\(\int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) [187]

Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 b \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-b^{2/3}} d}-\frac {2 b \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}+\frac {2 b \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

-1/2*arctanh(cos(d*x+c))/a/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d-2/3*b*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(
a^(2/3)-b^(2/3))^(1/2))/a^(5/3)/d/(a^(2/3)-b^(2/3))^(1/2)-2/3*b*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*tan(1/2*d*x
+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/a^(5/3)/d/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)+2/3*b*arctan((-1)^(1
/3)*(b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/a^(5/3)/d/(a^(2/3)-(-1
)^(2/3)*b^(2/3))^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3299, 3853, 3855, 3292, 2739, 632, 210} \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 b \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 b \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 b \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[In]

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

[Out]

(-2*b*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^(5/3)*Sqrt[a^(2/3) - b^(2/3)]
*d) - (2*b*ArcTan[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*a^(5
/3)*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) + (2*b*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x
)/2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a^(5/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*d) - ArcTanh[Cos[c +
d*x]]/(2*a*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3292

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\csc ^3(c+d x)}{a}-\frac {b}{a \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx \\ & = \frac {\int \csc ^3(c+d x) \, dx}{a}-\frac {b \int \frac {1}{a+b \sin ^3(c+d x)} \, dx}{a} \\ & = -\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\int \csc (c+d x) \, dx}{2 a}-\frac {b \int \left (-\frac {1}{3 a^{2/3} \left (-\sqrt [3]{a}-\sqrt [3]{b} \sin (c+d x)\right )}-\frac {1}{3 a^{2/3} \left (-\sqrt [3]{a}+\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)\right )}-\frac {1}{3 a^{2/3} \left (-\sqrt [3]{a}-(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)\right )}\right ) \, dx}{a} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {b \int \frac {1}{-\sqrt [3]{a}-\sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^{5/3}}+\frac {b \int \frac {1}{-\sqrt [3]{a}+\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^{5/3}}+\frac {b \int \frac {1}{-\sqrt [3]{a}-(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^{5/3}} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{-\sqrt [3]{a}-2 \sqrt [3]{b} x-\sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{-\sqrt [3]{a}+2 \sqrt [3]{-1} \sqrt [3]{b} x-\sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{-\sqrt [3]{a}-2 (-1)^{2/3} \sqrt [3]{b} x-\sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}-\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}-2 \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d}-\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^{2/3}+\sqrt [3]{-1} b^{2/3}\right )-x^2} \, dx,x,-2 (-1)^{2/3} \sqrt [3]{b}-2 \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d}-\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^{2/3}-(-1)^{2/3} b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{-1} \sqrt [3]{b}-2 \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^{5/3} d} \\ & = \frac {2 b \arctan \left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {2 b \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-b^{2/3}} d}-\frac {2 b \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.40 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.63 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {16 i b \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]-3 \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )}{24 a d} \]

[In]

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

[Out]

((16*I)*b*RootSum[-b + 3*b*#1^2 - (8*I)*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] -
#1)]*#1 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ] - 3*(Csc[(c + d*x)/
2]^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] - Sec[(c + d*x)/2]^2))/(24*a*d)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.49 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.47

method	result	size
derivativedivides	\(\frac {\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{d}\)	\(136\)
default	\(\frac {\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{d}\)	\(136\)
risch	\(\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-8 i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (191102976 a^{12} d^{6}-191102976 a^{10} b^{2} d^{6}\right ) \textit {\_Z}^{6}-995328 a^{8} b^{2} d^{4} \textit {\_Z}^{4}+1728 a^{4} b^{4} d^{2} \textit {\_Z}^{2}-b^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {15925248 i d^{5} a^{11}}{b^{6}}+\frac {15925248 i d^{5} a^{9}}{b^{4}}\right ) \textit {\_R}^{5}+\left (-\frac {331776 i d^{4} a^{9}}{b^{5}}+\frac {331776 i d^{4} a^{7}}{b^{3}}\right ) \textit {\_R}^{4}+\left (\frac {69120 i d^{3} a^{7}}{b^{4}}+\frac {13824 i d^{3} a^{5}}{b^{2}}\right ) \textit {\_R}^{3}+\frac {1728 i d^{2} a^{5} \textit {\_R}^{2}}{b^{3}}-\frac {72 i d \,a^{3} \textit {\_R}}{b^{2}}-\frac {2 i a}{b}\right )\right )+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}\)	\(279\)

