Integrand size = 24, antiderivative size = 177 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{9/4} d}-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{9/4} d}+\frac {(a+b) \cos (c+d x)}{b^2 d}-\frac {2 \cos ^3(c+d x)}{3 b d}+\frac {\cos ^5(c+d x)}{5 b d} \] Output:
-1/2*a^(3/2)*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/(a^(1/2)-b ^(1/2))^(1/2)/b^(9/4)/d-1/2*a^(3/2)*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^ (1/2))^(1/2))/(a^(1/2)+b^(1/2))^(1/2)/b^(9/4)/d+(a+b)*cos(d*x+c)/b^2/d-2/3 *cos(d*x+c)^3/b/d+1/5*cos(d*x+c)^5/b/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 4.82 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.29 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\cos (c+d x) (120 a+89 b-28 b \cos (2 (c+d x))+3 b \cos (4 (c+d x)))+60 i a^2 \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{-b-8 a \text {$\#$1}^2+3 b \text {$\#$1}^2-3 b \text {$\#$1}^4+b \text {$\#$1}^6}\&\right ]}{120 b^2 d} \] Input:
Integrate[Sin[c + d*x]^9/(a - b*Sin[c + d*x]^4),x]
Output:
(Cos[c + d*x]*(120*a + 89*b - 28*b*Cos[2*(c + d*x)] + 3*b*Cos[4*(c + d*x)] ) + (60*I)*a^2*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b* #1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 + I*Log[1 - 2*Cos [c + d*x]*#1 + #1^2]*#1 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3* b*#1^4 + b*#1^6) & ])/(120*b^2*d)
Time = 0.41 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3694, 1484, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^9}{a-b \sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3694 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^4}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 1484 |
\(\displaystyle -\frac {\int \left (-\frac {\cos ^4(c+d x)}{b}+\frac {2 \cos ^2(c+d x)}{b}-\frac {a+b}{b^2}+\frac {a^2}{b^2 \left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )}\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {a^{3/2} \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{9/4} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{9/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {(a+b) \cos (c+d x)}{b^2}-\frac {\cos ^5(c+d x)}{5 b}+\frac {2 \cos ^3(c+d x)}{3 b}}{d}\) |
Input:
Int[Sin[c + d*x]^9/(a - b*Sin[c + d*x]^4),x]
Output:
-(((a^(3/2)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqr t[Sqrt[a] - Sqrt[b]]*b^(9/4)) + (a^(3/2)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sq rt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(9/4)) - ((a + b)*Cos [c + d*x])/b^2 + (2*Cos[c + d*x]^3)/(3*b) - Cos[c + d*x]^5/(5*b))/d)
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 4.79 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {\frac {\frac {b \cos \left (d x +c \right )^{5}}{5}-\frac {2 b \cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right ) a +\cos \left (d x +c \right ) b}{b^{2}}+\frac {a^{2} \left (-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{b}}{d}\) | \(137\) |
default | \(\frac {\frac {\frac {b \cos \left (d x +c \right )^{5}}{5}-\frac {2 b \cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right ) a +\cos \left (d x +c \right ) b}{b^{2}}+\frac {a^{2} \left (-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{b}}{d}\) | \(137\) |
risch | \(\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{16 b d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a \,b^{9} d^{4}-b^{10} d^{4}\right ) \textit {\_Z}^{4}-32768 a^{3} b^{5} d^{2} \textit {\_Z}^{2}-268435456 a^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {i b^{7} d^{3}}{1048576 a^{4}}+\frac {i b^{8} d^{3}}{1048576 a^{5}}\right ) \textit {\_R}^{3}+\left (\frac {i b^{2} d}{64 a}+\frac {i b^{3} d}{64 a^{2}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{512}+\frac {\cos \left (5 d x +5 c \right )}{80 b d}-\frac {5 \cos \left (3 d x +3 c \right )}{48 b d}\) | \(231\) |
Input:
int(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^2*(1/5*b*cos(d*x+c)^5-2/3*b*cos(d*x+c)^3+cos(d*x+c)*a+cos(d*x+c)* b)+a^2/b*(-1/2/(a*b)^(1/2)/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(b*cos(d*x+c)/( ((a*b)^(1/2)-b)*b)^(1/2))-1/2/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arctan h(b*cos(d*x+c)/(((a*b)^(1/2)+b)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 872 vs. \(2 (133) = 266\).
