3.169 \(\int \cot ^6(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=291 \[ \frac{45 a^2 b \cos (c+d x)}{8 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac{15 a^2 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{45 a^2 b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}-a^3 x-\frac{5 a b^2 \cot ^3(c+d x)}{2 d}+\frac{15 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac{15}{2} a b^2 x-\frac{5 b^3 \cos ^3(c+d x)}{6 d}-\frac{5 b^3 \cos (c+d x)}{2 d}-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{5 b^3 \tanh ^{-1}(\cos (c+d x))}{2 d} \]

[Out]

-(a^3*x) + (15*a*b^2*x)/2 - (45*a^2*b*ArcTanh[Cos[c + d*x]])/(8*d) + (5*b^3*ArcTanh[Cos[c + d*x]])/(2*d) + (45
*a^2*b*Cos[c + d*x])/(8*d) - (5*b^3*Cos[c + d*x])/(2*d) - (5*b^3*Cos[c + d*x]^3)/(6*d) - (a^3*Cot[c + d*x])/d
+ (15*a*b^2*Cot[c + d*x])/(2*d) + (15*a^2*b*Cos[c + d*x]*Cot[c + d*x]^2)/(8*d) - (b^3*Cos[c + d*x]^3*Cot[c + d
*x]^2)/(2*d) + (a^3*Cot[c + d*x]^3)/(3*d) - (5*a*b^2*Cot[c + d*x]^3)/(2*d) + (3*a*b^2*Cos[c + d*x]^2*Cot[c + d
*x]^3)/(2*d) - (3*a^2*b*Cos[c + d*x]*Cot[c + d*x]^4)/(4*d) - (a^3*Cot[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.2309, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {2722, 2592, 288, 302, 206, 2591, 203, 321, 3473, 8} \[ \frac{45 a^2 b \cos (c+d x)}{8 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}+\frac{15 a^2 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{45 a^2 b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}-a^3 x-\frac{5 a b^2 \cot ^3(c+d x)}{2 d}+\frac{15 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac{15}{2} a b^2 x-\frac{5 b^3 \cos ^3(c+d x)}{6 d}-\frac{5 b^3 \cos (c+d x)}{2 d}-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{5 b^3 \tanh ^{-1}(\cos (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Sin[c + d*x])^3,x]

[Out]

-(a^3*x) + (15*a*b^2*x)/2 - (45*a^2*b*ArcTanh[Cos[c + d*x]])/(8*d) + (5*b^3*ArcTanh[Cos[c + d*x]])/(2*d) + (45
*a^2*b*Cos[c + d*x])/(8*d) - (5*b^3*Cos[c + d*x])/(2*d) - (5*b^3*Cos[c + d*x]^3)/(6*d) - (a^3*Cot[c + d*x])/d
+ (15*a*b^2*Cot[c + d*x])/(2*d) + (15*a^2*b*Cos[c + d*x]*Cot[c + d*x]^2)/(8*d) - (b^3*Cos[c + d*x]^3*Cot[c + d
*x]^2)/(2*d) + (a^3*Cot[c + d*x]^3)/(3*d) - (5*a*b^2*Cot[c + d*x]^3)/(2*d) + (3*a*b^2*Cos[c + d*x]^2*Cot[c + d
*x]^3)/(2*d) - (3*a^2*b*Cos[c + d*x]*Cot[c + d*x]^4)/(4*d) - (a^3*Cot[c + d*x]^5)/(5*d)

