Integrand size = 24, antiderivative size = 115 \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{3/4} d}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{3/4} d} \] Output:
-1/2*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^(3/4)/d+1/2*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^(3/4)/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 6.74 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.48 \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+6 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-6 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{8 d} \] Input:
Integrate[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]
Output:
((-1/8*I)*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)] + I*Log[1 - 2*Cos[c + d*x ]*#1 + #1^2] + 6*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - (3*I)*Log [1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 6*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + (3*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + 2*ArcTan[Sin[ c + d*x]/(Cos[c + d*x] - #1)]*#1^6 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*# 1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) & ])/d
Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3694, 1480, 218, 221}
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.
| \(\displaystyle \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3694 |
| \(\displaystyle -\frac {\int \frac {1-\cos ^2(c+d x)}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {-\frac {1}{2} \int \frac {1}{-b \cos ^2(c+d x)-\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {b}}d\cos (c+d x)-\frac {1}{2} \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {\arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {1}{2} \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}}}{d}\) |
Input:
Int[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]
Output:
-((ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]]/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/4)) - ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b ]]]/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/4)))/d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
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 = 0.71 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(\frac {b \left (\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\) | \(83\) |
| default | \(\frac {b \left (\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\) | \(83\) |
| risch | \(\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (-16+\left (a \,b^{3} d^{4}-b^{4} d^{4}\right ) \textit {\_Z}^{4}-8 b^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {1}{4} i a \,b^{2} d^{3}+\frac {1}{4} i b^{3} d^{3}\right ) \textit {\_R}^{3}+2 i b d \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{8}\) | \(97\) |
Input:
int(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/d*b*(1/2/b/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/(((a*b)^(1/2)+ b)*b)^(1/2))-1/2/b/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(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. 703 vs. \(2 (79) = 158\).
Time = 0.11 (sec) , antiderivative size = 703, normalized size of antiderivative = 6.11 \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
1/4*sqrt(-((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) + 1)/(( a*b - b^2)*d^2))*log(-((a*b^2 - b^3)*d^3*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5) *d^4)) - b*d)*sqrt(-((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4 )) + 1)/((a*b - b^2)*d^2)) + cos(d*x + c)) - 1/4*sqrt(-((a*b - b^2)*d^2*sq rt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) + 1)/((a*b - b^2)*d^2))*log(-((a*b^2 - b^3)*d^3*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) - b*d)*sqrt(-((a*b - b ^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) + 1)/((a*b - b^2)*d^2)) - cos(d*x + c)) - 1/4*sqrt(((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5 )*d^4)) - 1)/((a*b - b^2)*d^2))*log(-((a*b^2 - b^3)*d^3*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) + b*d)*sqrt(((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a* b^4 + b^5)*d^4)) - 1)/((a*b - b^2)*d^2)) + cos(d*x + c)) + 1/4*sqrt(((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) - 1)/((a*b - b^2)*d^2)) *log(-((a*b^2 - b^3)*d^3*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) + b*d)*sq rt(((a*b - b^2)*d^2*sqrt(a/((a^2*b^3 - 2*a*b^4 + b^5)*d^4)) - 1)/((a*b - b ^2)*d^2)) - cos(d*x + c))
Timed out. \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3/(a-b*sin(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
-integrate(sin(d*x + c)^3/(b*sin(d*x + c)^4 - a), x)
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (79) = 158\).
