3.1.18 \(\int \cot ^7(c+d x) (a+a \sin (c+d x))^2 \, dx\) [18]

Optimal. Leaf size=132 \[ -\frac {6 a^2 \csc (c+d x)}{d}+\frac {2 a^2 \csc ^3(c+d x)}{d}+\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^6(c+d x)}{6 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x)}{2 d} \]

[Out]

-6*a^2*csc(d*x+c)/d+2*a^2*csc(d*x+c)^3/d+1/2*a^2*csc(d*x+c)^4/d-2/5*a^2*csc(d*x+c)^5/d-1/6*a^2*csc(d*x+c)^6/d+
2*a^2*ln(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d-1/2*a^2*sin(d*x+c)^2/d

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Rubi [A]
time = 0.05, antiderivative size = 132, 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 = {2786, 90} \begin {gather*} -\frac {a^2 \sin ^2(c+d x)}{2 d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^6(c+d x)}{6 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}+\frac {a^2 \csc ^4(c+d x)}{2 d}+\frac {2 a^2 \csc ^3(c+d x)}{d}-\frac {6 a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 90

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 2786

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

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Mathematica [A]
time = 0.15, size = 86, normalized size = 0.65 \begin {gather*} -\frac {a^2 \left (180 \csc (c+d x)-60 \csc ^3(c+d x)-15 \csc ^4(c+d x)+12 \csc ^5(c+d x)+5 \csc ^6(c+d x)-60 \log (\sin (c+d x))+60 \sin (c+d x)+15 \sin ^2(c+d x)\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

-1/30*(a^2*(180*Csc[c + d*x] - 60*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 5*Csc[c + d*x]^6 -
60*Log[Sin[c + d*x]] + 60*Sin[c + d*x] + 15*Sin[c + d*x]^2))/d

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Maple [A]
time = 0.23, size = 228, normalized size = 1.73

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos ^{8}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos ^{8}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}+\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\cos ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos ^{8}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos ^{8}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{3}}-\frac {\cos ^{8}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\left (\cot ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(228\)
default	\(\frac {a^{2} \left (-\frac {\cos ^{8}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos ^{8}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}+\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\cos ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos ^{8}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos ^{8}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{3}}-\frac {\cos ^{8}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\left (\cot ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(228\)
risch	\(-2 i a^{2} x +\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {4 i a^{2} c}{d}-\frac {4 i a^{2} \left (45 \,{\mathrm e}^{11 i \left (d x +c \right )}+30 i {\mathrm e}^{8 i \left (d x +c \right )}-165 \,{\mathrm e}^{9 i \left (d x +c \right )}-20 i {\mathrm e}^{6 i \left (d x +c \right )}+318 \,{\mathrm e}^{7 i \left (d x +c \right )}+30 i {\mathrm e}^{4 i \left (d x +c \right )}-318 \,{\mathrm e}^{5 i \left (d x +c \right )}+165 \,{\mathrm e}^{3 i \left (d x +c \right )}-45 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^7*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-1/4/sin(d*x+c)^4*cos(d*x+c)^8+1/2/sin(d*x+c)^2*cos(d*x+c)^8+1/2*cos(d*x+c)^6+3/4*cos(d*x+c)^4+3/2*c
os(d*x+c)^2+3*ln(sin(d*x+c)))+2*a^2*(-1/5/sin(d*x+c)^5*cos(d*x+c)^8+1/5/sin(d*x+c)^3*cos(d*x+c)^8-1/sin(d*x+c)
*cos(d*x+c)^8-(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+a^2*(-1/6*cot(d*x+c)^6+1/4*cot
(d*x+c)^4-1/2*cot(d*x+c)^2-ln(sin(d*x+c))))

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Maxima [A]
time = 0.28, size = 107, normalized size = 0.81 \begin {gather*} -\frac {15 \, a^{2} \sin \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{2} \sin \left (d x + c\right ) + \frac {180 \, a^{2} \sin \left (d x + c\right )^{5} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(15*a^2*sin(d*x + c)^2 - 60*a^2*log(sin(d*x + c)) + 60*a^2*sin(d*x + c) + (180*a^2*sin(d*x + c)^5 - 60*a
^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 12*a^2*sin(d*x + c) + 5*a^2)/sin(d*x + c)^6)/d

