3.1121 \(\int \cot ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=194 \[ -\frac {a^3 \cot ^3(c+d x)}{3 d}+\frac {a^3 \cot (c+d x)}{d}+a^3 x-\frac {9 a^2 b \cos (c+d x)}{2 d}-\frac {3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac {9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {9 a b^2 \cot (c+d x)}{2 d}+\frac {3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {9}{2} a b^2 x+\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {b^3 \cos (c+d x)}{d}-\frac {b^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

a^3*x-9/2*a*b^2*x+9/2*a^2*b*arctanh(cos(d*x+c))/d-b^3*arctanh(cos(d*x+c))/d-9/2*a^2*b*cos(d*x+c)/d+b^3*cos(d*x
+c)/d+1/3*b^3*cos(d*x+c)^3/d+a^3*cot(d*x+c)/d-9/2*a*b^2*cot(d*x+c)/d+3/2*a*b^2*cos(d*x+c)^2*cot(d*x+c)/d-3/2*a
^2*b*cos(d*x+c)*cot(d*x+c)^2/d-1/3*a^3*cot(d*x+c)^3/d

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Rubi [A]  time = 0.22, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2722, 2592, 302, 206, 2591, 288, 321, 203, 3473, 8} \[ -\frac {9 a^2 b \cos (c+d x)}{2 d}-\frac {3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac {9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}+\frac {a^3 \cot (c+d x)}{d}+a^3 x-\frac {9 a b^2 \cot (c+d x)}{2 d}+\frac {3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {9}{2} a b^2 x+\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {b^3 \cos (c+d x)}{d}-\frac {b^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

a^3*x - (9*a*b^2*x)/2 + (9*a^2*b*ArcTanh[Cos[c + d*x]])/(2*d) - (b^3*ArcTanh[Cos[c + d*x]])/d - (9*a^2*b*Cos[c
+ d*x])/(2*d) + (b^3*Cos[c + d*x])/d + (b^3*Cos[c + d*x]^3)/(3*d) + (a^3*Cot[c + d*x])/d - (9*a*b^2*Cot[c + d
*x])/(2*d) + (3*a*b^2*Cos[c + d*x]^2*Cot[c + d*x])/(2*d) - (3*a^2*b*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d) - (a^3*
Cot[c + d*x]^3)/(3*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 ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \left (b^3 \cos ^3(c+d x) \cot (c+d x)+3 a b^2 \cos ^2(c+d x) \cot ^2(c+d x)+3 a^2 b \cos (c+d x) \cot ^3(c+d x)+a^3 \cot ^4(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^4(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos (c+d x) \cot ^3(c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^2(c+d x) \cot ^2(c+d x) \, dx+b^3 \int \cos ^3(c+d x) \cot (c+d x) \, dx\\ &=-\frac {a^3 \cot ^3(c+d x)}{3 d}-a^3 \int \cot ^2(c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 \cot (c+d x)}{d}+\frac {3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}+a^3 \int 1 \, dx+\frac {\left (9 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}-\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}-\frac {b^3 \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 x-\frac {9 a^2 b \cos (c+d x)}{2 d}+\frac {b^3 \cos (c+d x)}{d}+\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cot (c+d x)}{d}-\frac {9 a b^2 \cot (c+d x)}{2 d}+\frac {3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}+\frac {\left (9 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}+\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 x-\frac {9}{2} a b^2 x+\frac {9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {b^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {9 a^2 b \cos (c+d x)}{2 d}+\frac {b^3 \cos (c+d x)}{d}+\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cot (c+d x)}{d}-\frac {9 a b^2 \cot (c+d x)}{2 d}+\frac {3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 6.25, size = 355, normalized size = 1.83 \[ \frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (4 a^3 \cos \left (\frac {1}{2} (c+d x)\right )-9 a b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (9 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )-4 a^3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d}-\frac {a^3 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {a^3 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {\left (2 b^3-9 a^2 b\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {\left (9 a^2 b-2 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {a \left (2 a^2-9 b^2\right ) (c+d x)}{2 d}+\frac {b \left (5 b^2-12 a^2\right ) \cos (c+d x)}{4 d}-\frac {3 a^2 b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {3 a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {3 a b^2 \sin (2 (c+d x))}{4 d}+\frac {b^3 \cos (3 (c+d x))}{12 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(a*(2*a^2 - 9*b^2)*(c + d*x))/(2*d) + (b*(-12*a^2 + 5*b^2)*Cos[c + d*x])/(4*d) + (b^3*Cos[3*(c + d*x)])/(12*d)
+ ((4*a^3*Cos[(c + d*x)/2] - 9*a*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*d) - (3*a^2*b*Csc[(c + d*x)/2]^2)
/(8*d) - (a^3*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + ((9*a^2*b - 2*b^3)*Log[Cos[(c + d*x)/2]])/(2*d) +
((-9*a^2*b + 2*b^3)*Log[Sin[(c + d*x)/2]])/(2*d) + (3*a^2*b*Sec[(c + d*x)/2]^2)/(8*d) + (Sec[(c + d*x)/2]*(-4*
a^3*Sin[(c + d*x)/2] + 9*a*b^2*Sin[(c + d*x)/2]))/(6*d) - (3*a*b^2*Sin[2*(c + d*x)])/(4*d) + (a^3*Sec[(c + d*x
)/2]^2*Tan[(c + d*x)/2])/(24*d)

