\(\int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\) [44]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (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 = 88 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {2 a^4}{d (a-a \cos (c+d x))}+\frac {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]

[Out]

-2*a^4/d/(a-a*cos(d*x+c))+5*a^3*ln(1-cos(d*x+c))/d-5*a^3*ln(cos(d*x+c))/d+3*a^3*sec(d*x+c)/d+1/2*a^3*sec(d*x+c
)^2/d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 78} \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {2 a^4}{d (a-a \cos (c+d x))}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

(-2*a^4)/(d*(a - a*Cos[c + d*x])) + (5*a^3*Log[1 - Cos[c + d*x]])/d - (5*a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec
[c + d*x])/d + (a^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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2915

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[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 (-a-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 (-a+x)}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \frac {-a+x}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \left (-\frac {1}{a x^3}+\frac {3}{a^2 x^2}-\frac {5}{a^3 x}+\frac {2}{a^2 (a+x)^2}+\frac {5}{a^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {2 a^4}{d (a-a \cos (c+d x))}+\frac {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+10 \left (\log (\cos (c+d x))-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-6 \sec (c+d x)-\sec ^2(c+d x)\right )}{16 d} \]

[In]

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

[Out]

-1/16*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*Csc[(c + d*x)/2]^2 + 10*(Log[Cos[c + d*x]] - 2*Log[Sin[(
c + d*x)/2]]) - 6*Sec[c + d*x] - Sec[c + d*x]^2))/d

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.50

method result size
risch \(\frac {2 a^{3} \left (5 \,{\mathrm e}^{5 i \left (d x +c \right )}-5 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{3 i \left (d x +c \right )}-5 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}+\frac {10 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) \(132\)
norman \(\frac {\frac {8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {a^{3}}{d}-\frac {5 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {10 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) \(134\)
derivativedivides \(\frac {a^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )}{d}\) \(160\)
default \(\frac {a^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )}{d}\) \(160\)
parallelrisch \(\frac {a^{3} \left (9 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (2 d x +2 c \right )-56 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (d x +c \right )+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (2 d x +2 c \right )-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )+75 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-32 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) \(197\)

[In]

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

[Out]

2*a^3/d/(exp(2*I*(d*x+c))+1)^2/(exp(I*(d*x+c))-1)^2*(5*exp(5*I*(d*x+c))-5*exp(4*I*(d*x+c))+8*exp(3*I*(d*x+c))-
5*exp(2*I*(d*x+c))+5*exp(I*(d*x+c)))+10/d*a^3*ln(exp(I*(d*x+c))-1)-5/d*a^3*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.50 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3} - 10 \, {\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + 10 \, {\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/2*(10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a^3 - 10*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-cos(
d*x + c)) + 10*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^3 - d*c
os(d*x + c)^2)

Sympy [F]

\[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \csc ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \csc ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \csc ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(3*csc(c + d*x)**3*sec(c + d*x), x) + Integral(3*csc(c + d*x)**3*sec(c + d*x)**2, x) + Integral(
csc(c + d*x)**3*sec(c + d*x)**3, x) + Integral(csc(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {10 \, a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - 10 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac {10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(10*a^3*log(cos(d*x + c) - 1) - 10*a^3*log(cos(d*x + c)) + (10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a
^3)/(cos(d*x + c)^3 - cos(d*x + c)^2))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (87) = 174\).

Time = 0.38 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.15 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {10 \, a^{3} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 10 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} - \frac {5 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} + \frac {27 \, a^{3} + \frac {38 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {15 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(10*a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 10*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + 2*(a^3 - 5*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) +
(27*a^3 + 38*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 15*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {-5\,a^3\,{\cos \left (c+d\,x\right )}^2+\frac {5\,a^3\,\cos \left (c+d\,x\right )}{2}+\frac {a^3}{2}}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^3\right )}-\frac {10\,a^3\,\mathrm {atanh}\left (2\,\cos \left (c+d\,x\right )-1\right )}{d} \]

[In]

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

[Out]

((5*a^3*cos(c + d*x))/2 + a^3/2 - 5*a^3*cos(c + d*x)^2)/(d*(cos(c + d*x)^2 - cos(c + d*x)^3)) - (10*a^3*atanh(
2*cos(c + d*x) - 1))/d