[In]

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

[Out]

1/d*(1/8*tan(1/2*d*x+1/2*c)^2/a-1/3/a*b*sum((_R^4+2*_R^2+1)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2
*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))-1/8/a/tan(1/2*d*x+1/2*c)^2+1/2/a*ln(tan(1/2*d*x+1/2*c)
))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 14.83 (sec) , antiderivative size = 29431, normalized size of antiderivative = 102.55 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

1/4*(4*(cos(3*d*x + 3*c) + cos(d*x + c))*cos(4*d*x + 4*c) - 4*(2*cos(2*d*x + 2*c) - 1)*cos(3*d*x + 3*c) - 8*co
s(2*d*x + 2*c)*cos(d*x + c) + 32*(a*b*d*cos(4*d*x + 4*c)^2 + 4*a*b*d*cos(2*d*x + 2*c)^2 + a*b*d*sin(4*d*x + 4*
c)^2 - 4*a*b*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*a*b*d*sin(2*d*x + 2*c)^2 - 4*a*b*d*cos(2*d*x + 2*c) + a*b
*d - 2*(2*a*b*d*cos(2*d*x + 2*c) - a*b*d)*cos(4*d*x + 4*c))*integrate(-(8*a*cos(3*d*x + 3*c)^2 - b*cos(3*d*x +
3*c)*sin(6*d*x + 6*c) + 3*b*cos(3*d*x + 3*c)*sin(4*d*x + 4*c) + b*cos(6*d*x + 6*c)*sin(3*d*x + 3*c) - 3*b*cos
(4*d*x + 4*c)*sin(3*d*x + 3*c) + 8*a*sin(3*d*x + 3*c)^2 - 3*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + (3*b*cos(2*d
*x + 2*c) - b)*sin(3*d*x + 3*c))/(a*b^2*cos(6*d*x + 6*c)^2 + 9*a*b^2*cos(4*d*x + 4*c)^2 + 64*a^3*cos(3*d*x + 3
*c)^2 + 9*a*b^2*cos(2*d*x + 2*c)^2 + a*b^2*sin(6*d*x + 6*c)^2 + 9*a*b^2*sin(4*d*x + 4*c)^2 + 64*a^3*sin(3*d*x
+ 3*c)^2 - 48*a^2*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*a*b^2*sin(2*d*x + 2*c)^2 - 6*a*b^2*cos(2*d*x + 2*c)
+ a*b^2 - 2*(3*a*b^2*cos(4*d*x + 4*c) - 3*a*b^2*cos(2*d*x + 2*c) - 8*a^2*b*sin(3*d*x + 3*c) + a*b^2)*cos(6*d*x
+ 6*c) - 6*(3*a*b^2*cos(2*d*x + 2*c) + 8*a^2*b*sin(3*d*x + 3*c) - a*b^2)*cos(4*d*x + 4*c) - 2*(8*a^2*b*cos(3*
d*x + 3*c) + 3*a*b^2*sin(4*d*x + 4*c) - 3*a*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6*(8*a^2*b*cos(3*d*x + 3*
c) - 3*a*b^2*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16*(3*a^2*b*cos(2*d*x + 2*c) - a^2*b)*sin(3*d*x + 3*c)), x)
+ (2*(2*cos(2*d*x + 2*c) - 1)*cos(4*d*x + 4*c) - cos(4*d*x + 4*c)^2 - 4*cos(2*d*x + 2*c)^2 - sin(4*d*x + 4*c)^
2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) - 1)*log(cos(d*x)^2 + 2*co
s(d*x)*cos(c) + cos(c)^2 + sin(d*x)^2 - 2*sin(d*x)*sin(c) + sin(c)^2) - (2*(2*cos(2*d*x + 2*c) - 1)*cos(4*d*x
+ 4*c) - cos(4*d*x + 4*c)^2 - 4*cos(2*d*x + 2*c)^2 - sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c)
- 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) - 1)*log(cos(d*x)^2 - 2*cos(d*x)*cos(c) + cos(c)^2 + sin(d*x)^2 +
2*sin(d*x)*sin(c) + sin(c)^2) + 4*(sin(3*d*x + 3*c) + sin(d*x + c))*sin(4*d*x + 4*c) - 8*sin(3*d*x + 3*c)*sin(
2*d*x + 2*c) - 8*sin(2*d*x + 2*c)*sin(d*x + c) + 4*cos(d*x + c))/(a*d*cos(4*d*x + 4*c)^2 + 4*a*d*cos(2*d*x + 2
*c)^2 + a*d*sin(4*d*x + 4*c)^2 - 4*a*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*a*d*sin(2*d*x + 2*c)^2 - 4*a*d*co
s(2*d*x + 2*c) + a*d - 2*(2*a*d*cos(2*d*x + 2*c) - a*d)*cos(4*d*x + 4*c))