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Fricas [A]
time = 0.40, size = 206, normalized size = 1.56 \begin {gather*} \frac {30 \, a^{2} \cos \left (d x + c\right )^{8} - 105 \, a^{2} \cos \left (d x + c\right )^{6} + 135 \, a^{2} \cos \left (d x + c\right )^{4} - 45 \, a^{2} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 120 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 24 \, {\left (5 \, a^{2} \cos \left (d x + c\right )^{6} - 30 \, a^{2} \cos \left (d x + c\right )^{4} + 40 \, a^{2} \cos \left (d x + c\right )^{2} - 16 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(30*a^2*cos(d*x + c)^8 - 105*a^2*cos(d*x + c)^6 + 135*a^2*cos(d*x + c)^4 - 45*a^2*cos(d*x + c)^2 - 5*a^2
+ 120*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c)) - 24*(5*a
^2*cos(d*x + c)^6 - 30*a^2*cos(d*x + c)^4 + 40*a^2*cos(d*x + c)^2 - 16*a^2)*sin(d*x + c))/(d*cos(d*x + c)^6 -
3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \sin {\left (c + d x \right )} \cot ^{7}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{7}{\left (c + d x \right )}\, dx + \int \cot ^{7}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**7*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*cot(c + d*x)**7, x) + Integral(sin(c + d*x)**2*cot(c + d*x)**7, x) + Integral(co
t(c + d*x)**7, x))

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Giac [A]
time = 4.02, size = 121, normalized size = 0.92 \begin {gather*} -\frac {15 \, a^{2} \sin \left (d x + c\right )^{2} - 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{2} \sin \left (d x + c\right ) + \frac {147 \, a^{2} \sin \left (d x + c\right )^{6} + 180 \, a^{2} \sin \left (d x + c\right )^{5} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 15 \, a^{2} \sin \left (d x + c\right )^{2} + 12 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{\sin \left (d x + c\right )^{6}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(15*a^2*sin(d*x + c)^2 - 60*a^2*log(abs(sin(d*x + c))) + 60*a^2*sin(d*x + c) + (147*a^2*sin(d*x + c)^6 +
180*a^2*sin(d*x + c)^5 - 60*a^2*sin(d*x + c)^3 - 15*a^2*sin(d*x + c)^2 + 12*a^2*sin(d*x + c) + 5*a^2)/sin(d*x
+ c)^6)/d

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Mupad [B]
time = 11.18, size = 392, normalized size = 2.97 \begin {gather*} -\frac {a^2\,\left (24\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-312\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-220\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3864\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-360\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21000\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+3510\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+21000\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-360\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+3864\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-220\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-312\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+3840\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+7680\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+3840\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-3840\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-7680\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3840\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\right )}{1920\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^7*(a + a*sin(c + d*x))^2,x)

[Out]

-(a^2*(24*tan(c/2 + (d*x)/2) - 20*tan(c/2 + (d*x)/2)^2 - 312*tan(c/2 + (d*x)/2)^3 - 220*tan(c/2 + (d*x)/2)^4 +
3864*tan(c/2 + (d*x)/2)^5 - 360*tan(c/2 + (d*x)/2)^6 + 21000*tan(c/2 + (d*x)/2)^7 + 3510*tan(c/2 + (d*x)/2)^8
+ 21000*tan(c/2 + (d*x)/2)^9 - 360*tan(c/2 + (d*x)/2)^10 + 3864*tan(c/2 + (d*x)/2)^11 - 220*tan(c/2 + (d*x)/2
)^12 - 312*tan(c/2 + (d*x)/2)^13 - 20*tan(c/2 + (d*x)/2)^14 + 24*tan(c/2 + (d*x)/2)^15 + 5*tan(c/2 + (d*x)/2)^
16 + 3840*tan(c/2 + (d*x)/2)^6*log(tan(c/2 + (d*x)/2)^2 + 1) + 7680*tan(c/2 + (d*x)/2)^8*log(tan(c/2 + (d*x)/2
)^2 + 1) + 3840*tan(c/2 + (d*x)/2)^10*log(tan(c/2 + (d*x)/2)^2 + 1) - 3840*log(tan(c/2 + (d*x)/2))*tan(c/2 + (
d*x)/2)^6 - 7680*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^8 - 3840*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2
)^10 + 5))/(1920*d*tan(c/2 + (d*x)/2)^6*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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