Integrand size = 24, antiderivative size = 313 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {3 \left (2 \sqrt {a}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {3 \left (2 \sqrt {a}+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {b \tan (c+d x) \left (3 a+b+4 (a+b) \tan ^2(c+d x)\right )}{8 (a-b)^3 d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )^2}-\frac {\tan (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}+\frac {(17 a+3 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{32 a d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \]
3/64*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(2*a^(1/2)-b^(1/2) )/a^(7/4)/d/(a^(1/2)-b^(1/2))^(5/2)/b^(1/2)-3/64*arctan((a^(1/2)+b^(1/2))^ (1/2)*tan(d*x+c)/a^(1/4))*(2*a^(1/2)+b^(1/2))/a^(7/4)/d/b^(1/2)/(a^(1/2)+b ^(1/2))^(5/2)-1/8*b*tan(d*x+c)*(3*a+b+4*(a+b)*tan(d*x+c)^2)/(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)*((9*a^2-24*a*b-b^2)/ (a-b)^3+(17*a+3*b)*tan(d*x+c)^2/(a-b)^2)/a/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan (d*x+c)^4)
Time = 11.44 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.01 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {-\frac {3 \left (2 a^{3/2}-3 a \sqrt {b}+b^{3/2}\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{a^{3/2} \sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b}}-\frac {3 \left (2 a^{3/2}+3 a \sqrt {b}-b^{3/2}\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{a^{3/2} \sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b}}+\frac {8 (-7 a-2 b+(2 a+b) \cos (2 (c+d x))) \sin (2 (c+d x))}{a (8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x)))}+\frac {64 (a-b) (-6 \sin (2 (c+d x))+\sin (4 (c+d x)))}{(-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}}{64 (a-b)^2 d} \]
((-3*(2*a^(3/2) - 3*a*Sqrt[b] + b^(3/2))*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(a^(3/2)*Sqrt[a + Sqrt[a]*Sqrt[b]]*Sq rt[b]) - (3*(2*a^(3/2) + 3*a*Sqrt[b] - b^(3/2))*ArcTanh[((Sqrt[a] - Sqrt[b ])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(a^(3/2)*Sqrt[-a + Sqrt[a]*S qrt[b]]*Sqrt[b]) + (8*(-7*a - 2*b + (2*a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(a*(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c + d*x)])) + (64* (a - 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 - b)^2*d)
Time = 0.80 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.14, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{\left (a-b \sin (c+d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tan ^4(c+d x) \left (\tan ^2(c+d x)+1\right )^3}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1672 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {8 a^2 (a-3 b) b \tan ^4(c+d x)}{(a-b)^2}-\frac {4 a^2 (3 a-b) b^2 \tan ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (3 a+b)}{(a-b)^3}\right )}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{16 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {8 a^2 b \tan ^6(c+d x)}{a-b}+\frac {8 a^2 (a-3 b) b \tan ^4(c+d x)}{(a-b)^2}-\frac {4 a^2 (3 a-b) b^2 \tan ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (3 a+b)}{(a-b)^3}}{\left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{8 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {6 a^3 b^2 \left ((5 a-b) \tan ^2(c+d x)+3 a-b\right )}{(a-b)^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{8 a^2 b}-\frac {a b \tan (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}+\frac {(17 a+3 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 a b \int \frac {(5 a-b) \tan ^2(c+d x)+3 a-b}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 (a-b)^2}-\frac {a b \tan (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}+\frac {(17 a+3 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {3 a b \left (\frac {\left (2 \sqrt {a}-\sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^3 \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a} \sqrt {b}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^3 \left (2 \sqrt {a}+\sqrt {b}\right ) \int \frac {1}{(a-b) \tan ^2(c+d x)+\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tan (c+d x)}{2 \sqrt {a} \sqrt {b}}\right )}{4 (a-b)^2}-\frac {a b \tan (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}+\frac {(17 a+3 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {3 a b \left (\frac {\left (2 \sqrt {a}-\sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (2 \sqrt {a}+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} \sqrt {\sqrt {a}+\sqrt {b}}}\right )}{4 (a-b)^2}-\frac {a b \tan (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}+\frac {(17 a+3 b) \tan ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{8 a^2 b}-\frac {b \tan (c+d x) \left (4 (a+b) \tan ^2(c+d x)+3 a+b\right )}{8 (a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )^2}}{d}\) |