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 15.06 (sec) , antiderivative size = 1573, normalized size of antiderivative = 5.48 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

int(1/(sin(c + d*x)^3*(a + b*sin(c + d*x)^3)),x)

[Out]

symsum(log(-(65536*a*b^9 - 262144*b^10*tan(c/2 + (d*x)/2) - 131072*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*
a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)*a^2*b^9 - 61440*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z
^4 - 27*a^4*b^4*z^2 - b^6, z, k)*a^4*b^7 + 860160*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*
a^4*b^4*z^2 - b^6, z, k)^2*a^5*b^7 - 3244032*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b
^4*z^2 - b^6, z, k)^3*a^6*b^7 - 1105920*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^
2 - b^6, z, k)^3*a^8*b^5 + 3538944*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b
^6, z, k)^4*a^7*b^7 + 3870720*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z
, k)^4*a^9*b^5 + 663552*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^5
*a^10*b^5 - 4976640*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^5*a^1
2*b^3 - 7962624*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^11*b^
5 + 9953280*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^13*b^3 +
24576*a^2*b^8*tan(c/2 + (d*x)/2) + 540672*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*
z^2 - b^6, z, k)*a^3*b^8*tan(c/2 + (d*x)/2) - 7077888*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 -
27*a^4*b^4*z^2 - b^6, z, k)^2*a^4*b^8*tan(c/2 + (d*x)/2) + 442368*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*
a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^2*a^6*b^6*tan(c/2 + (d*x)/2) - 2359296*root(729*a^10*b^2*z^6 - 729*a
^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^3*a^5*b^8*tan(c/2 + (d*x)/2) + 7741440*root(729*a^10*b
^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^3*a^7*b^6*tan(c/2 + (d*x)/2) - 80953344*
root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^4*a^8*b^6*tan(c/2 + (d*x)
/2) + 1990656*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^4*a^10*b^4*
tan(c/2 + (d*x)/2) - 31850496*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z
, k)^5*a^9*b^6*tan(c/2 + (d*x)/2) + 26873856*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b
^4*z^2 - b^6, z, k)^5*a^11*b^4*tan(c/2 + (d*x)/2) + 254803968*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b
^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^10*b^6*tan(c/2 + (d*x)/2) - 252813312*root(729*a^10*b^2*z^6 - 729*a^1
2*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^12*b^4*tan(c/2 + (d*x)/2))/a^5)*root(729*a^10*b^2*z^
6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k), k, 1, 6)/d - cot(c/2 + (d*x)/2)^2/(8*a*d) +
tan(c/2 + (d*x)/2)^2/(8*a*d) + log(tan(c/2 + (d*x)/2))/(2*a*d)