3.531 \(\int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=148 \[ \frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}-\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {4 a^4 \csc (c+d x)}{d}-\frac {10 a^4 \log (\sin (c+d x))}{d} \]

[Out]

4*a^4*csc(d*x+c)/d-2*a^4*csc(d*x+c)^2/d-4/3*a^4*csc(d*x+c)^3/d-1/4*a^4*csc(d*x+c)^4/d-10*a^4*ln(sin(d*x+c))/d-
4*a^4*sin(d*x+c)/d+2*a^4*sin(d*x+c)^2/d+4/3*a^4*sin(d*x+c)^3/d+1/4*a^4*sin(d*x+c)^4/d

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Rubi [A]  time = 0.08, antiderivative size = 148, 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 = {2707, 88} \[ \frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}-\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {4 a^4 \csc (c+d x)}{d}-\frac {10 a^4 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^4,x]

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2707

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 - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+a \sin (c+d x))^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2 (a+x)^6}{x^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-4 a^3+\frac {a^8}{x^5}+\frac {4 a^7}{x^4}+\frac {4 a^6}{x^3}-\frac {4 a^5}{x^2}-\frac {10 a^4}{x}+4 a^2 x+4 a x^2+x^3\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {4 a^4 \csc (c+d x)}{d}-\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{4 d}-\frac {10 a^4 \log (\sin (c+d x))}{d}-\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 96, normalized size = 0.65 \[ \frac {a^4 \left (3 \sin ^4(c+d x)+16 \sin ^3(c+d x)+24 \sin ^2(c+d x)-48 \sin (c+d x)-3 \csc ^4(c+d x)-16 \csc ^3(c+d x)-24 \csc ^2(c+d x)+48 \csc (c+d x)-120 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(48*Csc[c + d*x] - 24*Csc[c + d*x]^2 - 16*Csc[c + d*x]^3 - 3*Csc[c + d*x]^4 - 120*Log[Sin[c + d*x]] - 48*
Sin[c + d*x] + 24*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/(12*d)

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fricas [A]  time = 0.97, size = 144, normalized size = 0.97 \[ \frac {24 \, a^{4} \cos \left (d x + c\right )^{8} - 128 \, a^{4} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 288 \, a^{4} \cos \left (d x + c\right )^{6} + 615 \, a^{4} \cos \left (d x + c\right )^{4} - 270 \, a^{4} \cos \left (d x + c\right )^{2} - 105 \, a^{4} - 960 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/96*(24*a^4*cos(d*x + c)^8 - 128*a^4*cos(d*x + c)^6*sin(d*x + c) - 288*a^4*cos(d*x + c)^6 + 615*a^4*cos(d*x +
c)^4 - 270*a^4*cos(d*x + c)^2 - 105*a^4 - 960*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(1/2*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 = 0.46, size = 134, normalized size = 0.91 \[ \frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 24 \, a^{4} \sin \left (d x + c\right )^{2} - 120 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 48 \, a^{4} \sin \left (d x + c\right ) + \frac {250 \, a^{4} \sin \left (d x + c\right )^{4} + 48 \, a^{4} \sin \left (d x + c\right )^{3} - 24 \, a^{4} \sin \left (d x + c\right )^{2} - 16 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 24*a^4*sin(d*x + c)^2 - 120*a^4*log(abs(sin(d*x + c))) -
48*a^4*sin(d*x + c) + (250*a^4*sin(d*x + c)^4 + 48*a^4*sin(d*x + c)^3 - 24*a^4*sin(d*x + c)^2 - 16*a^4*sin(d*x
+ c) - 3*a^4)/sin(d*x + c)^4)/d

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maple [A]  time = 0.44, size = 129, normalized size = 0.87 \[ -\frac {11 a^{4} \left (\cos ^{4}\left (d x +c \right )\right )}{4 d}-\frac {11 a^{4} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}-\frac {10 a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {3 a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {4 a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}-\frac {a^{4} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x)

[Out]

-11/4/d*a^4*cos(d*x+c)^4-11/2/d*a^4*cos(d*x+c)^2-10*a^4*ln(sin(d*x+c))/d-3/d*a^4/sin(d*x+c)^2*cos(d*x+c)^6-4/3
/d*a^4/sin(d*x+c)^3*cos(d*x+c)^6-1/4/d*a^4*cot(d*x+c)^4+1/2/d*a^4*cot(d*x+c)^2

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maxima [A]  time = 0.72, size = 120, normalized size = 0.81 \[ \frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 24 \, a^{4} \sin \left (d x + c\right )^{2} - 120 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) - 48 \, a^{4} \sin \left (d x + c\right ) + \frac {48 \, a^{4} \sin \left (d x + c\right )^{3} - 24 \, a^{4} \sin \left (d x + c\right )^{2} - 16 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 24*a^4*sin(d*x + c)^2 - 120*a^4*log(sin(d*x + c)) - 48*a^
4*sin(d*x + c) + (48*a^4*sin(d*x + c)^3 - 24*a^4*sin(d*x + c)^2 - 16*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c)^4)
/d

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mupad [B]  time = 9.07, size = 368, normalized size = 2.49 \[ \frac {3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {10\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {104\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-119\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+120\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {1135\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+80\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-73\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+48\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {75\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {40\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^4}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {9\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}+\frac {10\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^4)/sin(c + d*x)^5,x)

[Out]

(3*a^4*tan(c/2 + (d*x)/2))/(2*d) - (a^4*tan(c/2 + (d*x)/2)^3)/(6*d) - (a^4*tan(c/2 + (d*x)/2)^4)/(64*d) - (10*
a^4*log(tan(c/2 + (d*x)/2)))/d - (10*a^4*tan(c/2 + (d*x)/2)^2 - (40*a^4*tan(c/2 + (d*x)/2)^3)/3 + (75*a^4*tan(
c/2 + (d*x)/2)^4)/2 + 48*a^4*tan(c/2 + (d*x)/2)^5 - 73*a^4*tan(c/2 + (d*x)/2)^6 + 80*a^4*tan(c/2 + (d*x)/2)^7
- (1135*a^4*tan(c/2 + (d*x)/2)^8)/4 + 120*a^4*tan(c/2 + (d*x)/2)^9 - 119*a^4*tan(c/2 + (d*x)/2)^10 + 104*a^4*t
an(c/2 + (d*x)/2)^11 + a^4/4 + (8*a^4*tan(c/2 + (d*x)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 64*tan(c/2 + (d*x)/
2)^6 + 96*tan(c/2 + (d*x)/2)^8 + 64*tan(c/2 + (d*x)/2)^10 + 16*tan(c/2 + (d*x)/2)^12)) - (9*a^4*tan(c/2 + (d*x
)/2)^2)/(16*d) + (10*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**5*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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