Integrand size = 24, antiderivative size = 347 \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {\left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {b \tan (c+d x) \left (a (a+3 b)+\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )^2}-\frac {\tan (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}+\frac {5 \left (2 a^2+3 a b-b^2\right ) \tan ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \]
1/64*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(12*a+5*b-14*a^(1/ 2)*b^(1/2))/a^(9/4)/d/(a^(1/2)-b^(1/2))^(5/2)/b^(1/2)-1/64*arctan((a^(1/2) +b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(12*a+5*b+14*a^(1/2)*b^(1/2))/a^(9/4)/ d/b^(1/2)/(a^(1/2)+b^(1/2))^(5/2)-1/8*b*tan(d*x+c)*(a*(a+3*b)+(a^2+6*a*b+b ^2)*tan(d*x+c)^2)/a/(a-b)^3/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)^2-1/ 32*tan(d*x+c)*(2*a*(5*a^2-9*a*b-4*b^2)/(a-b)^3+5*(2*a^2+3*a*b-b^2)*tan(d*x +c)^2/(a-b)^2)/a^2/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)
Time = 7.17 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\left (-12 a^{5/2} \sqrt {b}+10 a^2 b+11 a^{3/2} b^{3/2}-4 a b^2-5 \sqrt {a} b^{5/2}\right ) \arctan \left (\frac {\left (\sqrt {a} \sqrt {b}+b\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b}}\right )}{64 a^{5/2} \sqrt {a+\sqrt {a} \sqrt {b}} (a-b)^2 b d}-\frac {\left (12 a^{5/2} \sqrt {b}+10 a^2 b-11 a^{3/2} b^{3/2}-4 a b^2+5 \sqrt {a} b^{5/2}\right ) \text {arctanh}\left (\frac {\left (\sqrt {a} \sqrt {b}-b\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b}}\right )}{64 a^{5/2} \sqrt {-a+\sqrt {a} \sqrt {b}} (a-b)^2 b d}+\frac {-4 a \sin (2 (c+d x))-2 b \sin (2 (c+d x))+b \sin (4 (c+d x))}{a (a-b) d (-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}+\frac {24 a^2 \sin (2 (c+d x))+22 a b \sin (2 (c+d x))-10 b^2 \sin (2 (c+d x))-11 a b \sin (4 (c+d x))+5 b^2 \sin (4 (c+d x))}{32 a^2 (a-b)^2 d (-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))} \]
((-12*a^(5/2)*Sqrt[b] + 10*a^2*b + 11*a^(3/2)*b^(3/2) - 4*a*b^2 - 5*Sqrt[a ]*b^(5/2))*ArcTan[((Sqrt[a]*Sqrt[b] + b)*Tan[c + d*x])/(Sqrt[a + Sqrt[a]*S qrt[b]]*Sqrt[b])])/(64*a^(5/2)*Sqrt[a + Sqrt[a]*Sqrt[b]]*(a - b)^2*b*d) - ((12*a^(5/2)*Sqrt[b] + 10*a^2*b - 11*a^(3/2)*b^(3/2) - 4*a*b^2 + 5*Sqrt[a] *b^(5/2))*ArcTanh[((Sqrt[a]*Sqrt[b] - b)*Tan[c + d*x])/(Sqrt[-a + Sqrt[a]* Sqrt[b]]*Sqrt[b])])/(64*a^(5/2)*Sqrt[-a + Sqrt[a]*Sqrt[b]]*(a - b)^2*b*d) + (-4*a*Sin[2*(c + d*x)] - 2*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])/(a*( a - b)*d*(-8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])^2) + (24 *a^2*Sin[2*(c + d*x)] + 22*a*b*Sin[2*(c + d*x)] - 10*b^2*Sin[2*(c + d*x)] - 11*a*b*Sin[4*(c + d*x)] + 5*b^2*Sin[4*(c + d*x)])/(32*a^2*(a - b)^2*d*(- 8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]))
Time = 0.85 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3696, 1672, 27, 2206, 27, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{\left (a-b \sin (c+d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tan ^2(c+d x) \left (\tan ^2(c+d x)+1\right )^4}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1672 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {16 a^2 (a-2 b) b \tan ^4(c+d x)}{(a-b)^2}+\frac {a b \left (8 a^3-29 b a^2+18 b^2 a-5 b^3\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (a+3 b)}{(a-b)^3}\right )}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{16 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {16 a^2 (a-2 b) b \tan ^4(c+d x)}{(a-b)^2}+\frac {a b \left (8 a^3-29 b a^2+18 b^2 a-5 b^3\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (a+3 b)}{(a-b)^3}}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{8 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {2 a^2 b^2 \left (\left (22 a^2-15 b a+5 b^2\right ) \tan ^2(c+d x)+2 a (5 a-2 b)\right )}{(a-b)^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{8 a^2 b}-\frac {b \tan (c+d x) \left (\frac {5 \left (2 a^2+3 a b-b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {b \int \frac {\left (22 a^2-15 b a+5 b^2\right ) \tan ^2(c+d x)+2 a (5 a-2 b)}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 (a-b)^2}-\frac {b \tan (c+d x) \left (\frac {5 \left (2 a^2+3 a b-b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {b \left (\frac {\left (\sqrt {a}+\sqrt {b}\right )^3 \left (-14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {b}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^3 \left (14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {b}}\right )}{4 (a-b)^2}-\frac {b \tan (c+d x) \left (\frac {5 \left (2 a^2+3 a b-b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {b \left (\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (-14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}+\sqrt {b}}}\right )}{4 (a-b)^2}-\frac {b \tan (c+d x) \left (\frac {5 \left (2 a^2+3 a b-b^2\right ) \tan ^2(c+d x)}{(a-b)^2}+\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (\left (a^2+6 a b+b^2\right ) \tan ^2(c+d x)+a (a+3 b)\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
