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

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

[Out]

-a^3*x+15/2*a*b^2*x-45/8*a^2*b*arctanh(cos(d*x+c))/d+5/2*b^3*arctanh(cos(d*x+c))/d+45/8*a^2*b*cos(d*x+c)/d-5/2
*b^3*cos(d*x+c)/d-5/6*b^3*cos(d*x+c)^3/d-a^3*cot(d*x+c)/d+15/2*a*b^2*cot(d*x+c)/d+15/8*a^2*b*cos(d*x+c)*cot(d*
x+c)^2/d-1/2*b^3*cos(d*x+c)^3*cot(d*x+c)^2/d+1/3*a^3*cot(d*x+c)^3/d-5/2*a*b^2*cot(d*x+c)^3/d+3/2*a*b^2*cos(d*x
+c)^2*cot(d*x+c)^3/d-3/4*a^2*b*cos(d*x+c)*cot(d*x+c)^4/d-1/5*a^3*cot(d*x+c)^5/d

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Rubi [A]  time = 0.23, antiderivative size = 291, normalized size of antiderivative = 1.00, 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 8

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

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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.57, size = 346, normalized size = 1.19 \[ \frac {-600 a \left (2 a^2-15 b^2\right ) (c+d x) \csc ^4(c+d x)+1200 b \left (4 b^2-9 a^2\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+5 \cot (c+d x) \csc ^4(c+d x) \left (-80 a^3+12 b \left (60 a^2-29 b^2\right ) \sin (c+d x)+285 a b^2\right )+\csc ^5(c+d x) \left (5 \left (40 a^3-489 a b^2\right ) \cos (3 (c+d x))+\left (1065 a b^2-184 a^3\right ) \cos (5 (c+d x))+5 \left (-24 a^3 c \sin (5 (c+d x))-24 a^3 d x \sin (5 (c+d x))+60 a \left (2 a^2-15 b^2\right ) (c+d x) \sin (3 (c+d x))-306 a^2 b \sin (4 (c+d x))+36 a^2 b \sin (6 (c+d x))+180 a b^2 c \sin (5 (c+d x))+180 a b^2 d x \sin (5 (c+d x))-9 a b^2 \cos (7 (c+d x))+122 b^3 \sin (4 (c+d x))-22 b^3 \sin (6 (c+d x))-b^3 \sin (8 (c+d x))\right )\right )}{1920 d} \]

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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fricas [A]  time = 0.57, size = 412, normalized size = 1.42 \[ -\frac {360 \, a b^{2} \cos \left (d x + c\right )^{7} + 184 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 280 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 75 \, {\left ({\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 9 \, a^{2} b - 4 \, b^{3} - 2 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 75 \, {\left ({\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 9 \, a^{2} b - 4 \, b^{3} - 2 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 120 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} \cos \left (d x + c\right ) + 10 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{7} + 12 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{4} - 8 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} - 24 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 25 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} d x - 15 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 1.05, size = 471, normalized size = 1.62 \[ \frac {6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 660 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3240 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 480 \, {\left (2 \, a^{3} - 15 \, a b^{2}\right )} {\left (d x + c\right )} + 600 \, {\left (9 \, a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {320 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b + 14 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} - \frac {12330 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5480 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3240 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

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

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maple [A]  time = 0.26, size = 415, normalized size = 1.43 \[ -\frac {a^{3} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{3} \cot \left (d x +c \right )}{d}-a^{3} x -\frac {a^{3} c}{d}-\frac {3 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {9 a^{2} b \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {9 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}+\frac {15 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {45 a^{2} b \cos \left (d x +c \right )}{8 d}+\frac {45 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a \,b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}+\frac {4 a \,b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {4 a \,b^{2} \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{d}+\frac {5 a \,b^{2} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {15 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {15 a \,b^{2} x}{2}+\frac {15 a \,b^{2} c}{2 d}-\frac {b^{3} \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {b^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d}-\frac {5 b^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {5 b^{3} \cos \left (d x +c \right )}{2 d}-\frac {5 b^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

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 = 1.89, size = 252, normalized size = 0.87 \[ -\frac {16 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{3} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a b^{2} + 20 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{3} + 45 \, a^{2} b {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

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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mupad [B]  time = 7.06, size = 507, normalized size = 1.74 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (12\,a\,b^2-22\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a\,b^2-\frac {26\,a^3}{15}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (96\,a\,b^2-\frac {78\,a^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (320\,a\,b^2-\frac {191\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (408\,a\,b^2-\frac {296\,a^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {39\,a^2\,b}{2}-4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (216\,a^2\,b-196\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {519\,a^2\,b}{2}-\frac {484\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {909\,a^2\,b}{2}-268\,b^3\right )-\frac {a^3}{5}-\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a\,b^2}{8}-\frac {7\,a^3}{96}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a^2\,b}{4}-\frac {b^3}{8}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {45\,a^2\,b}{8}-\frac {5\,b^3}{2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,\left (\frac {a\,b^2\,15{}\mathrm {i}}{2}-a^3\,1{}\mathrm {i}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {27\,a\,b^2}{8}-\frac {11\,a^3}{16}\right )}{d}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (2\,a^2-15\,b^2\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (tan(c/2 + (d*x)/2)^10*(12*a*b^2 - 22*a^3) - tan(c/2 + (d*x)/2)^2*(4*a*b^
2 - (26*a^3)/15) + tan(c/2 + (d*x)/2)^4*(96*a*b^2 - (78*a^3)/5) + tan(c/2 + (d*x)/2)^8*(320*a*b^2 - (191*a^3)/
3) + tan(c/2 + (d*x)/2)^6*(408*a*b^2 - (296*a^3)/5) + tan(c/2 + (d*x)/2)^3*((39*a^2*b)/2 - 4*b^3) + tan(c/2 +
(d*x)/2)^9*(216*a^2*b - 196*b^3) + tan(c/2 + (d*x)/2)^5*((519*a^2*b)/2 - (484*b^3)/3) + tan(c/2 + (d*x)/2)^7*(
(909*a^2*b)/2 - 268*b^3) - a^3/5 - (3*a^2*b*tan(c/2 + (d*x)/2))/2)/(d*(32*tan(c/2 + (d*x)/2)^5 + 96*tan(c/2 +
(d*x)/2)^7 + 96*tan(c/2 + (d*x)/2)^9 + 32*tan(c/2 + (d*x)/2)^11)) + (tan(c/2 + (d*x)/2)^3*((a*b^2)/8 - (7*a^3)
/96))/d - (tan(c/2 + (d*x)/2)^2*((3*a^2*b)/4 - b^3/8))/d + (log(tan(c/2 + (d*x)/2))*((45*a^2*b)/8 - (5*b^3)/2)
)/d - (log(tan(c/2 + (d*x)/2) - 1i)*((a*b^2*15i)/2 - a^3*1i))/d - (tan(c/2 + (d*x)/2)*((27*a*b^2)/8 - (11*a^3)
/16))/d + (3*a^2*b*tan(c/2 + (d*x)/2)^4)/(64*d) - (a*log(tan(c/2 + (d*x)/2) + 1i)*(2*a^2 - 15*b^2)*1i)/(2*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(cot(d*x+c)**6*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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