Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(2 a-b) x}{2 b^2}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^2 \sqrt {a+b} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \]
-1/2*(2*a-b)*x/b^2-1/2*cos(d*x+c)*sin(d*x+c)/b/d+a^(3/2)*arctan((a+b)^(1/2 )*tan(d*x+c)/a^(1/2))/b^2/d/(a+b)^(1/2)
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {2 (2 a-b) (c+d x)-\frac {4 a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+b \sin (2 (c+d x))}{4 b^2 d} \]
-1/4*(2*(2*a - b)*(c + d*x) - (4*a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x]) /Sqrt[a]])/Sqrt[a + b] + b*Sin[2*(c + d*x)])/(b^2*d)
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3666, 372, 397, 216, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{a+b \sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2 \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {a-(a-b) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 b}-\frac {\tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {(2 a-b) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{b}}{2 b}-\frac {\tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {(2 a-b) \arctan (\tan (c+d x))}{b}}{2 b}-\frac {\tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a+b}}-\frac {(2 a-b) \arctan (\tan (c+d x))}{b}}{2 b}-\frac {\tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}}{d}\) |
((-(((2*a - b)*ArcTan[Tan[c + d*x]])/b) + (2*a^(3/2)*ArcTan[(Sqrt[a + b]*T an[c + d*x])/Sqrt[a]])/(b*Sqrt[a + b]))/(2*b) - Tan[c + d*x]/(2*b*(1 + Tan [c + d*x]^2)))/d
3.1.87.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Time = 0.71 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\tan \left (d x +c \right ) b}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(81\) |
| default | \(\frac {-\frac {\frac {\tan \left (d x +c \right ) b}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{b^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(81\) |
| risch | \(-\frac {a x}{b^{2}}+\frac {x}{2 b}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right ) d \,b^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right ) d \,b^{2}}\) | \(160\) |
1/d*(-1/b^2*(1/2*b*tan(d*x+c)/(1+tan(d*x+c)^2)+1/2*(2*a-b)*arctan(tan(d*x+ c)))+a^2/b^2/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
Time = 0.28 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.96 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {2 \, {\left (2 \, a - b\right )} d x + 2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{4 \, b^{2} d}, -\frac {{\left (2 \, a - b\right )} d x + b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, b^{2} d}\right ] \]
[-1/4*(2*(2*a - b)*d*x + 2*b*cos(d*x + c)*sin(d*x + c) - a*sqrt(-a/(a + b) )*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos( d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2) *cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos (d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)))/(b^2*d), -1/2*((2*a - b)*d*x + b*cos(d*x + c)*sin(d*x + c) + a*sqrt(a/(a + b))*arc tan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c) *sin(d*x + c))))/(b^2*d)]
Timed out. \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\text {Timed out} \]
Time = 0.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, a^{2} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{2}} - \frac {{\left (d x + c\right )} {\left (2 \, a - b\right )}}{b^{2}} - \frac {\tan \left (d x + c\right )}{b \tan \left (d x + c\right )^{2} + b}}{2 \, d} \]
1/2*(2*a^2*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*b ^2) - (d*x + c)*(2*a - b)/b^2 - tan(d*x + c)/(b*tan(d*x + c)^2 + b))/d
Time = 0.43 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{2}}{\sqrt {a^{2} + a b} b^{2}} - \frac {{\left (d x + c\right )} {\left (2 \, a - b\right )}}{b^{2}} - \frac {\tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} b}}{2 \, d} \]
1/2*(2*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))*a^2/(sqrt(a^2 + a*b)*b^2) - (d*x + c)*(2*a - b)/b^2 - tan(d*x + c)/((tan(d*x + c)^2 + 1)*b))/d
Time = 14.14 (sec) , antiderivative size = 481, normalized size of antiderivative = 6.25 \[ \int \frac {\sin ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {\sin \left (c+d\,x\right )}{\cos \left (c+d\,x\right )}\right )}{d\,\left (2\,b^3+2\,a\,b^2\right )}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (c+d\,x\right )}{\cos \left (c+d\,x\right )}\right )}{d\,\left (2\,b^3+2\,a\,b^2\right )}-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\left (2\,b^3+2\,a\,b^2\right )}-\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\left (2\,b^3+2\,a\,b^2\right )}-\frac {a\,b\,\mathrm {atan}\left (\frac {\sin \left (c+d\,x\right )}{\cos \left (c+d\,x\right )}\right )}{d\,\left (2\,b^3+2\,a\,b^2\right )}-\frac {\mathrm {atan}\left (\frac {a\,\sin \left (c+d\,x\right )\,{\left (-a^4-b\,a^3\right )}^{3/2}\,8{}\mathrm {i}+b\,\sin \left (c+d\,x\right )\,{\left (-a^4-b\,a^3\right )}^{3/2}\,4{}\mathrm {i}+a^5\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,8{}\mathrm {i}+b^5\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,1{}\mathrm {i}-a\,b^4\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,1{}\mathrm {i}+a^4\,b\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,12{}\mathrm {i}-a^2\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,5{}\mathrm {i}+a^3\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {-a^4-b\,a^3}\,1{}\mathrm {i}}{3\,\cos \left (c+d\,x\right )\,a^5\,b^2+5\,\cos \left (c+d\,x\right )\,a^4\,b^3+\cos \left (c+d\,x\right )\,a^3\,b^4-\cos \left (c+d\,x\right )\,a^2\,b^5}\right )\,\sqrt {-a^4-b\,a^3}\,2{}\mathrm {i}}{d\,\left (2\,b^3+2\,a\,b^2\right )} \]
(b^2*atan(sin(c + d*x)/cos(c + d*x)))/(d*(2*a*b^2 + 2*b^3)) - (2*a^2*atan( sin(c + d*x)/cos(c + d*x)))/(d*(2*a*b^2 + 2*b^3)) - (b^2*sin(2*c + 2*d*x)) /(2*d*(2*a*b^2 + 2*b^3)) - (atan((a*sin(c + d*x)*(- a^3*b - a^4)^(3/2)*8i + b*sin(c + d*x)*(- a^3*b - a^4)^(3/2)*4i + a^5*sin(c + d*x)*(- a^3*b - a^ 4)^(1/2)*8i + b^5*sin(c + d*x)*(- a^3*b - a^4)^(1/2)*1i - a*b^4*sin(c + d* x)*(- a^3*b - a^4)^(1/2)*1i + a^4*b*sin(c + d*x)*(- a^3*b - a^4)^(1/2)*12i - a^2*b^3*sin(c + d*x)*(- a^3*b - a^4)^(1/2)*5i + a^3*b^2*sin(c + d*x)*(- a^3*b - a^4)^(1/2)*1i)/(a^3*b^4*cos(c + d*x) - a^2*b^5*cos(c + d*x) + 5*a ^4*b^3*cos(c + d*x) + 3*a^5*b^2*cos(c + d*x)))*(- a^3*b - a^4)^(1/2)*2i)/( d*(2*a*b^2 + 2*b^3)) - (a*b*sin(2*c + 2*d*x))/(2*d*(2*a*b^2 + 2*b^3)) - (a *b*atan(sin(c + d*x)/cos(c + d*x)))/(d*(2*a*b^2 + 2*b^3))