(-1/8*(b*Tan[c + d*x]*(a*(a + 3*b) + (a^2 + 6*a*b + b^2)*Tan[c + d*x]^2))/ (a*(a - b)^3*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)^2) + ((b*(( (Sqrt[a] + Sqrt[b])^2*(12*a - 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTan[(Sqrt[Sqrt[ a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*S qrt[b]) - ((Sqrt[a] - Sqrt[b])^2*(12*a + 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTan[ (Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(1/4)*Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[b])))/(4*(a - b)^2) - (b*Tan[c + d*x]*((2*a*(5*a^2 - 9*a*b - 4*b^2))/(a - b)^3 + (5*(2*a^2 + 3*a*b - b^2)*Tan[c + d*x]^2)/(a - b)^2)) /(4*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)))/(8*a^2*b))/d
3.3.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_), x_Symbol] :> With[{f = Coeff[PolynomialRemainder[x^m*(d + e*x^2)^q , a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a *b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c*x^4)^(p + 1)* Simp[ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x], x]] /; FreeQ[{a, b, c, d, e}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] && IGtQ[m/2, 0]
Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Poly nomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^ 4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b ^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c *x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[Px, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4* p + 7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x ^2] && Expon[Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) ^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & & IntegerQ[m/2] && IntegerQ[p]
Time = 5.49 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {5 \left (2 a^{2}+3 a b -b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 a^{2} \left (a -b \right )}-\frac {3 \left (5 a^{2}+2 a b -3 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (10 a^{2}+a b -3 b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \tan \left (d x +c \right )}{16 \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (22 a^{2} \sqrt {a b}-15 a b \sqrt {a b}+5 b^{2} \sqrt {a b}-12 a^{3}+a^{2} b -a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (22 a^{2} \sqrt {a b}-15 a b \sqrt {a b}+5 b^{2} \sqrt {a b}+12 a^{3}-a^{2} b +a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(422\) |
| default | \(\frac {\frac {-\frac {5 \left (2 a^{2}+3 a b -b^{2}\right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 a^{2} \left (a -b \right )}-\frac {3 \left (5 a^{2}+2 a b -3 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (10 a^{2}+a b -3 b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \tan \left (d x +c \right )}{16 \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (22 a^{2} \sqrt {a b}-15 a b \sqrt {a b}+5 b^{2} \sqrt {a b}-12 a^{3}+a^{2} b -a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (22 a^{2} \sqrt {a b}-15 a b \sqrt {a b}+5 b^{2} \sqrt {a b}+12 a^{3}-a^{2} b +a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(422\) |
| risch | \(\text {Expression too large to display}\) | \(2554\) |
1/d*((-5/32*(2*a^2+3*a*b-b^2)/a^2/(a-b)*tan(d*x+c)^7-3/16*(5*a^2+2*a*b-3*b ^2)/a/(a^2-2*a*b+b^2)*tan(d*x+c)^5-3/32*(10*a^2+a*b-3*b^2)/a/(a^2-2*a*b+b^ 2)*tan(d*x+c)^3-1/16*(5*a-2*b)/(a^2-2*a*b+b^2)*tan(d*x+c))/(tan(d*x+c)^4*a -b*tan(d*x+c)^4+2*a*tan(d*x+c)^2+a)^2+1/32/a^2/(a^2-2*a*b+b^2)*(a-b)*(1/2* (22*a^2*(a*b)^(1/2)-15*a*b*(a*b)^(1/2)+5*b^2*(a*b)^(1/2)-12*a^3+a^2*b-a*b^ 2)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+ c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/2*(22*a^2*(a*b)^(1/2)-15*a*b*(a*b)^(1/ 2)+5*b^2*(a*b)^(1/2)+12*a^3-a^2*b+a*b^2)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a )*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 6215 vs. \(2 (295) = 590\).
