Integrand size = 23, antiderivative size = 93 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {x}{b^2}-\frac {\sqrt {a} (2 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^2 (a+b)^{3/2} d}+\frac {a \tan (c+d x)}{2 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
x/b^2-1/2*(2*a+3*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))*a^(1/2)/b^2/(a+ b)^(3/2)/d+1/2*a*tan(d*x+c)/b/(a+b)/d/(a+(a+b)*tan(d*x+c)^2)
Time = 11.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {2 (c+d x)-\frac {\sqrt {a} (2 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2}}+\frac {a b \sin (2 (c+d x))}{(a+b) (2 a+b-b \cos (2 (c+d x)))}}{2 b^2 d} \]
(2*(c + d*x) - (Sqrt[a]*(2*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt [a]])/(a + b)^(3/2) + (a*b*Sin[2*(c + d*x)])/((a + b)*(2*a + b - b*Cos[2*( c + d*x)])))/(2*b^2*d)
Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{\left (a+b \sin (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {a \tan (c+d x)}{2 b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\int \frac {a-(a+2 b) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 b (a+b)}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {a \tan (c+d x)}{2 b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\frac {a (2 a+3 b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {2 (a+b) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{b}}{2 b (a+b)}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {a \tan (c+d x)}{2 b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\frac {a (2 a+3 b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {2 (a+b) \arctan (\tan (c+d x))}{b}}{2 b (a+b)}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {a \tan (c+d x)}{2 b (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\frac {\sqrt {a} (2 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a+b}}-\frac {2 (a+b) \arctan (\tan (c+d x))}{b}}{2 b (a+b)}}{d}\) |
(-1/2*((-2*(a + b)*ArcTan[Tan[c + d*x]])/b + (Sqrt[a]*(2*a + 3*b)*ArcTan[( Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b*Sqrt[a + b]))/(b*(a + b)) + (a*Tan[ c + d*x])/(2*b*(a + b)*(a + (a + b)*Tan[c + d*x]^2)))/d
3.2.1.3.1 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 = 0.70 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {-\frac {a \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{2}}}{d}\) | \(101\) |
| default | \(\frac {-\frac {a \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{2}}}{d}\) | \(101\) |
| risch | \(\frac {x}{b^{2}}-\frac {i a \left (2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}{b^{2} \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}+\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{2} d \,b^{2}}+\frac {3 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{4 \left (a +b \right )^{2} d b}-\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{2} d \,b^{2}}-\frac {3 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{4 \left (a +b \right )^{2} d b}\) | \(307\) |
1/d*(-a/b^2*(-1/2*b/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+1/2 *(2*a+3*b)/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))) +1/b^2*arctan(tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (81) = 162\).
Time = 0.33 (sec) , antiderivative size = 492, normalized size of antiderivative = 5.29 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [\frac {8 \, {\left (a b + b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 5 \, a b - 3 \, b^{2}\right )} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{8 \, {\left ({\left (a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d\right )}}, \frac {4 \, {\left (a b + b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 5 \, a b - 3 \, b^{2}\right )} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left ({\left (a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d\right )}}\right ] \]
[1/8*(8*(a*b + b^2)*d*x*cos(d*x + c)^2 - 4*a*b*cos(d*x + c)*sin(d*x + c) - 8*(a^2 + 2*a*b + b^2)*d*x + ((2*a*b + 3*b^2)*cos(d*x + c)^2 - 2*a^2 - 5*a *b - 3*b^2)*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)))/((a*b^3 + b^4)*d*cos(d*x + c)^2 - (a^2*b^2 + 2*a*b^3 + b^4)*d), 1/4*(4*(a*b + b^2)*d*x*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d* x + c) - 4*(a^2 + 2*a*b + b^2)*d*x + ((2*a*b + 3*b^2)*cos(d*x + c)^2 - 2*a ^2 - 5*a*b - 3*b^2)*sqrt(a/(a + b))*arctan(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))))/((a*b^3 + b^4)*d*c os(d*x + c)^2 - (a^2*b^2 + 2*a*b^3 + b^4)*d)]
Timed out. \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {a \tan \left (d x + c\right )}{a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {{\left (2 \, a^{2} + 3 \, a b\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {2 \, {\left (d x + c\right )}}{b^{2}}}{2 \, d} \]
1/2*(a*tan(d*x + c)/(a^2*b + a*b^2 + (a^2*b + 2*a*b^2 + b^3)*tan(d*x + c)^ 2) - (2*a^2 + 3*a*b)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a*b^2 + b^3)*sqrt((a + b)*a)) + 2*(d*x + c)/b^2)/d
Time = 0.39 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (2 \, a^{2} + 3 \, a b\right )}}{{\left (a b^{2} + b^{3}\right )} \sqrt {a^{2} + a b}} - \frac {a \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a b + b^{2}\right )}} - \frac {2 \, {\left (d x + c\right )}}{b^{2}}}{2 \, d} \]
-1/2*((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)))*(2*a^2 + 3*a*b)/((a*b^2 + b^3)*sqrt( a^2 + a*b)) - a*tan(d*x + c)/((a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a *b + b^2)) - 2*(d*x + c)/b^2)/d
Time = 15.21 (sec) , antiderivative size = 1959, normalized size of antiderivative = 21.06 \[ \int \frac {\sin ^4(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
(a*tan(c + d*x))/(2*d*(a + tan(c + d*x)^2*(a + b))*(a*b + b^2)) - atan(((( ((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) - (tan(c + d*x)*( 80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b^2 + b^3)))/(2*b^2) + (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 3 3*a^2*b^2))/(4*(a*b^2 + b^3)))/b^2 - ((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)* 1i)/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 1 12*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b^2 + b^3)))/(2*b^2) - (tan(c + d*x)*( 16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(4*(a*b^2 + b^3)))/b^2) /((3*a*b^2 + (7*a^2*b)/2 + a^3)/(a*b^3 + b^4) + (((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) - (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144 *a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b^2 + b^3)))*1i)/(2*b^2) + (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2)*1i)/(4*( a*b^2 + b^3)))/b^2 + (((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32 *a^4*b^4))/(8*b^2*(a*b^2 + b^3)))*1i)/(2*b^2) - (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2)*1i)/(4*(a*b^2 + b^3)))/b^2))/(b^2*d ) - (atan((((-a*(a + b)^3)^(1/2)*((tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a ^4 + 4*b^4 + 33*a^2*b^2))/(2*(a*b^2 + b^3)) - (((2*a*b^6 + 4*a^2*b^5 + 2*a ^3*b^4)/(a*b^3 + b^4) - (tan(c + d*x)*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*(80 *a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*(a*b^2 + ...