(-1/8*(b*Tan[c + d*x]*(3*a + b + 4*(a + b)*Tan[c + d*x]^2))/((a - b)^3*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4)^2) + ((3*a*b*(((2*Sqrt[a] - Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d* x])/a^(1/4)])/(2*a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]*Sqrt[b]) - ((Sqrt[a] - Sq rt[b])^2*(2*Sqrt[a] + Sqrt[b])*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x ])/a^(1/4)])/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[b])))/(4*(a - b)^2) - (a*b*Tan[c + d*x]*((9*a^2 - 24*a*b - b^2)/(a - b)^3 + ((17*a + 3*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
3.3.32.3.1 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 = 4.44 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (17 a +3 b \right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 a \left (a -b \right )}-\frac {\left (43 a^{2}-18 a b -b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (35 a -11 b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (3 a -b \right ) \tan \left (d x +c \right )}{32 \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {3 \left (a -b \right ) \left (\frac {\left (5 a \sqrt {a b}-\sqrt {a b}\, b -2 a^{2}-3 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (5 a \sqrt {a b}-\sqrt {a b}\, b +2 a^{2}+3 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(372\) |
| default | \(\frac {\frac {-\frac {\left (17 a +3 b \right ) \left (\tan ^{7}\left (d x +c \right )\right )}{32 a \left (a -b \right )}-\frac {\left (43 a^{2}-18 a b -b^{2}\right ) \left (\tan ^{5}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (35 a -11 b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{32 \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (3 a -b \right ) \tan \left (d x +c \right )}{32 \left (a^{2}-2 a b +b^{2}\right )}}{{\left (\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}^{2}}+\frac {3 \left (a -b \right ) \left (\frac {\left (5 a \sqrt {a b}-\sqrt {a b}\, b -2 a^{2}-3 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (5 a \sqrt {a b}-\sqrt {a b}\, b +2 a^{2}+3 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(372\) |
| risch | \(\text {Expression too large to display}\) | \(1716\) |
1/d*((-1/32*(17*a+3*b)/a/(a-b)*tan(d*x+c)^7-1/32*(43*a^2-18*a*b-b^2)/a/(a^ 2-2*a*b+b^2)*tan(d*x+c)^5-1/32*(35*a-11*b)/(a^2-2*a*b+b^2)*tan(d*x+c)^3-3/ 32*(3*a-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+3/32/a/(a^2-2*a*b+b^2)*(a-b)*(1/2*(5*a*(a*b)^(1/2)-(a*b) ^(1/2)*b-2*a^2-3*a*b+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*(5*a*(a*b)^(1 /2)-(a*b)^(1/2)*b+2*a^2+3*a*b-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. 5510 vs. \(2 (261) = 522\).
Time = 1.44 (sec) , antiderivative size = 5510, normalized size of antiderivative = 17.60 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{4}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
-1/8*(3*a*b^3*sin(2*d*x + 2*c) - 12*(8*a^2*b^2 + 13*a*b^3 - 2*b^4)*cos(4*d *x + 4*c)*sin(2*d*x + 2*c) - (3*a*b^3*sin(14*d*x + 14*c) - 3*(10*a*b^3 - b ^4)*sin(12*d*x + 12*c) - (80*a^2*b^2 - 111*a*b^3 + 16*b^4)*sin(10*d*x + 10 *c) + (256*a^3*b - 64*a^2*b^2 - 26*a*b^3 + 35*b^4)*sin(8*d*x + 8*c) + (336 *a^2*b^2 - 95*a*b^3 - 40*b^4)*sin(6*d*x + 6*c) - (64*a^2*b^2 - 54*a*b^3 - 25*b^4)*sin(4*d*x + 4*c) - (19*a*b^3 + 8*b^4)*sin(2*d*x + 2*c))*cos(16*d*x + 16*c) - 2*(6*(8*a^2*b^2 + 13*a*b^3 - 2*b^4)*sin(12*d*x + 12*c) + 8*(16* a^2*b^2 - 45*a*b^3 + 8*b^4)*sin(10*d*x + 10*c) - (1408*a^3*b - 544*a^2*b^2 + a*b^3 + 140*b^4)*sin(8*d*x + 8*c) - 16*(96*a^2*b^2 - 29*a*b^3 - 10*b^4) *sin(6*d*x + 6*c) + 2*(152*a^2*b^2 - 129*a*b^3 - 50*b^4)*sin(4*d*x + 4*c) + 8*(11*a*b^3 + 4*b^4)*sin(2*d*x + 2*c))*cos(14*d*x + 14*c) - 2*(2*(640*a^ 3*b - 488*a^2*b^2 + 389*a*b^3 - 70*b^4)*sin(10*d*x + 10*c) - (4096*a^4 - 8 448*a^3*b + 3744*a^2*b^2 - 414*a*b^3 - 385*b^4)*sin(8*d*x + 8*c) - 2*(2688 *a^3*b - 4072*a^2*b^2 + 861*a*b^3 + 238*b^4)*sin(6*d*x + 6*c) + 4*(256*a^3 *b - 560*a^2*b^2 + 206*a*b^3 + 77*b^4)*sin(4*d*x + 4*c) + 2*(152*a^2*b^2 - 129*a*b^3 - 50*b^4)*sin(2*d*x + 2*c))*cos(12*d*x + 12*c) - 2*((26624*a^4 - 33152*a^3*b + 15632*a^2*b^2 - 2453*a*b^3 - 420*b^4)*sin(8*d*x + 8*c) + 8 *(3328*a^3*b - 3104*a^2*b^2 + 529*a*b^3 + 84*b^4)*sin(6*d*x + 6*c) - 2*(26 88*a^3*b - 4072*a^2*b^2 + 861*a*b^3 + 238*b^4)*sin(4*d*x + 4*c) - 16*(96*a ^2*b^2 - 29*a*b^3 - 10*b^4)*sin(2*d*x + 2*c))*cos(10*d*x + 10*c) - 2*((...
