\(\int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) [190]

Optimal result

Integrand size = 23, antiderivative size = 281 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 (-1)^{2/3} a^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b^{4/3} d}+\frac {2 a^{2/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{4/3} d}-\frac {2 \sqrt [3]{-1} a^{2/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b^{4/3} d}-\frac {\cos (c+d x)}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3299, 2718, 2739, 632, 210} \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 (-1)^{2/3} a^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 a^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{-1} a^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {\cos (c+d x)}{b d} \]

[In]

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

[Out]

(-2*(-1)^(2/3)*a^(2/3)*ArcTan[((-1)^(1/3)*b^(1/3) - a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3
)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(4/3)*d) + (2*a^(2/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])
/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b^(4/3)*d) - (2*(-1)^(1/3)*a^(2/3)*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*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b
^(4/3)*d) - Cos[c + d*x]/(b*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 2718

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

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]))

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

Mathematica [C] (verified)

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

Time = 0.24 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.66 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-3 \cos (c+d x)+a \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 b d} \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 1.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.37

method	result	size
derivativedivides	\(\frac {-\frac {2}{b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\)	\(104\)
default	\(\frac {-\frac {2}{b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\)	\(104\)
risch	\(-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{8} d^{6}-729 b^{10} d^{6}\right ) \textit {\_Z}^{6}+62208 a^{2} b^{6} d^{4} \textit {\_Z}^{4}[phone] a^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {243 a^{2} b^{7} d^{5}}{1048576 a^{4}+1048576 a^{2} b^{2}}+\frac {243 b^{9} d^{5}}{1048576 a^{4}+1048576 a^{2} b^{2}}\right ) \textit {\_R}^{5}+\left (\frac {1296 i d^{4} b^{5} a^{3}}{1048576 a^{4}+1048576 a^{2} b^{2}}-\frac {1296 i d^{4} b^{7} a}{1048576 a^{4}+1048576 a^{2} b^{2}}\right ) \textit {\_R}^{4}+\left (-\frac {13824 a^{2} b^{5} d^{3}}{1048576 a^{4}+1048576 a^{2} b^{2}}-\frac {6912 d^{3} b^{7}}{1048576 a^{4}+1048576 a^{2} b^{2}}\right ) \textit {\_R}^{3}+\left (\frac {73728 i d^{2} b^{3} a^{3}}{1048576 a^{4}+1048576 a^{2} b^{2}}+\frac {36864 i d^{2} b^{5} a}{1048576 a^{4}+1048576 a^{2} b^{2}}\right ) \textit {\_R}^{2}+\left (\frac {196608 d b \,a^{4}}{1048576 a^{4}+1048576 a^{2} b^{2}}+\frac {393216 d \,b^{3} a^{2}}{1048576 a^{4}+1048576 a^{2} b^{2}}\right ) \textit {\_R} -\frac {1048576 i a^{3} b}{1048576 a^{4}+1048576 a^{2} b^{2}}\right )\right )}{16}\)	\(403\)

[In]

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

[Out]

1/d*(-2/b/(1+tan(1/2*d*x+1/2*c)^2)-2/3/b*a*sum((_R^3+_R)/(_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 = 1.33 (sec) , antiderivative size = 21185, normalized size of antiderivative = 75.39 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.32 (sec) , antiderivative size = 665, normalized size of antiderivative = 2.37 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\sum _{k=1}^6\ln \left (8192\,a^8\,b^5-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^2\,a^6\,b^9\,294912+{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^3\,a^6\,b^{10}\,1548288-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^4\,a^6\,b^{11}\,1990656-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^5\,a^4\,b^{14}\,7962624+{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^5\,a^6\,b^{12}\,5971968-65536\,a^7\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )\,a^7\,b^7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^2\,a^7\,b^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^3\,a^5\,b^{11}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472+{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^3\,a^7\,b^9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,221184-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^4\,a^5\,b^{12}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,2654208-{\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}^5\,a^5\,b^{13}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656\right )\,\mathrm {root}\left (729\,a^2\,b^8\,d^6-729\,b^{10}\,d^6+243\,a^2\,b^6\,d^4+a^4,d,k\right )}{d}-\frac {2}{b\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,d} \]

[In]

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

[Out]

symsum(log(8192*a^8*b^5 - 294912*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d, k)^2*a^6*b^9
+ 1548288*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d, k)^3*a^6*b^10 - 1990656*root(729*a^2
*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d, k)^4*a^6*b^11 - 7962624*root(729*a^2*b^8*d^6 - 729*b^10*d^
6 + 243*a^2*b^6*d^4 + a^4, d, k)^5*a^4*b^14 + 5971968*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 +
a^4, d, k)^5*a^6*b^12 - 65536*a^7*b^6*tan(c/2 + (d*x)/2) + 196608*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^
2*b^6*d^4 + a^4, d, k)*a^7*b^7*tan(c/2 + (d*x)/2) - 294912*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d
^4 + a^4, d, k)^2*a^7*b^8*tan(c/2 + (d*x)/2) - 1769472*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 +
a^4, d, k)^3*a^5*b^11*tan(c/2 + (d*x)/2) + 221184*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4
, d, k)^3*a^7*b^9*tan(c/2 + (d*x)/2) - 2654208*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d,
k)^4*a^5*b^12*tan(c/2 + (d*x)/2) - 1990656*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d, k)
^5*a^5*b^13*tan(c/2 + (d*x)/2))*root(729*a^2*b^8*d^6 - 729*b^10*d^6 + 243*a^2*b^6*d^4 + a^4, d, k), k, 1, 6)/d
- 2/(b*d + b*d*tan(c/2 + (d*x)/2)^2)