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fricas [A]  time = 0.84, size = 293, normalized size = 1.51 \[ \frac {18 \, a b^{2} \cos \left (d x + c\right )^{5} + 8 \, {\left (2 \, a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (9 \, a^{2} b - 2 \, b^{3} - {\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (9 \, a^{2} b - 2 \, b^{3} - {\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 6 \, {\left (2 \, a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right ) + 2 \, {\left (2 \, b^{3} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, a^{3} - 9 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{3} - 9 \, a b^{2}\right )} d x + 3 \, {\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(18*a*b^2*cos(d*x + c)^5 + 8*(2*a^3 - 9*a*b^2)*cos(d*x + c)^3 - 3*(9*a^2*b - 2*b^3 - (9*a^2*b - 2*b^3)*co
s(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(9*a^2*b - 2*b^3 - (9*a^2*b - 2*b^3)*cos(d*x + c)^2
)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 6*(2*a^3 - 9*a*b^2)*cos(d*x + c) + 2*(2*b^3*cos(d*x + c)^5 + 3*(
2*a^3 - 9*a*b^2)*d*x*cos(d*x + c)^2 - 2*(9*a^2*b - 2*b^3)*cos(d*x + c)^3 - 3*(2*a^3 - 9*a*b^2)*d*x + 3*(9*a^2*
b - 2*b^3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [B]  time = 0.32, size = 421, normalized size = 2.17 \[ \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, {\left (2 \, a^{3} - 9 \, a b^{2}\right )} {\left (d x + c\right )} - 36 \, {\left (9 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {198 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 108 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 135 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 156 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 132 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 324 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 351 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 156 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 126 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 540 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 315 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 148 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 108 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3}}}{72 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/72*(3*a^3*tan(1/2*d*x + 1/2*c)^3 + 27*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 45*a^3*tan(1/2*d*x + 1/2*c) + 108*a*b^2
*tan(1/2*d*x + 1/2*c) + 36*(2*a^3 - 9*a*b^2)*(d*x + c) - 36*(9*a^2*b - 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) +
(198*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 44*b^3*tan(1/2*d*x + 1/2*c)^9 + 45*a^3*tan(1/2*d*x + 1/2*c)^8 + 108*a*b^2
*tan(1/2*d*x + 1/2*c)^8 + 135*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 156*b^3*tan(1/2*d*x + 1/2*c)^7 + 132*a^3*tan(1/2*
d*x + 1/2*c)^6 - 324*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 351*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 156*b^3*tan(1/2*d*x + 1
/2*c)^5 + 126*a^3*tan(1/2*d*x + 1/2*c)^4 - 540*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 315*a^2*b*tan(1/2*d*x + 1/2*c)^3
+ 148*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*a^3*tan(1/2*d*x + 1/2*c)^2 - 108*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 27*a^2*
b*tan(1/2*d*x + 1/2*c) - 3*a^3)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^3)/d