Time = 2.37 (sec) , antiderivative size = 6215, normalized size of antiderivative = 17.91 \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
1/16*(4*(96*a^3*b^2 + 36*a^2*b^3 - 53*a*b^4 + 35*b^5)*cos(4*d*x + 4*c)*sin (2*d*x + 2*c) + ((12*a^2*b^3 - 11*a*b^4 + 5*b^5)*sin(14*d*x + 14*c) - (104 *a^2*b^3 - 85*a*b^4 + 35*b^5)*sin(12*d*x + 12*c) - (320*a^3*b^2 - 652*a^2* b^3 + 407*a*b^4 - 105*b^5)*sin(10*d*x + 10*c) + (1408*a^3*b^2 - 1696*a^2*b ^3 + 865*a*b^4 - 175*b^5)*sin(8*d*x + 8*c) + (320*a^3*b^2 + 756*a^2*b^3 - 849*a*b^4 + 175*b^5)*sin(6*d*x + 6*c) - (248*a^2*b^3 - 383*a*b^4 + 105*b^5 )*sin(4*d*x + 4*c) - (12*a^2*b^3 + 77*a*b^4 - 35*b^5)*sin(2*d*x + 2*c))*co s(16*d*x + 16*c) + 2*(2*(96*a^3*b^2 + 36*a^2*b^3 - 53*a*b^4 + 35*b^5)*sin( 12*d*x + 12*c) + 8*(64*a^3*b^2 - 196*a^2*b^3 + 125*a*b^4 - 35*b^5)*sin(10* d*x + 10*c) - 3*(512*a^4*b + 1024*a^3*b^2 - 1556*a^2*b^3 + 865*a*b^4 - 175 *b^5)*sin(8*d*x + 8*c) - 16*(128*a^3*b^2 + 124*a^2*b^3 - 173*a*b^4 + 35*b^ 5)*sin(6*d*x + 6*c) + 2*(96*a^3*b^2 + 324*a^2*b^3 - 649*a*b^4 + 175*b^5)*s in(4*d*x + 4*c) + 24*(4*a^2*b^3 + 11*a*b^4 - 5*b^5)*sin(2*d*x + 2*c))*cos( 14*d*x + 14*c) + 2*(2*(2560*a^4*b - 4128*a^3*b^2 + 3644*a^2*b^3 - 1379*a*b ^4 + 245*b^5)*sin(10*d*x + 10*c) - (9216*a^4*b - 25984*a^3*b^2 + 21304*a^2 *b^3 - 8575*a*b^4 + 1225*b^5)*sin(8*d*x + 8*c) - 2*(2560*a^4*b + 480*a^3*b ^2 - 7908*a^2*b^3 + 5033*a*b^4 - 735*b^5)*sin(6*d*x + 6*c) + 4*(576*a^3*b^ 2 - 1696*a^2*b^3 + 1323*a*b^4 - 245*b^5)*sin(4*d*x + 4*c) + 2*(96*a^3*b^2 + 324*a^2*b^3 - 649*a*b^4 + 175*b^5)*sin(2*d*x + 2*c))*cos(12*d*x + 12*c) + 2*((40960*a^5 - 24064*a^4*b - 22080*a^3*b^2 + 27516*a^2*b^3 - 11095*a...
Leaf count of result is larger than twice the leaf count of optimal. 1184 vs. \(2 (295) = 590\).
