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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 165, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 1.04, size = 144, normalized size = 0.87 \[ \frac {-3 a^3 \csc ^4(c+d x)+6 a \left (2 a^2-3 b^2\right ) \csc ^2(c+d x)-12 b \left (b^2-6 a^2\right ) \csc (c+d x)+2 \left (6 b \left (3 a^2-2 b^2\right ) \sin (c+d x)+6 a \left (a^2-6 b^2\right ) \log (\sin (c+d x))+9 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)\right )-12 a^2 b \csc ^3(c+d x)}{12 d} \]

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)

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fricas [A]  time = 0.50, size = 225, normalized size = 1.36 \[ -\frac {18 \, a b^{2} \cos \left (d x + c\right )^{6} - 45 \, a b^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{3} + 9 \, a b^{2} + 6 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left ({\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 6 \, a b^{2} - 2 \, {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (b^{3} \cos \left (d x + c\right )^{6} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 24 \, a^{2} b + 8 \, b^{3} + 12 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

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)

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giac [A]  time = 2.45, size = 185, normalized size = 1.12 \[ \frac {4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 36 \, a^{2} b \sin \left (d x + c\right ) - 24 \, b^{3} \sin \left (d x + c\right ) + 12 \, {\left (a^{3} - 6 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {25 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a b^{2} \sin \left (d x + c\right )^{4} - 72 \, a^{2} b \sin \left (d x + c\right )^{3} + 12 \, b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{3} \sin \left (d x + c\right )^{2} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} b \sin \left (d x + c\right ) + 3 \, a^{3}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

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

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maple [B]  time = 0.23, size = 316, normalized size = 1.92 \[ -\frac {a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{2} b \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}+\frac {3 a^{2} b \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {8 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a^{2} b \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{d}+\frac {4 a^{2} b \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {3 a \,b^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{2 d}-\frac {3 a \,b^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {6 a \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {b^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {8 b^{3} \sin \left (d x +c \right )}{3 d}-\frac {b^{3} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{d}-\frac {4 b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d} \]

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*cos(d*x+c)^2*sin(
d*x+c)-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)

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maxima [A]  time = 0.64, size = 142, normalized size = 0.86 \[ \frac {4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 12 \, {\left (a^{3} - 6 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + 12 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right ) - \frac {3 \, {\left (4 \, a^{2} b \sin \left (d x + c\right ) - 4 \, {\left (6 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} + a^{3} - 2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{2}\right )}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

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

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mupad [B]  time = 6.97, size = 424, normalized size = 2.57 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (6\,a\,b^2-a^3\right )}{d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a\,b^2}{8}-\frac {3\,a^3}{16}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,a\,b^2-a^3\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (3\,a^3+90\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (18\,a\,b^2-\frac {33\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a\,b^2-\frac {9\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {35\,a^3}{4}+78\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (36\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (138\,a^2\,b-72\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (216\,a^2\,b-88\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (316\,a^2\,b-\frac {328\,b^3}{3}\right )-\frac {a^3}{4}-2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {21\,a^2\,b}{8}-\frac {b^3}{2}\right )}{d}-\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2)^2 + 1)*(6*a*b^2 - a^3))/d - (a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (tan(c/2 + (d*x)/2)^2*
((3*a*b^2)/8 - (3*a^3)/16))/d - (log(tan(c/2 + (d*x)/2))*(6*a*b^2 - a^3))/d + (tan(c/2 + (d*x)/2)^8*(90*a*b^2
+ 3*a^3) - tan(c/2 + (d*x)/2)^4*(18*a*b^2 - (33*a^3)/4) - tan(c/2 + (d*x)/2)^2*(6*a*b^2 - (9*a^3)/4) + tan(c/2
+ (d*x)/2)^6*(78*a*b^2 + (35*a^3)/4) + tan(c/2 + (d*x)/2)^3*(36*a^2*b - 8*b^3) + tan(c/2 + (d*x)/2)^9*(138*a^
2*b - 72*b^3) + tan(c/2 + (d*x)/2)^5*(216*a^2*b - 88*b^3) + tan(c/2 + (d*x)/2)^7*(316*a^2*b - (328*b^3)/3) - a
^3/4 - 2*a^2*b*tan(c/2 + (d*x)/2))/(d*(16*tan(c/2 + (d*x)/2)^4 + 48*tan(c/2 + (d*x)/2)^6 + 48*tan(c/2 + (d*x)/
2)^8 + 16*tan(c/2 + (d*x)/2)^10)) + (tan(c/2 + (d*x)/2)*((21*a^2*b)/8 - b^3/2))/d - (a^2*b*tan(c/2 + (d*x)/2)^
3)/(8*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cot ^{5}{\left (c + d x \right )}\, dx \]

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]

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

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