Integrand size = 11, antiderivative size = 83 \[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\frac {\arctan \left (\sqrt {1+\sqrt {\frac {b}{a}}} \tan (x)\right )}{2 a \sqrt {1+\sqrt {\frac {b}{a}}}}+\frac {\text {arctanh}\left (\sqrt {-1+\sqrt {\frac {b}{a}}} \tan (x)\right )}{2 a \sqrt {-1+\sqrt {\frac {b}{a}}}} \] Output:
1/2*arctan((1+(b/a)^(1/2))^(1/2)*tan(x))/a/(1+(b/a)^(1/2))^(1/2)+1/2*arcta nh((-1+(b/a)^(1/2))^(1/2)*tan(x))/a/(-1+(b/a)^(1/2))^(1/2)
Time = 0.78 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.59 \[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\frac {\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {a+\sqrt {a} \sqrt {b}} \arctan \left (\frac {\sqrt {a+\sqrt {a} \sqrt {b}} \tan (x)}{\sqrt {a}}\right )-\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {-a+\sqrt {a} \sqrt {b}} \text {arctanh}\left (\frac {\sqrt {-a+\sqrt {a} \sqrt {b}} \tan (x)}{\sqrt {a}}\right )}{2 a (a-b)} \] Input:
Integrate[(a - b*Sin[x]^4)^(-1),x]
Output:
((Sqrt[a] - Sqrt[b])*Sqrt[a + Sqrt[a]*Sqrt[b]]*ArcTan[(Sqrt[a + Sqrt[a]*Sq rt[b]]*Tan[x])/Sqrt[a]] - (Sqrt[a] + Sqrt[b])*Sqrt[-a + Sqrt[a]*Sqrt[b]]*A rcTanh[(Sqrt[-a + Sqrt[a]*Sqrt[b]]*Tan[x])/Sqrt[a]])/(2*a*(a - b))
Time = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3688, 1480, 218}
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.
| \(\displaystyle \int \frac {1}{a-b \sin ^4(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a-b \sin (x)^4}dx\) |
| \(\Big \downarrow \) 3688 |
| \(\displaystyle \int \frac {\tan ^2(x)+1}{(a-b) \tan ^4(x)+2 a \tan ^2(x)+a}d\tan (x)\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {1}{2} \left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \int \frac {1}{(a-b) \tan ^2(x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (x)+\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \int \frac {1}{(a-b) \tan ^2(x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (x)\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \left (\sqrt {a}+\sqrt {b}\right )}+\frac {\left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {\sqrt {a}+\sqrt {b}}}\) |
Input:
Int[(a - b*Sin[x]^4)^(-1),x]
Output:
((1 + Sqrt[b]/Sqrt[a])*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[x])/a^(1/4)])/( 2*a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*(Sqrt[a] + Sqrt[b])) + ((1 - Sqrt[b]/Sqr t[a])*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[x])/a^(1/4)])/(2*a^(1/4)*(Sqrt[a ] - Sqrt[b])*Sqrt[Sqrt[a] + Sqrt[b]])
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[((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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + 2*a*ff^2*x^2 + ( a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x]] / ; FreeQ[{a, b, e, f}, x] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22
| method | result | size |
| risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (256 a^{4}-256 a^{3} b \right ) \textit {\_Z}^{4}+32 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (\frac {128 i a^{4}}{b}-128 i a^{3}\right ) \textit {\_R}^{3}+\left (-\frac {32 a^{3}}{b}+32 a^{2}\right ) \textit {\_R}^{2}+\left (\frac {8 i a^{2}}{b}+8 i a \right ) \textit {\_R} -\frac {2 a}{b}-1\right )\) | \(101\) |
| default | \(\left (a -b \right ) \left (\frac {\left (b +\sqrt {b a}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (x \right )}{\sqrt {\left (\sqrt {b a}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {b a}\, \left (a -b \right ) \sqrt {\left (\sqrt {b a}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {b a}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (x \right )}{\sqrt {\left (\sqrt {b a}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {b a}\, \left (a -b \right ) \sqrt {\left (\sqrt {b a}-a \right ) \left (a -b \right )}}\right )\) | \(134\) |
Input:
int(1/(a-b*sin(x)^4),x,method=_RETURNVERBOSE)
Output:
sum(_R*ln(exp(2*I*x)+(128*I/b*a^4-128*I*a^3)*_R^3+(-32/b*a^3+32*a^2)*_R^2+ (8*I/b*a^2+8*I*a)*_R-2/b*a-1),_R=RootOf(1+(256*a^4-256*a^3*b)*_Z^4+32*a^2* _Z^2))
Leaf count of result is larger than twice the leaf count of optimal. 835 vs. \(2 (63) = 126\).
