Integrand size = 15, antiderivative size = 115 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}} d} \]
1/2*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/d/(a^(1/2)- b^(1/2))^(1/2)+1/2*arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3 /4)/d/(a^(1/2)+b^(1/2))^(1/2)
Time = 0.97 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {\arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}}{2 \sqrt {a} d} \]
(ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]]/Sqrt [a + Sqrt[a]*Sqrt[b]] - ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/Sqrt[-a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[a]*d)
Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, 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.
| \(\displaystyle \int \frac {1}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3688 |
| \(\displaystyle \frac {\int \frac {\tan ^2(c+d x)+1}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {1}{2} \left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)+\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \left (\sqrt {a}+\sqrt {b}\right )}+\frac {\left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {\sqrt {a}+\sqrt {b}}}}{d}\) |
(((1 + Sqrt[b]/Sqrt[a])*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1 /4)])/(2*a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*(Sqrt[a] + Sqrt[b])) + ((1 - Sqrt [b]/Sqrt[a])*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^ (1/4)*(Sqrt[a] - Sqrt[b])*Sqrt[Sqrt[a] + Sqrt[b]]))/d
3.3.7.3.1 Defintions of rubi rules used
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 = 0.57 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11
| method | result | size |
| risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (256 a^{4} d^{4}-256 a^{3} b \,d^{4}\right ) \textit {\_Z}^{4}+32 a^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {128 i d^{3} a^{4}}{b}-128 i a^{3} d^{3}\right ) \textit {\_R}^{3}+\left (-\frac {32 d^{2} a^{3}}{b}+32 a^{2} d^{2}\right ) \textit {\_R}^{2}+\left (\frac {8 i a^{2} d}{b}+8 i a d \right ) \textit {\_R} -\frac {2 a}{b}-1\right )\) | \(128\) |
| derivativedivides | \(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\) | \(145\) |
| default | \(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\) | \(145\) |
sum(_R*ln(exp(2*I*(d*x+c))+(128*I/b*d^3*a^4-128*I*a^3*d^3)*_R^3+(-32/b*d^2 *a^3+32*a^2*d^2)*_R^2+(8*I/b*a^2*d+8*I*a*d)*_R-2/b*a-1),_R=RootOf(1+(256*a ^4*d^4-256*a^3*b*d^4)*_Z^4+32*a^2*d^2*_Z^2))
Leaf count of result is larger than twice the leaf count of optimal. 1079 vs. \(2 (79) = 158\).
Time = 0.41 (sec) , antiderivative size = 1079, normalized size of antiderivative = 9.38 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
1/8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/(( a^2 - a*b)*d^2))*log(1/4*b*cos(d*x + c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/ ((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^ 2)*d^4)) + 1)/((a^2 - a*b)*d^2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4*b) - 1 /8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a ^2 - a*b)*d^2))*log(1/4*b*cos(d*x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/( (a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2 )*d^4)) + 1)/((a^2 - a*b)*d^2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4*b) + 1/ 8*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*cos(d*x + c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/(( a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) + a*b*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)* d^4)) - 1)/((a^2 - a*b)*d^2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1/4*b) - 1/8* sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*cos(d*x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/(...
Timed out. \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {1}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (79) = 158\).
Time = 0.45 (sec) , antiderivative size = 361, normalized size of antiderivative = 3.14 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}}}{2 \, d} \]
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((d* x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/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) + (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*flo or((d*x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/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))/d
Time = 15.42 (sec) , antiderivative size = 671, normalized size of antiderivative = 5.83 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a^3\,\mathrm {tan}\left (c+d\,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 (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-a^2\,b\,\mathrm {tan}\left (c+d\,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}}{d}+\frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a^3\,\mathrm {tan}\left (c+d\,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 (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,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}}{d} \]
(atan((a^3*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*4i + a^5*ta n(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*64i + a^3*tan(c + d*x)*( -1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i + a^2*b*tan(c + d* x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*4i - a^4*b*tan(c + d*x)*(-1/(16* a^2 + 16*(a^3*b)^(1/2)))^(3/2)*64i + a*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3* b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i + b*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b )^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - a^2*b*tan(c + d*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)/d + (atan((a^3*tan(c + d*x)*(-1/(16*a^2 - 16 *(a^3*b)^(1/2)))^(1/2)*4i + a^5*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2 )))^(3/2)*64i - a^3*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*(a ^3*b)^(1/2)*64i + a^2*b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2 )*4i - a^4*b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*64i - a*t an(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - b*ta n(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i + a^2*b *tan(c + d*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)/d