Integrand size = 15, antiderivative size = 319 \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\left (32 a-50 \sqrt {a} \sqrt {b}+21 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} d}+\frac {\left (32 a+50 \sqrt {a} \sqrt {b}+21 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} d}-\frac {b^2 \tan (c+d x) \left (3 a+b+4 (a+b) \tan ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )^2}-\frac {b \tan (c+d x) \left (\frac {17 a^2-40 a b+7 b^2}{(a-b)^3}+\frac {(33 a-13 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \]
1/64*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(32*a+21*b-50*a^(1 /2)*b^(1/2))/a^(11/4)/d/(a^(1/2)-b^(1/2))^(5/2)+1/64*arctan((a^(1/2)+b^(1/ 2))^(1/2)*tan(d*x+c)/a^(1/4))*(32*a+21*b+50*a^(1/2)*b^(1/2))/a^(11/4)/d/(a ^(1/2)+b^(1/2))^(5/2)-1/8*b^2*tan(d*x+c)*(3*a+b+4*(a+b)*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*b*tan(d*x+c)*((17*a ^2-40*a*b+7*b^2)/(a-b)^3+(33*a-13*b)*tan(d*x+c)^2/(a-b)^2)/a^2/d/(a+2*a*ta n(d*x+c)^2+(a-b)*tan(d*x+c)^4)
Time = 8.21 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (32 a+50 \sqrt {a} \sqrt {b}+21 b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (32 a-50 \sqrt {a} \sqrt {b}+21 b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {8 \sqrt {a} b (-19 a+10 b+(6 a-3 b) \cos (2 (c+d x))) \sin (2 (c+d x))}{8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))}+\frac {64 a^{3/2} (a-b) b (-6 \sin (2 (c+d x))+\sin (4 (c+d x)))}{(-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}}{64 a^{5/2} (a-b)^2 d} \]
(((Sqrt[a] - Sqrt[b])^2*(32*a + 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTan[((Sqrt[a ] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sq rt[b]] - ((Sqrt[a] + Sqrt[b])^2*(32*a - 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTanh [((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] + (8*Sqrt[a]*b*(-19*a + 10*b + (6*a - 3*b)*Cos[2*(c + d* x)])*Sin[2*(c + d*x)])/(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c + d* x)]) + (64*a^(3/2)*(a - b)*b*(-6*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(-8 *a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])^2)/(64*a^(5/2)*(a - b)^2*d)
Time = 0.81 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3688, 1517, 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.
| \(\displaystyle \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-b \sin (c+d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3688 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^5}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1517 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {8 a^2 (3 a-5 b) b \tan ^4(c+d x)}{(a-b)^2}+\frac {4 a b \left (6 a^3-18 b a^2+15 b^2 a-5 b^3\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a b \left (8 a^3-24 b a^2+27 b^2 a-7 b^3\right )}{(a-b)^3}\right )}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{16 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {8 a^2 (3 a-5 b) b \tan ^4(c+d x)}{(a-b)^2}+\frac {4 a b \left (6 a^3-18 b a^2+15 b^2 a-5 b^3\right ) \tan ^2(c+d x)}{(a-b)^3}+\frac {a b \left (8 a^3-24 b a^2+27 b^2 a-7 b^3\right )}{(a-b)^3}}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {2 a^2 b^2 \left (32 a^2-47 b a+21 b^2+\left (32 a^2-33 b a+13 b^2\right ) \tan ^2(c+d x)\right )}{(a-b)^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (\frac {17 a^2-40 a b+7 b^2}{(a-b)^3}+\frac {(33 a-13 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {b \int \frac {32 a^2-47 b a+21 b^2+\left (32 a^2-33 b a+13 b^2\right ) \tan ^2(c+d x)}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 (a-b)^2}-\frac {b^2 \tan (c+d x) \left (\frac {17 a^2-40 a b+7 b^2}{(a-b)^3}+\frac {(33 a-13 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {b \left (\frac {\left (50 \sqrt {a} \sqrt {b}+32 a+21 b\right ) \left (\sqrt {a}-\sqrt {b}\right )^3 \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^3 \left (-50 \sqrt {a} \sqrt {b}+32 a+21 b\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a}}\right )}{4 (a-b)^2}-\frac {b^2 \tan (c+d x) \left (\frac {17 a^2-40 a b+7 b^2}{(a-b)^3}+\frac {(33 a-13 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {b \left (\frac {\left (50 \sqrt {a} \sqrt {b}+32 a+21 b\right ) \left (\sqrt {a}-\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (-50 \sqrt {a} \sqrt {b}+32 a+21 b\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}\right )}{4 (a-b)^2}-\frac {b^2 \tan (c+d x) \left (\frac {17 a^2-40 a b+7 b^2}{(a-b)^3}+\frac {(33 a-13 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b^2 \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 a (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