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cot ^6(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \left (b^3 \cos ^3(c+d x) \cot ^3(c+d x)+3 a b^2 \cos ^2(c+d x) \cot ^4(c+d x)+3 a^2 b \cos (c+d x) \cot ^5(c+d x)+a^3 \cot ^6(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos (c+d x) \cot ^5(c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^2(c+d x) \cot ^4(c+d x) \, dx+b^3 \int \cos ^3(c+d x) \cot ^3(c+d x) \, dx\\ &=-\frac{a^3 \cot ^5(c+d x)}{5 d}-a^3 \int \cot ^4(c+d x) \, dx-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}+a^3 \int \cot ^2(c+d x) \, dx+\frac{\left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{4 d}-\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac{a^3 \cot (c+d x)}{d}+\frac{15 a^2 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-a^3 \int 1 \, dx-\frac{\left (45 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 d}-\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{2 d}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-a^3 x+\frac{45 a^2 b \cos (c+d x)}{8 d}-\frac{5 b^3 \cos (c+d x)}{2 d}-\frac{5 b^3 \cos ^3(c+d x)}{6 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{15 a b^2 \cot (c+d x)}{2 d}+\frac{15 a^2 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{5 a b^2 \cot ^3(c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{\left (45 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 d}-\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-a^3 x+\frac{15}{2} a b^2 x-\frac{45 a^2 b \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{5 b^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac{45 a^2 b \cos (c+d x)}{8 d}-\frac{5 b^3 \cos (c+d x)}{2 d}-\frac{5 b^3 \cos ^3(c+d x)}{6 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{15 a b^2 \cot (c+d x)}{2 d}+\frac{15 a^2 b \cos (c+d x) \cot ^2(c+d x)}{8 d}-\frac{b^3 \cos ^3(c+d x) \cot ^2(c+d x)}{2 d}+\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{5 a b^2 \cot ^3(c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^4(c+d x)}{4 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 2.57765, size = 346, normalized size = 1.19 \[ \frac{-600 a \left (2 a^2-15 b^2\right ) (c+d x) \csc ^4(c+d x)+1200 b \left (4 b^2-9 a^2\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+\csc ^5(c+d x) \left (5 \left (40 a^3-489 a b^2\right ) \cos (3 (c+d x))+\left (1065 a b^2-184 a^3\right ) \cos (5 (c+d x))+5 \left (60 a \left (2 a^2-15 b^2\right ) (c+d x) \sin (3 (c+d x))-306 a^2 b \sin (4 (c+d x))+36 a^2 b \sin (6 (c+d x))-24 a^3 c \sin (5 (c+d x))-24 a^3 d x \sin (5 (c+d x))+180 a b^2 c \sin (5 (c+d x))+180 a b^2 d x \sin (5 (c+d x))-9 a b^2 \cos (7 (c+d x))+122 b^3 \sin (4 (c+d x))-22 b^3 \sin (6 (c+d x))-b^3 \sin (8 (c+d x))\right )\right )+5 \cot (c+d x) \csc ^4(c+d x) \left (12 b \left (60 a^2-29 b^2\right ) \sin (c+d x)-80 a^3+285 a b^2\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Sin[c + d*x])^3,x]

[Out]

(-600*a*(2*a^2 - 15*b^2)*(c + d*x)*Csc[c + d*x]^4 + 1200*b*(-9*a^2 + 4*b^2)*(Log[Cos[(c + d*x)/2]] - Log[Sin[(
c + d*x)/2]]) + 5*Cot[c + d*x]*Csc[c + d*x]^4*(-80*a^3 + 285*a*b^2 + 12*b*(60*a^2 - 29*b^2)*Sin[c + d*x]) + Cs
c[c + d*x]^5*(5*(40*a^3 - 489*a*b^2)*Cos[3*(c + d*x)] + (-184*a^3 + 1065*a*b^2)*Cos[5*(c + d*x)] + 5*(-9*a*b^2
*Cos[7*(c + d*x)] + 60*a*(2*a^2 - 15*b^2)*(c + d*x)*Sin[3*(c + d*x)] - 306*a^2*b*Sin[4*(c + d*x)] + 122*b^3*Si
n[4*(c + d*x)] - 24*a^3*c*Sin[5*(c + d*x)] + 180*a*b^2*c*Sin[5*(c + d*x)] - 24*a^3*d*x*Sin[5*(c + d*x)] + 180*
a*b^2*d*x*Sin[5*(c + d*x)] + 36*a^2*b*Sin[6*(c + d*x)] - 22*b^3*Sin[6*(c + d*x)] - b^3*Sin[8*(c + d*x)])))/(19
20*d)

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Maple [A]  time = 0.063, size = 415, normalized size = 1.4 \begin{align*} -{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-{a}^{3}x-{\frac{{a}^{3}c}{d}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{9\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{9\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d}}+{\frac{15\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{45\,{a}^{2}b\cos \left ( dx+c \right ) }{8\,d}}+{\frac{45\,{a}^{2}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+4\,{\frac{a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{d\sin \left ( dx+c \right ) }}+4\,{\frac{a{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d}}+5\,{\frac{a{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{15\,a{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{15\,a{b}^{2}x}{2}}+{\frac{15\,a{b}^{2}c}{2\,d}}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d}}-{\frac{5\,{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{5\,{b}^{3}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{b}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*sin(d*x+c))^3,x)

[Out]

