3.164 \(\int \cot ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=165 \[ \frac{b \left (3 a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac{a \left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 d}+\frac{b \left (6 a^2-b^2\right ) \csc (c+d x)}{d}+\frac{a \left (a^2-6 b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{a^3 \csc ^4(c+d x)}{4 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14103, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 948} \[ \frac{b \left (3 a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac{a \left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 d}+\frac{b \left (6 a^2-b^2\right ) \csc (c+d x)}{d}+\frac{a \left (a^2-6 b^2\right ) \log (\sin (c+d x))}{d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{a^3 \csc ^4(c+d x)}{4 d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 2721

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

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 \cot ^5(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+x)^3 \left (b^2-x^2\right )^2}{x^5} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (3 a^2 \left (1-\frac{2 b^2}{3 a^2}\right )+\frac{a^3 b^4}{x^5}+\frac{3 a^2 b^4}{x^4}+\frac{-2 a^3 b^2+3 a b^4}{x^3}+\frac{-6 a^2 b^2+b^4}{x^2}+\frac{a^3-6 a b^2}{x}+3 a x+x^2\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b \left (6 a^2-b^2\right ) \csc (c+d x)}{d}+\frac{a \left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 d}-\frac{a^2 b \csc ^3(c+d x)}{d}-\frac{a^3 \csc ^4(c+d x)}{4 d}+\frac{a \left (a^2-6 b^2\right ) \log (\sin (c+d x))}{d}+\frac{b \left (3 a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac{3 a b^2 \sin ^2(c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.16383, size = 144, normalized size = 0.87 \[ \frac{6 a \left (2 a^2-3 b^2\right ) \csc ^2(c+d x)-12 b \left (b^2-6 a^2\right ) \csc (c+d x)+2 \left (6 b \left (3 a^2-2 b^2\right ) \sin (c+d x)+6 a \left (a^2-6 b^2\right ) \log (\sin (c+d x))+9 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)\right )-12 a^2 b \csc ^3(c+d x)-3 a^3 \csc ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

(-12*b*(-6*a^2 + b^2)*Csc[c + d*x] + 6*a*(2*a^2 - 3*b^2)*Csc[c + d*x]^2 - 12*a^2*b*Csc[c + d*x]^3 - 3*a^3*Csc[
c + d*x]^4 + 2*(6*a*(a^2 - 6*b^2)*Log[Sin[c + d*x]] + 6*b*(3*a^2 - 2*b^2)*Sin[c + d*x] + 9*a*b^2*Sin[c + d*x]^
2 + 2*b^3*Sin[c + d*x]^3))/(12*d)

________________________________________________________________________________________

Maple [B]  time = 0.062, size = 316, normalized size = 1.9 \begin{align*} -{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+3\,{\frac{{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}+8\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}+4\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{3\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-3\,{\frac{a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-6\,{\frac{a{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{8\,{b}^{3}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{{b}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{4\,{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*sin(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.35412, size = 192, normalized size = 1.16 \begin{align*} \frac{4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 12 \,{\left (a^{3} - 6 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + 12 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right ) - \frac{3 \,{\left (4 \, a^{2} b \sin \left (d x + c\right ) - 4 \,{\left (6 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} + a^{3} - 2 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{2}\right )}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(4*b^3*sin(d*x + c)^3 + 18*a*b^2*sin(d*x + c)^2 + 12*(a^3 - 6*a*b^2)*log(sin(d*x + c)) + 12*(3*a^2*b - 2*
b^3)*sin(d*x + c) - 3*(4*a^2*b*sin(d*x + c) - 4*(6*a^2*b - b^3)*sin(d*x + c)^3 + a^3 - 2*(2*a^3 - 3*a*b^2)*sin
(d*x + c)^2)/sin(d*x + c)^4)/d

________________________________________________________________________________________

Fricas [A]  time = 1.60026, size = 536, normalized size = 3.25 \begin{align*} -\frac{18 \, a b^{2} \cos \left (d x + c\right )^{6} - 45 \, a b^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{3} + 9 \, a b^{2} + 6 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left ({\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 6 \, a b^{2} - 2 \,{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 4 \,{\left (b^{3} \cos \left (d x + c\right )^{6} - 3 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 24 \, a^{2} b + 8 \, b^{3} + 12 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(18*a*b^2*cos(d*x + c)^6 - 45*a*b^2*cos(d*x + c)^4 - 9*a^3 + 9*a*b^2 + 6*(2*a^3 + 3*a*b^2)*cos(d*x + c)^
2 - 12*((a^3 - 6*a*b^2)*cos(d*x + c)^4 + a^3 - 6*a*b^2 - 2*(a^3 - 6*a*b^2)*cos(d*x + c)^2)*log(1/2*sin(d*x + c
)) + 4*(b^3*cos(d*x + c)^6 - 3*(3*a^2*b - b^3)*cos(d*x + c)^4 - 24*a^2*b + 8*b^3 + 12*(3*a^2*b - b^3)*cos(d*x
+ c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

________________________________________________________________________________________

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(cot(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.1406, size = 250, normalized size = 1.52 \begin{align*} \frac{4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 36 \, a^{2} b \sin \left (d x + c\right ) - 24 \, b^{3} \sin \left (d x + c\right ) + 12 \,{\left (a^{3} - 6 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{25 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a b^{2} \sin \left (d x + c\right )^{4} - 72 \, a^{2} b \sin \left (d x + c\right )^{3} + 12 \, b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{3} \sin \left (d x + c\right )^{2} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} b \sin \left (d x + c\right ) + 3 \, a^{3}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/12*(4*b^3*sin(d*x + c)^3 + 18*a*b^2*sin(d*x + c)^2 + 36*a^2*b*sin(d*x + c) - 24*b^3*sin(d*x + c) + 12*(a^3 -
6*a*b^2)*log(abs(sin(d*x + c))) - (25*a^3*sin(d*x + c)^4 - 150*a*b^2*sin(d*x + c)^4 - 72*a^2*b*sin(d*x + c)^3
+ 12*b^3*sin(d*x + c)^3 - 12*a^3*sin(d*x + c)^2 + 18*a*b^2*sin(d*x + c)^2 + 12*a^2*b*sin(d*x + c) + 3*a^3)/si
n(d*x + c)^4)/d