Time = 0.18 (sec) , antiderivative size = 835, normalized size of antiderivative = 10.06 \[ \int \frac {1}{a-b \sin ^4(x)} \, dx =\text {Too large to display} \] Input:
integrate(1/(a-b*sin(x)^4),x, algorithm="fricas")
Output:
-1/8*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1)/(a^2 - a*b) )*log(1/4*b*cos(x)^2 + 1/2*(a*b*cos(x)*sin(x) - (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^ 4*b + a^3*b^2)) + 1)/(a^2 - a*b)) + 1/4*(a^3 - a^2*b - 2*(a^3 - a^2*b)*cos (x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1/4*b) + 1/8*sqrt(-((a^2 - a*b) *sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1)/(a^2 - a*b))*log(1/4*b*cos(x)^2 - 1/2*(a*b*cos(x)*sin(x) - (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*c os(x)*sin(x))*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1)/(a ^2 - a*b)) + 1/4*(a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2* a^4*b + a^3*b^2)) - 1/4*b) + 1/8*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1)/(a^2 - a*b))*log(-1/4*b*cos(x)^2 + 1/2*(a*b*cos(x)*sin(x) + (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(((a^ 2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1)/(a^2 - a*b)) + 1/4*(a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1/4* b) - 1/8*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1)/(a^2 - a *b))*log(-1/4*b*cos(x)^2 - 1/2*(a*b*cos(x)*sin(x) + (a^4 - a^3*b)*sqrt(b/( a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2 *a^4*b + a^3*b^2)) - 1)/(a^2 - a*b)) + 1/4*(a^3 - a^2*b - 2*(a^3 - a^2*b)* cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1/4*b)
Timed out. \[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\text {Timed out} \] Input:
integrate(1/(a-b*sin(x)**4),x)
Output:
Timed out
\[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\int { -\frac {1}{b \sin \left (x\right )^{4} - a} \,d x } \] Input:
integrate(1/(a-b*sin(x)^4),x, algorithm="maxima")
Output:
-integrate(1/(b*sin(x)^4 - a), x)
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (63) = 126\).
Time = 0.50 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.12 \[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{2 \, {\left (3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}\right )}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{2 \, {\left (3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}\right )}} \] Input:
integrate(1/(a-b*sin(x)^4),x, algorithm="giac")
Output:
1/2*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2 - 6*sqrt(a^2 - a*b + sqrt(a *b)*(a - b))*a*b - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*b^2)*(pi*floor(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*a + sqrt(-16*(a - b)*a + 16*a^2))/(a - b ))))*abs(a - b)/(3*a^5 - 12*a^4*b + 14*a^3*b^2 - 4*a^2*b^3 - a*b^4) + 1/2* (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2 - 6*sqrt(a^2 - a*b - sqrt(a*b)* (a - b))*a*b - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^2)*(pi*floor(x/pi + 1 /2) + arctan(2*tan(x)/sqrt((4*a - sqrt(-16*(a - b)*a + 16*a^2))/(a - b)))) *abs(a - b)/(3*a^5 - 12*a^4*b + 14*a^3*b^2 - 4*a^2*b^3 - a*b^4)
Time = 37.81 (sec) , antiderivative size = 601, normalized size of antiderivative = 7.24 \[ \int \frac {1}{a-b \sin ^4(x)} \, dx=\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+b\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+a^3\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}-a^2\,b\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b+\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-b\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-a^3\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b-\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i} \] Input:
int(1/(a - b*sin(x)^4),x)
Output:
atan((a^3*tan(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*4i + a^5*tan(x)*(- 1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*64i + a^2*b*tan(x)*(-1/(16*a^2 + 16*( a^3*b)^(1/2)))^(1/2)*4i - a^4*b*tan(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3 /2)*64i + a*tan(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i + b*tan(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i + a^3* tan(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i - a^2*b*ta n(x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i)/(a*b + (a^3 *b)^(1/2)))*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*2i + atan((a^3*tan(x)*( -1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*4i + a^5*tan(x)*(-1/(16*a^2 - 16*(a^ 3*b)^(1/2)))^(3/2)*64i + a^2*b*tan(x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/ 2)*4i - a^4*b*tan(x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*64i - a*tan(x) *(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - b*tan(x)*(-1/(1 6*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - a^3*tan(x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i + a^2*b*tan(x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i)/(a*b - (a^3*b)^(1/2)))*(-1/(1 6*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*2i
\[ \int \frac {1}{a-b \sin ^4(x)} \, dx=-\left (\int \frac {1}{\sin \left (x \right )^{4} b -a}d x \right ) \] Input:
int(1/(a-b*sin(x)^4),x)
Output:
- int(1/(sin(x)**4*b - a),x)