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maple [A]  time = 0.60, size = 264, normalized size = 1.36 \[ -\frac {a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} \cot \left (d x +c \right )}{d}+a^{3} x +\frac {a^{3} c}{d}-\frac {3 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{2 d}-\frac {9 a^{2} b \cos \left (d x +c \right )}{2 d}-\frac {9 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a \,b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {3 a \,b^{2} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{d}-\frac {9 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {9 a \,b^{2} x}{2}-\frac {9 a \,b^{2} c}{2 d}+\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {b^{3} \cos \left (d x +c \right )}{d}+\frac {b^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^4*(a+b*sin(d*x+c))^3,x)

[Out]

-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-3/2/d*a^2*b/sin(d*x+c)^2*cos(d*x+c)^5-3/2/d*a^2*b*cos
(d*x+c)^3-9/2*a^2*b*cos(d*x+c)/d-9/2/d*a^2*b*ln(csc(d*x+c)-cot(d*x+c))-3/d*a*b^2/sin(d*x+c)*cos(d*x+c)^5-3/d*a
*b^2*sin(d*x+c)*cos(d*x+c)^3-9/2*a*b^2*cos(d*x+c)*sin(d*x+c)/d-9/2*a*b^2*x-9/2/d*a*b^2*c+1/3*b^3*cos(d*x+c)^3/
d+b^3*cos(d*x+c)/d+1/d*b^3*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.47, size = 187, normalized size = 0.96 \[ \frac {4 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{3} - 18 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a b^{2} + 2 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{3} + 9 \, a^{2} b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(4*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^3 - 18*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(
tan(d*x + c)^3 + tan(d*x + c)))*a*b^2 + 2*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log
(cos(d*x + c) - 1))*b^3 + 9*a^2*b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) +
1) - 3*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 9.65, size = 405, normalized size = 2.09 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {9\,a^2\,b}{2}-b^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (\frac {a\,b^2\,9{}\mathrm {i}}{2}-a^3\,1{}\mathrm {i}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{2}-\frac {5\,a^3}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a\,b^2-4\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (5\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (60\,a\,b^2-14\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (36\,a\,b^2-\frac {44\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (51\,a^2\,b-32\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (57\,a^2\,b-\frac {64\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (105\,a^2\,b-32\,b^3\right )+\frac {a^3}{3}+3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,\left (2\,a^2-9\,b^2\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + b*sin(c + d*x))^3)/sin(c + d*x)^4,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - (log(tan(c/2 + (d*x)/2))*((9*a^2*b)/2 - b^3))/d - (log(tan(c/2 + (d*x)/2)
+ 1i)*((a*b^2*9i)/2 - a^3*1i))/d + (tan(c/2 + (d*x)/2)*((3*a*b^2)/2 - (5*a^3)/8))/d - (tan(c/2 + (d*x)/2)^2*(1
2*a*b^2 - 4*a^3) - tan(c/2 + (d*x)/2)^8*(12*a*b^2 + 5*a^3) + tan(c/2 + (d*x)/2)^4*(60*a*b^2 - 14*a^3) + tan(c/
2 + (d*x)/2)^6*(36*a*b^2 - (44*a^3)/3) + tan(c/2 + (d*x)/2)^7*(51*a^2*b - 32*b^3) + tan(c/2 + (d*x)/2)^3*(57*a
^2*b - (64*b^3)/3) + tan(c/2 + (d*x)/2)^5*(105*a^2*b - 32*b^3) + a^3/3 + 3*a^2*b*tan(c/2 + (d*x)/2))/(d*(8*tan
(c/2 + (d*x)/2)^3 + 24*tan(c/2 + (d*x)/2)^5 + 24*tan(c/2 + (d*x)/2)^7 + 8*tan(c/2 + (d*x)/2)^9)) + (3*a^2*b*ta
n(c/2 + (d*x)/2)^2)/(8*d) - (a*log(tan(c/2 + (d*x)/2) - 1i)*(2*a^2 - 9*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(cos(d*x+c)**4*csc(d*x+c)**4*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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