Time = 1.98 (sec) , antiderivative size = 1184, normalized size of antiderivative = 3.41 \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
1/64*((30*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4*b - 72*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3*b^2 + 14*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^ 3 + 4*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^4 + 36*sqrt(a^2 - a*b + sqrt (a*b)*(a - b))*sqrt(a*b)*a^4 - 105*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqr t(a*b)*a^3*b + 69*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 - 19*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 - 5*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^4)*(pi*floor((d*x + c)/pi + 1/2) + arctan (tan(d*x + c)/sqrt((a^5 - 2*a^4*b + a^3*b^2 + sqrt((a^5 - 2*a^4*b + a^3*b^ 2)^2 - (a^5 - 2*a^4*b + a^3*b^2)*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)))/( a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3))))*abs(-a + b)/(3*a^9*b - 18*a^8*b^2 + 41*a^7*b^3 - 44*a^6*b^4 + 21*a^5*b^5 - 2*a^4*b^6 - a^3*b^7) + (30*sqrt(a ^2 - a*b - sqrt(a*b)*(a - b))*a^4*b - 72*sqrt(a^2 - a*b - sqrt(a*b)*(a - b ))*a^3*b^2 + 14*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^3 + 4*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^4 - 36*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*s qrt(a*b)*a^4 + 105*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b - 6 9*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 + 19*sqrt(a^2 - a* b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 + 5*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^4)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sq rt((a^5 - 2*a^4*b + a^3*b^2 - sqrt((a^5 - 2*a^4*b + a^3*b^2)^2 - (a^5 - 2* a^4*b + a^3*b^2)*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)))/(a^5 - 3*a^4*b...
Time = 19.03 (sec) , antiderivative size = 6646, normalized size of antiderivative = 19.15 \[ \int \frac {\sin ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
- ((tan(c + d*x)*(5*a - 2*b))/(16*(a^2 - 2*a*b + b^2)) + (3*tan(c + d*x)^3 *(a*b + 10*a^2 - 3*b^2))/(32*a*(a^2 - 2*a*b + b^2)) + (5*tan(c + d*x)^7*(3 *a*b + 2*a^2 - b^2))/(32*a^2*(a - b)) + (3*tan(c + d*x)^5*(2*a*b + 5*a^2 - 3*b^2))/(16*a*(a - b)^2))/(d*(tan(c + d*x)^8*(a^2 - 2*a*b + b^2) + a^2 - tan(c + d*x)^4*(2*a*b - 6*a^2) - tan(c + d*x)^6*(4*a*b - 4*a^2) + 4*a^2*ta n(c + d*x)^2)) - (atan(((((163840*a^9*b + 65536*a^5*b^5 - 360448*a^6*b^4 + 688128*a^7*b^3 - 557056*a^8*b^2)/(32768*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6* b^2)) - (tan(c + d*x)*(-(384*a^4*(a^9*b^3)^(1/2) + 25*b^4*(a^9*b^3)^(1/2) - 144*a^9*b + 15*a^5*b^5 - 94*a^6*b^4 + 155*a^7*b^3 - 76*a^8*b^2 + 349*a^2 *b^2*(a^9*b^3)^(1/2) - 134*a*b^3*(a^9*b^3)^(1/2) - 480*a^3*b*(a^9*b^3)^(1/ 2))/(16384*(a^9*b^7 - 5*a^10*b^6 + 10*a^11*b^5 - 10*a^12*b^4 + 5*a^13*b^3 - a^14*b^2)))^(1/2)*(16384*a^10*b - 16384*a^5*b^6 + 81920*a^6*b^5 - 163840 *a^7*b^4 + 163840*a^8*b^3 - 81920*a^9*b^2))/(256*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))*(-(384*a^4*(a^9*b^3)^(1/2) + 25*b^4*(a^9*b^3)^(1/2) - 144*a ^9*b + 15*a^5*b^5 - 94*a^6*b^4 + 155*a^7*b^3 - 76*a^8*b^2 + 349*a^2*b^2*(a ^9*b^3)^(1/2) - 134*a*b^3*(a^9*b^3)^(1/2) - 480*a^3*b*(a^9*b^3)^(1/2))/(16 384*(a^9*b^7 - 5*a^10*b^6 + 10*a^11*b^5 - 10*a^12*b^4 + 5*a^13*b^3 - a^14* b^2)))^(1/2) - (tan(c + d*x)*(460*a^4*b - 149*a*b^4 + 144*a^5 + 25*b^5 + 4 43*a^2*b^3 - 635*a^3*b^2))/(256*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))*(- (384*a^4*(a^9*b^3)^(1/2) + 25*b^4*(a^9*b^3)^(1/2) - 144*a^9*b + 15*a^5*...