\(\int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\) [189]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-4 b) (a-b)^2 \log (1+\cos (c+d x))}{4 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

-1/2*a^2*(b*(3+b^2/a^2)+a*(1+3*b^2/a^2)*cos(d*x+c))*csc(d*x+c)^2/d+1/4*(a+b)^2*(a+4*b)*ln(1-cos(d*x+c))/d-b*(3
*a^2+2*b^2)*ln(cos(d*x+c))/d-1/4*(a-4*b)*(a-b)^2*ln(1+cos(d*x+c))/d+3*a*b^2*sec(d*x+c)/d+1/2*b^3*sec(d*x+c)^2/
d

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2916, 12, 1819, 1816} \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {a^2 \csc ^2(c+d x) \left (a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )\right )}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {(a-4 b) (a-b)^2 \log (\cos (c+d x)+1)}{4 d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \]

[In]

Int[Csc[c + d*x]^3*(a + b*Sec[c + d*x])^3,x]

[Out]

-1/2*(a^2*(b*(3 + b^2/a^2) + a*(1 + (3*b^2)/a^2)*Cos[c + d*x])*Csc[c + d*x]^2)/d + ((a + b)^2*(a + 4*b)*Log[1
- Cos[c + d*x]])/(4*d) - (b*(3*a^2 + 2*b^2)*Log[Cos[c + d*x]])/d - ((a - 4*b)*(a - b)^2*Log[1 + Cos[c + d*x]])
/(4*d) + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
 a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
 x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int (-b-a \cos (c+d x))^3 \csc ^3(c+d x) \sec ^3(c+d x) \, dx \\ & = \frac {a^3 \text {Subst}\left (\int \frac {a^3 (-b+x)^3}{x^3 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \frac {(-b+x)^3}{x^3 \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}-\frac {a^4 \text {Subst}\left (\int \frac {2 b^3-6 b^2 x+2 b \left (3+\frac {b^2}{a^2}\right ) x^2-\left (1+\frac {3 b^2}{a^2}\right ) x^3}{x^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{2 d} \\ & = -\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}-\frac {a^4 \text {Subst}\left (\int \left (-\frac {(a-4 b) (a-b)^2}{2 a^4 (a-x)}+\frac {2 b^3}{a^2 x^3}-\frac {6 b^2}{a^2 x^2}+\frac {2 \left (3 a^2 b+2 b^3\right )}{a^4 x}-\frac {(a+b)^2 (a+4 b)}{2 a^4 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{2 d} \\ & = -\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-4 b) (a-b)^2 \log (1+\cos (c+d x))}{4 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(669\) vs. \(2(162)=324\).

Time = 7.22 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.13 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3}+\frac {\left (-a^3-3 a^2 b-3 a b^2-b^3\right ) \cos ^3(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{8 d (b+a \cos (c+d x))^3}+\frac {\left (-a^3+6 a^2 b-9 a b^2+4 b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac {\left (-3 a^2 b-2 b^3\right ) \cos ^3(c+d x) \log (\cos (c+d x)) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3}+\frac {\left (a^3+6 a^2 b+9 a b^2+4 b^3\right ) \cos ^3(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac {\left (a^3-3 a^2 b+3 a b^2-b^3\right ) \cos ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{8 d (b+a \cos (c+d x))^3}+\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 a b^2 \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin \left (\frac {1}{2} (c+d x)\right )}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Csc[c + d*x]^3*(a + b*Sec[c + d*x])^3,x]

[Out]

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

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00

method result size
derivativedivides \(\frac {a^{3} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) \(162\)
default \(\frac {a^{3} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) \(162\)
norman \(\frac {-\frac {a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}{8 d}+\frac {\left (a^{3}+15 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d}+\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 d}-\frac {\left (2 a^{3}-3 a^{2} b +30 a \,b^{2}-9 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (a^{3}+6 a^{2} b +9 a \,b^{2}+4 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) \(250\)
parallelrisch \(\frac {-192 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {2 b^{2}}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-192 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {2 b^{2}}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (4 b +a \right ) \left (a +b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (-16 \cos \left (2 d x +2 c \right ) b^{3}+\left (-a^{3}-15 a \,b^{2}+2 b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (-4 a^{3}-36 a \,b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (-12 a^{3}-12 a \,b^{2}\right ) \cos \left (d x +c \right )+a^{3}-48 a^{2} b +15 a \,b^{2}-2 b^{3}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+192 a^{2} b \right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-192 a^{2} b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) \(274\)
risch \(\frac {a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+9 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+3 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{3} {\mathrm e}^{i \left (d x +c \right )}+9 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a^{2} b}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a \,b^{2}}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a^{2} b}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a \,b^{2}}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{3}}{d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) \(425\)

[In]

int(csc(d*x+c)^3*(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.79 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {12 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + {\left ({\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \csc ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**3*(a+b*sec(d*x+c))**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*csc(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )}}{\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2}}}{4 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (154) = 308\).

Time = 0.38 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.98 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {\frac {a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) + 8 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} - \frac {2 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {18 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac {4 \, {\left (9 \, a^{2} b + 12 \, a b^{2} + 6 \, b^{3} + \frac {18 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{8 \, d} \]

[In]

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

[Out]

-1/8*(a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a*b^2*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) - b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*(a^3 + 6*a^2*b + 9*a*b^2 + 4
*b^3)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + 8*(3*a^2*b + 2*b^3)*log(abs(-(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) - 1)) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a^2*b
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*b^3*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - 4*(9*a^2*b + 12*a*b^2 + 6*b^3 + 18*a^2*b*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b^3*(cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 9*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 6*b^3*(cos(d*x + c) - 1)^2/(cos(d*x +
 c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 13.96 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{4\,d}-\frac {\ln \left (\cos \left (c+d\,x\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3}{2}+\frac {9\,a\,b^2}{2}\right )-\frac {b^3}{2}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}+b^3\right )-3\,a\,b^2\,\cos \left (c+d\,x\right )}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^4\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2\,\left (a-4\,b\right )}{4\,d} \]

[In]

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

[Out]

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