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{ab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,{b}^{2}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,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))^2,x)

[Out]

-1/d*a^2/sin(d*x+c)*cos(d*x+c)^6-8/3*a^2*sin(d*x+c)/d-1/d*a^2*sin(d*x+c)*cos(d*x+c)^4-4/3/d*a^2*sin(d*x+c)*cos
(d*x+c)^2+1/2/d*a*b*cos(d*x+c)^4+1/d*a*b*cos(d*x+c)^2+2*a*b*ln(sin(d*x+c))/d+8/15*b^2*sin(d*x+c)/d+1/5/d*sin(d
*x+c)*b^2*cos(d*x+c)^4+4/15/d*sin(d*x+c)*b^2*cos(d*x+c)^2

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Maxima [A]  time = 0.979506, size = 142, normalized size = 1.14 \begin{align*} \frac{6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} - 60 \, a b \sin \left (d x + c\right )^{2} + 10 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{3} + 60 \, a b \log \left (\sin \left (d x + c\right )\right ) - 30 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right ) - \frac{30 \, a^{2}}{\sin \left (d x + c\right )}}{30 \, 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))^2,x, algorithm="maxima")

[Out]

1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 - 60*a*b*sin(d*x + c)^2 + 10*(a^2 - 2*b^2)*sin(d*x + c)^3 +
60*a*b*log(sin(d*x + c)) - 30*(2*a^2 - b^2)*sin(d*x + c) - 30*a^2/sin(d*x + c))/d

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Fricas [A]  time = 1.78321, size = 346, normalized size = 2.77 \begin{align*} -\frac{48 \, b^{2} \cos \left (d x + c\right )^{6} - 16 \,{\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 480 \, a b \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \,{\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 640 \, a^{2} - 128 \, b^{2} - 15 \,{\left (8 \, a b \cos \left (d x + c\right )^{4} + 16 \, a b \cos \left (d x + c\right )^{2} - 11 \, a b\right )} \sin \left (d x + c\right )}{240 \, 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))^2,x, algorithm="fricas")

[Out]

-1/240*(48*b^2*cos(d*x + c)^6 - 16*(5*a^2 - b^2)*cos(d*x + c)^4 - 480*a*b*log(1/2*sin(d*x + c))*sin(d*x + c) -
64*(5*a^2 - b^2)*cos(d*x + c)^2 + 640*a^2 - 128*b^2 - 15*(8*a*b*cos(d*x + c)^4 + 16*a*b*cos(d*x + c)^2 - 11*a
*b)*sin(d*x + c))/(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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24842, size = 171, normalized size = 1.37 \begin{align*} \frac{6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3} - 20 \, b^{2} \sin \left (d x + c\right )^{3} - 60 \, a b \sin \left (d x + c\right )^{2} + 60 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 60 \, a^{2} \sin \left (d x + c\right ) + 30 \, b^{2} \sin \left (d x + c\right ) - \frac{30 \,{\left (2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )}}{30 \, 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))^2,x, algorithm="giac")

[Out]

1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 + 10*a^2*sin(d*x + c)^3 - 20*b^2*sin(d*x + c)^3 - 60*a*b*si
n(d*x + c)^2 + 60*a*b*log(abs(sin(d*x + c))) - 60*a^2*sin(d*x + c) + 30*b^2*sin(d*x + c) - 30*(2*a*b*sin(d*x +
c) + a^2)/sin(d*x + c))/d