3.1217 \(\int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=130 \[ \frac{\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}-\frac{\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}+\frac{a^2 \log (\sin (c+d x))}{d}+\frac{2 a b \sin ^5(c+d x)}{5 d}-\frac{4 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d}+\frac{b^2 \sin ^6(c+d x)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.133519, antiderivative size = 130, 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, 948} \[ \frac{\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}-\frac{\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}+\frac{a^2 \log (\sin (c+d x))}{d}+\frac{2 a b \sin ^5(c+d x)}{5 d}-\frac{4 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d}+\frac{b^2 \sin ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(a^2*Log[Sin[c + d*x]])/d + (2*a*b*Sin[c + d*x])/d - ((2*a^2 - b^2)*Sin[c + d*x]^2)/(2*d) - (4*a*b*Sin[c + d*x
]^3)/(3*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^4)/(4*d) + (2*a*b*Sin[c + d*x]^5)/(5*d) + (b^2*Sin[c + d*x]^6)/(6*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 ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b (a+x)^2 \left (b^2-x^2\right )^2}{x} \, 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} \, dx,x,b \sin (c+d x)\right )}{b^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a b^4+\frac{a^2 b^4}{x}-b^2 \left (2 a^2-b^2\right ) x-4 a b^2 x^2+\left (a^2-2 b^2\right ) x^3+2 a x^4+x^5\right ) \, dx,x,b \sin (c+d x)\right )}{b^4 d}\\ &=\frac{a^2 \log (\sin (c+d x))}{d}+\frac{2 a b \sin (c+d x)}{d}-\frac{\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}-\frac{4 a b \sin ^3(c+d x)}{3 d}+\frac{\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}+\frac{2 a b \sin ^5(c+d x)}{5 d}+\frac{b^2 \sin ^6(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.156253, size = 105, normalized size = 0.81 \[ \frac{15 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+30 \left (b^2-2 a^2\right ) \sin ^2(c+d x)+60 a^2 \log (\sin (c+d x))+24 a b \sin ^5(c+d x)-80 a b \sin ^3(c+d x)+120 a b \sin (c+d x)+10 b^2 \sin ^6(c+d x)}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(60*a^2*Log[Sin[c + d*x]] + 120*a*b*Sin[c + d*x] + 30*(-2*a^2 + b^2)*Sin[c + d*x]^2 - 80*a*b*Sin[c + d*x]^3 +
15*(a^2 - 2*b^2)*Sin[c + d*x]^4 + 24*a*b*Sin[c + d*x]^5 + 10*b^2*Sin[c + d*x]^6)/(60*d)

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Maple [A]  time = 0.08, size = 119, normalized size = 0.9 \begin{align*}{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{16\,ab\sin \left ( dx+c \right ) }{15\,d}}+{\frac{2\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{8\,ab\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}{b}^{2}}{6\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*a^2*cos(d*x+c)^4+1/2/d*a^2*cos(d*x+c)^2+a^2*ln(sin(d*x+c))/d+16/15*a*b*sin(d*x+c)/d+2/5/d*sin(d*x+c)*a*b
*cos(d*x+c)^4+8/15/d*sin(d*x+c)*a*b*cos(d*x+c)^2-1/6/d*cos(d*x+c)^6*b^2

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Maxima [A]  time = 0.966875, size = 142, normalized size = 1.09 \begin{align*} \frac{10 \, b^{2} \sin \left (d x + c\right )^{6} + 24 \, a b \sin \left (d x + c\right )^{5} - 80 \, a b \sin \left (d x + c\right )^{3} + 15 \,{\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 120 \, a b \sin \left (d x + c\right ) - 30 \,{\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*b^2*sin(d*x + c)^6 + 24*a*b*sin(d*x + c)^5 - 80*a*b*sin(d*x + c)^3 + 15*(a^2 - 2*b^2)*sin(d*x + c)^4
+ 60*a^2*log(sin(d*x + c)) + 120*a*b*sin(d*x + c) - 30*(2*a^2 - b^2)*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.82381, size = 247, normalized size = 1.9 \begin{align*} -\frac{10 \, b^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (3 \, a b \cos \left (d x + c\right )^{4} + 4 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(10*b^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c)^2 - 60*a^2*log(1/2*sin(d*x + c)) -
8*(3*a*b*cos(d*x + c)^4 + 4*a*b*cos(d*x + c)^2 + 8*a*b)*sin(d*x + c))/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)**5*csc(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27565, size = 159, normalized size = 1.22 \begin{align*} \frac{10 \, b^{2} \sin \left (d x + c\right )^{6} + 24 \, a b \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 30 \, b^{2} \sin \left (d x + c\right )^{4} - 80 \, a b \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 30 \, b^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 120 \, a b \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*b^2*sin(d*x + c)^6 + 24*a*b*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 30*b^2*sin(d*x + c)^4 - 80*a*b*s
in(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 30*b^2*sin(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) + 120*a*b*sin(d*
x + c))/d