Time = 0.66 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\sqrt {-b^{2} - \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} + \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (b + \sqrt {a b}\right )} d {\left | b \right |}} + \frac {\sqrt {-b^{2} + \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} - \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (b - \sqrt {a b}\right )} d {\left | b \right |}} \] Input:
integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
1/2*sqrt(-b^2 - sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(-(b*d^2 + sqrt((a - b)*b*d^4 + b^2*d^4))/(b*d^4))))/((b + sqrt(a*b))*d*abs(b)) + 1/2*sqrt(- b^2 + sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(-(b*d^2 - sqrt((a - b)*b*d^ 4 + b^2*d^4))/(b*d^4))))/((b - sqrt(a*b))*d*abs(b))
Time = 0.52 (sec) , antiderivative size = 976, normalized size of antiderivative = 8.49 \[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {8\,a\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b+\frac {2\,a\,b^5}{a\,b^3-b^4}-\frac {2\,a\,b^3\,\sqrt {a\,b^3}}{a\,b^3-b^4}}-\frac {8\,a\,b^6\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}+\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}-\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {a\,b^3}\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}+\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}-\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}\right )\,\sqrt {-\frac {b^2-\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{d}-\frac {2\,\mathrm {atanh}\left (\frac {8\,a\,b^6\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}-\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}+\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}-\frac {8\,a\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b+\frac {2\,a\,b^5}{a\,b^3-b^4}+\frac {2\,a\,b^3\,\sqrt {a\,b^3}}{a\,b^3-b^4}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {a\,b^3}\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}-\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}+\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}\right )\,\sqrt {-\frac {b^2+\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{d} \] Input:
int(sin(c + d*x)^3/(a - b*sin(c + d*x)^4),x)
Output:
(2*atanh((8*a*b^2*cos(c + d*x)*((a*b^3)^(1/2)/(16*(a*b^3 - b^4)) - b^2/(16 *(a*b^3 - b^4)))^(1/2))/(2*a*b + (2*a*b^5)/(a*b^3 - b^4) - (2*a*b^3*(a*b^3 )^(1/2))/(a*b^3 - b^4)) - (8*a*b^6*cos(c + d*x)*((a*b^3)^(1/2)/(16*(a*b^3 - b^4)) - b^2/(16*(a*b^3 - b^4)))^(1/2))/(2*a*b^5 - 2*a^2*b^4 - (2*a^2*b^8 )/(a*b^3 - b^4) + (2*a*b^9)/(a*b^3 - b^4) + (2*a^2*b^6*(a*b^3)^(1/2))/(a*b ^3 - b^4) - (2*a*b^7*(a*b^3)^(1/2))/(a*b^3 - b^4)) + (8*a*b^4*cos(c + d*x) *(a*b^3)^(1/2)*((a*b^3)^(1/2)/(16*(a*b^3 - b^4)) - b^2/(16*(a*b^3 - b^4))) ^(1/2))/(2*a*b^5 - 2*a^2*b^4 - (2*a^2*b^8)/(a*b^3 - b^4) + (2*a*b^9)/(a*b^ 3 - b^4) + (2*a^2*b^6*(a*b^3)^(1/2))/(a*b^3 - b^4) - (2*a*b^7*(a*b^3)^(1/2 ))/(a*b^3 - b^4)))*(-(b^2 - (a*b^3)^(1/2))/(16*(a*b^3 - b^4)))^(1/2))/d - (2*atanh((8*a*b^6*cos(c + d*x)*(- b^2/(16*(a*b^3 - b^4)) - (a*b^3)^(1/2)/( 16*(a*b^3 - b^4)))^(1/2))/(2*a*b^5 - 2*a^2*b^4 - (2*a^2*b^8)/(a*b^3 - b^4) + (2*a*b^9)/(a*b^3 - b^4) - (2*a^2*b^6*(a*b^3)^(1/2))/(a*b^3 - b^4) + (2* a*b^7*(a*b^3)^(1/2))/(a*b^3 - b^4)) - (8*a*b^2*cos(c + d*x)*(- b^2/(16*(a* b^3 - b^4)) - (a*b^3)^(1/2)/(16*(a*b^3 - b^4)))^(1/2))/(2*a*b + (2*a*b^5)/ (a*b^3 - b^4) + (2*a*b^3*(a*b^3)^(1/2))/(a*b^3 - b^4)) + (8*a*b^4*cos(c + d*x)*(a*b^3)^(1/2)*(- b^2/(16*(a*b^3 - b^4)) - (a*b^3)^(1/2)/(16*(a*b^3 - b^4)))^(1/2))/(2*a*b^5 - 2*a^2*b^4 - (2*a^2*b^8)/(a*b^3 - b^4) + (2*a*b^9) /(a*b^3 - b^4) - (2*a^2*b^6*(a*b^3)^(1/2))/(a*b^3 - b^4) + (2*a*b^7*(a*b^3 )^(1/2))/(a*b^3 - b^4)))*(-(b^2 + (a*b^3)^(1/2))/(16*(a*b^3 - b^4)))^(1...
\[ \int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\sin \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:
int(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x)
Output:
- int(sin(c + d*x)**3/(sin(c + d*x)**4*b - a),x)