(-1/8*(b^2*Tan[c + d*x]*(3*a + b + 4*(a + b)*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] + S qrt[b])^2*(32*a - 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b ]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]) + ((Sqrt[a] - Sqrt[b])^2*(32*a + 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTan[(Sqrt[Sqrt[a] + Sq rt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]])))/(4*(a - b)^2) - (b^2*Tan[c + d*x]*((17*a^2 - 40*a*b + 7*b^2)/(a - b)^3 + ((33*a - 13*b)*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
3.3.34.3.1 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[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(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)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(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]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ [c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
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[((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]
Time = 5.66 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.30
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (33 a -13 b \right ) b \left (\tan ^{7}\left (d x +c \right )\right )}{32 a^{2} \left (a -b \right )}-\frac {b \left (83 a^{2}-66 a b +7 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (67 a -43 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {b \left (17 a -11 b \right ) \tan \left (d x +c \right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (32 a^{2} \sqrt {a b}-33 a b \sqrt {a b}+13 b^{2} \sqrt {a b}-46 a^{2} b +55 a \,b^{2}-21 b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (32 a^{2} \sqrt {a b}-33 a b \sqrt {a b}+13 b^{2} \sqrt {a b}+46 a^{2} b -55 a \,b^{2}+21 b^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(416\) |
| default | \(\frac {\frac {-\frac {\left (33 a -13 b \right ) b \left (\tan ^{7}\left (d x +c \right )\right )}{32 a^{2} \left (a -b \right )}-\frac {b \left (83 a^{2}-66 a b +7 b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (67 a -43 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {b \left (17 a -11 b \right ) \tan \left (d x +c \right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {\left (a -b \right ) \left (\frac {\left (32 a^{2} \sqrt {a b}-33 a b \sqrt {a b}+13 b^{2} \sqrt {a b}-46 a^{2} b +55 a \,b^{2}-21 b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (32 a^{2} \sqrt {a b}-33 a b \sqrt {a b}+13 b^{2} \sqrt {a b}+46 a^{2} b -55 a \,b^{2}+21 b^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(416\) |
| risch | \(\text {Expression too large to display}\) | \(2537\) |
1/d*((-1/32*(33*a-13*b)/a^2*b/(a-b)*tan(d*x+c)^7-1/32/a^2*b*(83*a^2-66*a*b +7*b^2)/(a^2-2*a*b+b^2)*tan(d*x+c)^5-1/32*(67*a-43*b)/a*b/(a^2-2*a*b+b^2)* tan(d*x+c)^3-1/32*b*(17*a-11*b)/a/(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*(32*a^2*(a*b)^(1/2)-33*a*b*(a*b)^(1/2)+13*b^2*(a*b)^(1/2)-46*a^2*b+55*a *b^2-21*b^3)/(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*(32*a^2*(a*b)^(1/2)-33*a*b *(a*b)^(1/2)+13*b^2*(a*b)^(1/2)+46*a^2*b-55*a*b^2+21*b^3)/(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. 6152 vs. \(2 (271) = 542\).
Time = 2.51 (sec) , antiderivative size = 6152, normalized size of antiderivative = 19.29 \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {1}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
1/8*(4*(120*a^2*b^3 - 77*a*b^4 + 14*b^5)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) + ((7*a*b^4 - 4*b^5)*sin(14*d*x + 14*c) - (32*a^2*b^3 + 2*a*b^4 - 7*b^5)* sin(12*d*x + 12*c) - (16*a^2*b^3 - 3*a*b^4 - 28*b^5)*sin(10*d*x + 10*c) + 3*(256*a^3*b^2 - 320*a^2*b^3 + 166*a*b^4 - 35*b^5)*sin(8*d*x + 8*c) + (784 *a^2*b^3 - 723*a*b^4 + 140*b^5)*sin(6*d*x + 6*c) - (160*a^2*b^3 - 266*a*b^ 4 + 91*b^5)*sin(4*d*x + 4*c) - (55*a*b^4 - 28*b^5)*sin(2*d*x + 2*c))*cos(1 6*d*x + 16*c) + 2*(2*(120*a^2*b^3 - 77*a*b^4 + 14*b^5)*sin(12*d*x + 12*c) - 8*(48*a^2*b^3 - 55*a*b^4 + 28*b^5)*sin(10*d*x + 10*c) - (3968*a^3*b^2 - 5024*a^2*b^3 + 2621*a*b^4 - 560*b^5)*sin(8*d*x + 8*c) - 16*(224*a^2*b^3 - 209*a*b^4 + 42*b^5)*sin(6*d*x + 6*c) + 2*(376*a^2*b^3 - 613*a*b^4 + 210*b^ 5)*sin(4*d*x + 4*c) + 8*(31*a*b^4 - 16*b^5)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) + 2*(2*(1152*a^3*b^2 - 520*a^2*b^3 - 455*a*b^4 + 294*b^5)*sin(10*d* x + 10*c) - (8192*a^4*b - 23296*a^3*b^2 + 21376*a^2*b^3 - 9394*a*b^4 + 171 5*b^5)*sin(8*d*x + 8*c) - 2*(5248*a^3*b^2 - 10888*a^2*b^3 + 6433*a*b^4 - 1 078*b^5)*sin(6*d*x + 6*c) + 4*(512*a^3*b^2 - 1520*a^2*b^3 + 1330*a*b^4 - 3 43*b^5)*sin(4*d*x + 4*c) + 2*(376*a^2*b^3 - 613*a*b^4 + 210*b^5)*sin(2*d*x + 2*c))*cos(12*d*x + 12*c) + 2*((51200*a^4*b - 84864*a^3*b^2 + 56016*a^2* b^3 - 18081*a*b^4 + 1960*b^5)*sin(8*d*x + 8*c) + 8*(6400*a^3*b^2 - 8608*a^ 2*b^3 + 3437*a*b^4 - 392*b^5)*sin(6*d*x + 6*c) - 2*(5248*a^3*b^2 - 10888*a ^2*b^3 + 6433*a*b^4 - 1078*b^5)*sin(4*d*x + 4*c) - 16*(224*a^2*b^3 - 20...