-1/5*a^3*cot(d*x+c)^5/d+1/3*a^3*cot(d*x+c)^3/d-a^3*cot(d*x+c)/d-a^3*x-1/d*a^3*c-3/4/d*a^2*b/sin(d*x+c)^4*cos(d
*x+c)^7+9/8/d*a^2*b/sin(d*x+c)^2*cos(d*x+c)^7+9/8/d*a^2*b*cos(d*x+c)^5+15/8/d*a^2*b*cos(d*x+c)^3+45/8*a^2*b*co
s(d*x+c)/d+45/8/d*a^2*b*ln(csc(d*x+c)-cot(d*x+c))-1/d*a*b^2/sin(d*x+c)^3*cos(d*x+c)^7+4/d*a*b^2/sin(d*x+c)*cos
(d*x+c)^7+4/d*a*b^2*sin(d*x+c)*cos(d*x+c)^5+5/d*a*b^2*sin(d*x+c)*cos(d*x+c)^3+15/2*a*b^2*cos(d*x+c)*sin(d*x+c)
/d+15/2*a*b^2*x+15/2/d*a*b^2*c-1/2/d*b^3/sin(d*x+c)^2*cos(d*x+c)^7-1/2/d*b^3*cos(d*x+c)^5-5/6*b^3*cos(d*x+c)^3
/d-5/2*b^3*cos(d*x+c)/d-5/2/d*b^3*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 2.64117, size = 340, normalized size = 1.17 \begin{align*} -\frac{16 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{3} - 120 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a b^{2} + 20 \,{\left (4 \, \cos \left (d x + c\right )^{3} - \frac{6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{3} + 45 \, a^{2} b{\left (\frac{2 \,{\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/240*(16*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a^3 - 120*(15*d*x + 15*
c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^5 + tan(d*x + c)^3))*a*b^2 + 20*(4*cos(d*x + c)^
3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1
))*b^3 + 45*a^2*b*(2*(9*cos(d*x + c)^3 - 7*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x
+ c) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.83246, size = 1014, normalized size = 3.48 \begin{align*} -\frac{360 \, a b^{2} \cos \left (d x + c\right )^{7} + 184 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 280 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 75 \,{\left ({\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 9 \, a^{2} b - 4 \, b^{3} - 2 \,{\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 75 \,{\left ({\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 9 \, a^{2} b - 4 \, b^{3} - 2 \,{\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 120 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right ) + 10 \,{\left (8 \, b^{3} \cos \left (d x + c\right )^{7} + 12 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{4} - 8 \,{\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} - 24 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 25 \,{\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 12 \,{\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x - 15 \,{\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/240*(360*a*b^2*cos(d*x + c)^7 + 184*(2*a^3 - 15*a*b^2)*cos(d*x + c)^5 - 280*(2*a^3 - 15*a*b^2)*cos(d*x + c)
^3 + 75*((9*a^2*b - 4*b^3)*cos(d*x + c)^4 + 9*a^2*b - 4*b^3 - 2*(9*a^2*b - 4*b^3)*cos(d*x + c)^2)*log(1/2*cos(
d*x + c) + 1/2)*sin(d*x + c) - 75*((9*a^2*b - 4*b^3)*cos(d*x + c)^4 + 9*a^2*b - 4*b^3 - 2*(9*a^2*b - 4*b^3)*co
s(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 120*(2*a^3 - 15*a*b^2)*cos(d*x + c) + 10*(8*b^3*cos(
d*x + c)^7 + 12*(2*a^3 - 15*a*b^2)*d*x*cos(d*x + c)^4 - 8*(9*a^2*b - 4*b^3)*cos(d*x + c)^5 - 24*(2*a^3 - 15*a*
b^2)*d*x*cos(d*x + c)^2 + 25*(9*a^2*b - 4*b^3)*cos(d*x + c)^3 + 12*(2*a^3 - 15*a*b^2)*d*x - 15*(9*a^2*b - 4*b^
3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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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(cot(d*x+c)**6*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 2.01324, size = 636, normalized size = 2.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(6*a^3*tan(1/2*d*x + 1/2*c)^5 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 70*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*a*
b^2*tan(1/2*d*x + 1/2*c)^3 - 720*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 120*b^3*tan(1/2*d*x + 1/2*c)^2 + 660*a^3*tan(1
/2*d*x + 1/2*c) - 3240*a*b^2*tan(1/2*d*x + 1/2*c) - 480*(2*a^3 - 15*a*b^2)*(d*x + c) + 600*(9*a^2*b - 4*b^3)*l
og(abs(tan(1/2*d*x + 1/2*c))) - 320*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 18*b^3
*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 24*b^3*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*
x + 1/2*c) - 18*a^2*b + 14*b^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 - (12330*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 5480*b^
3*tan(1/2*d*x + 1/2*c)^5 + 660*a^3*tan(1/2*d*x + 1/2*c)^4 - 3240*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 720*a^2*b*tan(
1/2*d*x + 1/2*c)^3 + 120*b^3*tan(1/2*d*x + 1/2*c)^3 - 70*a^3*tan(1/2*d*x + 1/2*c)^2 + 120*a*b^2*tan(1/2*d*x +
1/2*c)^2 + 45*a^2*b*tan(1/2*d*x + 1/2*c) + 6*a^3)/tan(1/2*d*x + 1/2*c)^5)/d