\(\int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) [194]

Optimal result

Integrand size = 23, antiderivative size = 296 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {2 b^{4/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^2 \sqrt {a^{2/3}-b^{2/3}} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} d}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} d}-\frac {\cot (c+d x)}{a d}-\frac {\cot ^3(c+d x)}{3 a d} \]

[Out]

b*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a/d-1/3*cot(d*x+c)^3/a/d+2/3*b^(4/3)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*
x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/a^2/d/(a^(2/3)-b^(2/3))^(1/2)-2/3*b^(4/3)*arctanh((b^(1/3)+(-1)^(2/3)*a^(1/
3)*tan(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2))/a^2/d/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)-2/3*b^(4/3
)*arctanh((b^(1/3)-(-1)^(1/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2))/a^2/d/(-(-1)^(2
/3)*a^(2/3)+b^(2/3))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3299, 3855, 3852, 2739, 632, 210, 212} \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 b^{4/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 a^2 d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}-\frac {2 b^{4/3} \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d} \]

[In]

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

[Out]

(2*b^(4/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^2*Sqrt[a^(2/3) - b^(2/3)
]*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - (2*b^(4/3)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/
Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*a^2*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*d) - (2*b^(4/3)*ArcTanh[(
b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*a^2*Sqrt[(-1)^(1/3)*a^(
2/3) + b^(2/3)]*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

Mathematica [C] (verified)

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

Time = 1.45 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-8 a \cot \left (\frac {1}{2} (c+d x)\right )+24 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-24 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 i b^2 \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]+8 a \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+8 a \tan \left (\frac {1}{2} (c+d x)\right )}{24 a^2 d} \]

[In]

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

[Out]

(-8*a*Cot[(c + d*x)/2] + 24*b*Log[Cos[(c + d*x)/2]] - 24*b*Log[Sin[(c + d*x)/2]] + (4*I)*b^2*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)] - I*Log[1 - 2*Cos[c +
d*x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1
^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)
*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] + 8*a*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - (a*Csc[(c + d*x)/2]^4*Sin[c + d*x])
/2 + 8*a*Tan[(c + d*x)/2])/(24*a^2*d)

Maple [C] (verified)

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

Time = 1.48 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.55

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {4 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a^{2}}}{d}\)	\(162\)
default	\(\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {4 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a^{2}}}{d}\)	\(162\)
risch	\(\frac {4 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (12230590464 a^{14} d^{6}-12230590464 a^{12} b^{2} d^{6}\right ) \textit {\_Z}^{6}[phone] a^{8} b^{4} d^{4} \textit {\_Z}^{4}-6912 a^{4} b^{6} d^{2} \textit {\_Z}^{2}+b^{8}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {254803968 a^{12} d^{5}}{b^{7}}-\frac {254803968 a^{10} d^{5}}{b^{5}}\right ) \textit {\_R}^{5}+\left (-\frac {5308416 i d^{4} a^{9}}{b^{5}}+\frac {5308416 i d^{4} a^{7}}{b^{3}}\right ) \textit {\_R}^{4}+\frac {331776 a^{6} d^{3} \textit {\_R}^{3}}{b^{3}}+\left (-\frac {2304 i a^{5} d^{2}}{b^{3}}-\frac {4608 i a^{3} d^{2}}{b}\right ) \textit {\_R}^{2}-\frac {144 d \,a^{2} \textit {\_R}}{b}+\frac {i b}{a}\right )\right )+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(269\)

[In]

int(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/8/a*(1/3*tan(1/2*d*x+1/2*c)^3+3*tan(1/2*d*x+1/2*c))-1/24/a/tan(1/2*d*x+1/2*c)^3-3/8/a/tan(1/2*d*x+1/2*c
)-1/a^2*b*ln(tan(1/2*d*x+1/2*c))+4/3*b^2/a^2*sum(_R^2/(_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)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.05 (sec) , antiderivative size = 29423, normalized size of antiderivative = 99.40 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

[In]

integrate(csc(d*x+c)**4/(a+b*sin(d*x+c)**3),x)

[Out]

Integral(csc(c + d*x)**4/(a + b*sin(c + d*x)**3), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 16.27 (sec) , antiderivative size = 1503, normalized size of antiderivative = 5.08 \[ \int \frac {\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

int(1/(sin(c + d*x)^4*(a + b*sin(c + d*x)^3)),x)

[Out]

symsum(log((98304*b^11 + 589824*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8,
z, k)*a^2*b^10 - 98304*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^2
*a^4*b^9 - 5898240*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^3*a^6*
b^8 - 7962624*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^4*a^8*b^7 -
663552*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^4*a^10*b^5 + 5308
416*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^10*b^6 - 10616832
*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^12*b^4 + 7962624*roo
t(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^6*a^12*b^5 - 9953280*root(72
9*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^6*a^14*b^3 + 24576*root(729*a^12
*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)*a^3*b^9*tan(c/2 + (d*x)/2) - 3145728*r
oot(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^2*a^3*b^10*tan(c/2 + (d*x)
/2) + 466944*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^2*a^5*b^8*ta
n(c/2 + (d*x)/2) + 18874368*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z,
k)^3*a^5*b^9*tan(c/2 + (d*x)/2) + 3981312*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*
z^2 - b^8, z, k)^3*a^7*b^7*tan(c/2 + (d*x)/2) + 56623104*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^
4 + 27*a^4*b^6*z^2 - b^8, z, k)^4*a^7*b^8*tan(c/2 + (d*x)/2) + 20791296*root(729*a^12*b^2*z^6 - 729*a^14*z^6 -
243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^4*a^9*b^6*tan(c/2 + (d*x)/2) - 84934656*root(729*a^12*b^2*z^6 -
729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^9*b^7*tan(c/2 + (d*x)/2) + 78962688*root(729
*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^5*a^11*b^5*tan(c/2 + (d*x)/2) - 2
54803968*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)^6*a^11*b^6*tan(c
/2 + (d*x)/2) + 252813312*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z^2 - b^8, z, k)
^6*a^13*b^4*tan(c/2 + (d*x)/2) - 1048576*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^4*b^6*z
^2 - b^8, z, k)*a*b^11*tan(c/2 + (d*x)/2))/a^6)*root(729*a^12*b^2*z^6 - 729*a^14*z^6 - 243*a^8*b^4*z^4 + 27*a^
4*b^6*z^2 - b^8, z, k), k, 1, 6)/d - (a*((3*cot(c/2 + (d*x)/2))/8 - (3*tan(c/2 + (d*x)/2))/8 + cot(c/2 + (d*x)
/2)^3/24 - tan(c/2 + (d*x)/2)^3/24) + b*log(tan(c/2 + (d*x)/2)))/(a^2*d)