Leaf count of result is larger than twice the leaf count of optimal. 1131 vs. \(2 (271) = 542\).
Time = 0.71 (sec) , antiderivative size = 1131, normalized size of antiderivative = 3.55 \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
1/64*((96*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4 - 333*sqrt(a^2 - a*b + s qrt(a*b)*(a - b))*a^3*b + 313*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^2 - 79*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^3 - 21*sqrt(a^2 - a*b + sqrt( a*b)*(a - b))*b^4 + 42*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 - 108*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b + 34*sqrt(a^2 - a *b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 + 8*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/s qrt((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 - 18*a^8*b + 41*a^7*b^2 - 44*a ^6*b^3 + 21*a^5*b^4 - 2*a^4*b^5 - a^3*b^6) + (96*sqrt(a^2 - a*b - sqrt(a*b )*(a - b))*a^4 - 333*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^3*b + 313*sqrt( a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^2 - 79*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^3 - 21*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^4 - 42*sqrt(a^2 - a *b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 + 108*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b - 34*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a *b^2 - 8*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(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 ...
Time = 19.24 (sec) , antiderivative size = 6267, normalized size of antiderivative = 19.65 \[ \int \frac {1}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
- ((tan(c + d*x)^5*(83*a^2*b - 66*a*b^2 + 7*b^3))/(32*a^2*(a - b)^2) + (ta n(c + d*x)^7*(33*a*b - 13*b^2))/(32*a^2*(a - b)) + (tan(c + d*x)*(17*a*b - 11*b^2))/(32*a*(a^2 - 2*a*b + b^2)) + (tan(c + d*x)^3*(67*a*b - 43*b^2))/ (32*a*(a^2 - 2*a*b + 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*t an(c + d*x)^2)) - (atan(((((524288*a^10*b - 344064*a^5*b^6 + 1802240*a^6*b ^5 - 3866624*a^7*b^4 + 4227072*a^8*b^3 - 2342912*a^9*b^2)/(32768*(3*a^8*b - a^9 + a^6*b^3 - 3*a^7*b^2)) - (tan(c + d*x)*((1920*a^4*(a^11*b)^(1/2) + 441*b^4*(a^11*b)^(1/2) - 1916*a^9*b + 1024*a^10 + 105*a^6*b^4 - 570*a^7*b^ 3 + 1501*a^8*b^2 - 2246*a*b^3*(a^11*b)^(1/2) - 4640*a^3*b*(a^11*b)^(1/2) + 4669*a^2*b^2*(a^11*b)^(1/2))/(16384*(5*a^15*b - a^16 + a^11*b^5 - 5*a^12* b^4 + 10*a^13*b^3 - 10*a^14*b^2)))^(1/2)*(16384*a^11*b - 16384*a^6*b^6 + 8 1920*a^7*b^5 - 163840*a^8*b^4 + 163840*a^9*b^3 - 81920*a^10*b^2))/(256*(3* a^6*b - a^7 + a^4*b^3 - 3*a^5*b^2)))*((1920*a^4*(a^11*b)^(1/2) + 441*b^4*( a^11*b)^(1/2) - 1916*a^9*b + 1024*a^10 + 105*a^6*b^4 - 570*a^7*b^3 + 1501* a^8*b^2 - 2246*a*b^3*(a^11*b)^(1/2) - 4640*a^3*b*(a^11*b)^(1/2) + 4669*a^2 *b^2*(a^11*b)^(1/2))/(16384*(5*a^15*b - a^16 + a^11*b^5 - 5*a^12*b^4 + 10* a^13*b^3 - 10*a^14*b^2)))^(1/2) - (tan(c + d*x)*(1024*a^5*b - 2141*a*b^5 + 441*b^6 + 4099*a^2*b^4 - 3139*a^3*b^3 + 4*a^4*b^2))/(256*(3*a^6*b - a^7 + a^4*b^3 - 3*a^5*b^2)))*((1920*a^4*(a^11*b)^(1/2) + 441*b^4*(a^11*b)^(1...