Time = 0.15 (sec) , antiderivative size = 872, normalized size of antiderivative = 4.93 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
1/60*(12*b*cos(d*x + c)^5 - 15*b^2*d*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b ^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))*log(a^5*cos(d* x + c) + (a^4*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)* d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4) )*d^2 + a^3)/((a*b^4 - b^5)*d^2))) + 15*b^2*d*sqrt(((a*b^4 - b^5)*sqrt(a^7 /((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*log(a^ 5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^1 1)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))) + 15*b^2*d*sqrt(-((a*b^4 - b^5)* sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2) )*log(-a^5*cos(d*x + c) + (a^4*b^2*d - (a*b^7 - b^8)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(-((a*b^4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a* b^10 + b^11)*d^4))*d^2 + a^3)/((a*b^4 - b^5)*d^2))) - 15*b^2*d*sqrt(((a*b^ 4 - b^5)*sqrt(a^7/((a^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))*log(-a^5*cos(d*x + c) - (a^4*b^2*d + (a*b^7 - b^8)*sqrt(a^7/((a ^2*b^9 - 2*a*b^10 + b^11)*d^4))*d^3)*sqrt(((a*b^4 - b^5)*sqrt(a^7/((a^2*b^ 9 - 2*a*b^10 + b^11)*d^4))*d^2 - a^3)/((a*b^4 - b^5)*d^2))) - 40*b*cos(d*x + c)^3 + 60*(a + b)*cos(d*x + c))/(b^2*d)
Timed out. \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**9/(a-b*sin(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{9}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
1/240*(240*b^2*d*integrate(8*(4*a^2*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 2*(8*a^3 - 3*a^2*b)*cos(3*d*x + 3*c)*sin(4*d*x + 4*c) - 2*(8*a^3 - 3*a^2*b )*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) - (a^2*b*sin(5*d*x + 5*c) - a^2*b*sin( 3*d*x + 3*c))*cos(8*d*x + 8*c) + 4*(a^2*b*sin(5*d*x + 5*c) - a^2*b*sin(3*d *x + 3*c))*cos(6*d*x + 6*c) - 2*(2*a^2*b*sin(2*d*x + 2*c) + (8*a^3 - 3*a^2 *b)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c) + (a^2*b*cos(5*d*x + 5*c) - a^2*b*c os(3*d*x + 3*c))*sin(8*d*x + 8*c) - 4*(a^2*b*cos(5*d*x + 5*c) - a^2*b*cos( 3*d*x + 3*c))*sin(6*d*x + 6*c) + (4*a^2*b*cos(2*d*x + 2*c) - a^2*b + 2*(8* a^3 - 3*a^2*b)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) - (4*a^2*b*cos(2*d*x + 2 *c) - a^2*b)*sin(3*d*x + 3*c))/(b^4*cos(8*d*x + 8*c)^2 + 16*b^4*cos(6*d*x + 6*c)^2 + 16*b^4*cos(2*d*x + 2*c)^2 + b^4*sin(8*d*x + 8*c)^2 + 16*b^4*sin (6*d*x + 6*c)^2 + 16*b^4*sin(2*d*x + 2*c)^2 - 8*b^4*cos(2*d*x + 2*c) + b^4 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*sin(4*d*x + 4*c)^2 + 16*(8*a*b^3 - 3*b^4)*sin(4*d*x + 4* c)*sin(2*d*x + 2*c) - 2*(4*b^4*cos(6*d*x + 6*c) + 4*b^4*cos(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^4*c os(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) - 4*(8*a*b^3 - 3*b^4 - 4*(8*a*b^3 - 3*b^4)*cos(2*d*x + 2*c))*cos(4*d* x + 4*c) - 4*(2*b^4*sin(6*d*x + 6*c) + 2*b^4*sin(2*d*x + 2*c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^4*sin(2*d*x + 2*c)...
\[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{9}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
sage0*x
Time = 38.48 (sec) , antiderivative size = 1067, normalized size of antiderivative = 6.03 \[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(c + d*x)^9/(a - b*sin(c + d*x)^4),x)
Output:
(atan((a^4*b^9*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3* b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7* b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) - (2*a^ 4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*cos(c + d*x)*(- (a^7*b^9)^( 1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^ 6*b^7)/(a*b^9 - b^10) + (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) + (a*b ^4*cos(c + d*x)*(- (a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a* b^9 - b^10)))^(1/2)*(a^7*b^9)^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2* a^7*b^15)/(a*b^9 - b^10) + (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) - ( 2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10)))*(-((a^7*b^9)^(1/2) + a^3*b^5) /(16*(a*b^9 - b^10)))^(1/2)*2i)/d - (atan((a^4*cos(c + d*x)*((a^7*b^9)^(1/ 2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*8i)/((2*a^6* b^7)/(a*b^9 - b^10) - (2*a^3*b^2*(a^7*b^9)^(1/2))/(a*b^9 - b^10)) - (a^4*b ^9*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^9 - b^10)) - (a^3*b^5)/(16*(a*b^ 9 - b^10)))^(1/2)*8i)/((2*a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7*b^9)^(1/2))/(a*b^9 - b^10) + (2*a^4*b^10*(a^7*b^ 9)^(1/2))/(a*b^9 - b^10)) + (a*b^4*cos(c + d*x)*((a^7*b^9)^(1/2)/(16*(a*b^ 9 - b^10)) - (a^3*b^5)/(16*(a*b^9 - b^10)))^(1/2)*(a^7*b^9)^(1/2)*8i)/((2* a^6*b^16)/(a*b^9 - b^10) - (2*a^7*b^15)/(a*b^9 - b^10) - (2*a^3*b^11*(a^7* b^9)^(1/2))/(a*b^9 - b^10) + (2*a^4*b^10*(a^7*b^9)^(1/2))/(a*b^9 - b^10...
\[ \int \frac {\sin ^9(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\sin \left (d x +c \right )^{9}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:
int(sin(d*x+c)^9/(a-b*sin(d*x+c)^4),x)
Output:
- int(sin(c + d*x)**9/(sin(c + d*x)**4*b - a),x)