Leaf count of result is larger than twice the leaf count of optimal. 1986 vs. \(2 (261) = 522\).
Time = 1.41 (sec) , antiderivative size = 1986, normalized size of antiderivative = 6.35 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
-1/64*(3*((15*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b - 33*sqr t(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 + sqrt(a^2 - a*b + sqrt (a*b)*(a - b))*sqrt(a*b)*a*b^3 + sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt( a*b)*b^4)*(a^3 - 2*a^2*b + a*b^2)^2*abs(-a + b) - (9*sqrt(a^2 - a*b + sqrt (a*b)*(a - b))*a^7*b - 48*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^6*b^2 + 93 *sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^5*b^3 - 80*sqrt(a^2 - a*b + sqrt(a* b)*(a - b))*a^4*b^4 + 27*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3*b^5 - sqr t(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^7)*abs(a^3 - 2*a^2*b + a*b^2)*abs(-a + b) - (6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^10 - 27*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^9*b + 25*sqrt(a^2 - a*b + sqrt(a*b )*(a - b))*sqrt(a*b)*a^8*b^2 + 53*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt (a*b)*a^7*b^3 - 131*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^6*b^4 + 103*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^5*b^5 - 29*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4*b^6 - sqrt(a^2 - a*b + sqrt(a*b)* (a - b))*sqrt(a*b)*a^3*b^7 + sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b) *a^2*b^8)*abs(-a + b))*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c) /sqrt((a^4 - 2*a^3*b + a^2*b^2 + sqrt((a^4 - 2*a^3*b + a^2*b^2)^2 - (a^4 - 2*a^3*b + a^2*b^2)*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)))/(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3))))/((3*a^12*b - 27*a^11*b^2 + 104*a^10*b^3 - 224*a^9*b ^4 + 294*a^8*b^5 - 238*a^7*b^6 + 112*a^6*b^7 - 24*a^5*b^8 - a^4*b^9 + a...
Time = 18.73 (sec) , antiderivative size = 5892, normalized size of antiderivative = 18.82 \[ \int \frac {\sin ^4(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
- (atan(((((3*(49152*a^7*b + 16384*a^3*b^5 - 98304*a^4*b^4 + 196608*a^5*b^ 3 - 163840*a^6*b^2))/(32768*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)) - (tan( c + d*x)*((9*(16*a^3*(a^7*b^3)^(1/2) + b^3*(a^7*b^3)^(1/2) + 4*a^7*b + a^4 *b^4 - 10*a^5*b^3 + 21*a^6*b^2 - 6*a*b^2*(a^7*b^3)^(1/2) + 5*a^2*b*(a^7*b^ 3)^(1/2)))/(16384*(a^7*b^7 - 5*a^8*b^6 + 10*a^9*b^5 - 10*a^10*b^4 + 5*a^11 *b^3 - a^12*b^2)))^(1/2)*(16384*a^9*b - 16384*a^4*b^6 + 81920*a^5*b^5 - 16 3840*a^6*b^4 + 163840*a^7*b^3 - 81920*a^8*b^2))/(256*(3*a^4*b - a^5 + a^2* b^3 - 3*a^3*b^2)))*((9*(16*a^3*(a^7*b^3)^(1/2) + b^3*(a^7*b^3)^(1/2) + 4*a ^7*b + a^4*b^4 - 10*a^5*b^3 + 21*a^6*b^2 - 6*a*b^2*(a^7*b^3)^(1/2) + 5*a^2 *b*(a^7*b^3)^(1/2)))/(16384*(a^7*b^7 - 5*a^8*b^6 + 10*a^9*b^5 - 10*a^10*b^ 4 + 5*a^11*b^3 - a^12*b^2)))^(1/2) - (tan(c + d*x)*(333*a^3*b - 45*a*b^3 + 36*a^4 + 9*b^4 - 45*a^2*b^2))/(256*(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2)) )*((9*(16*a^3*(a^7*b^3)^(1/2) + b^3*(a^7*b^3)^(1/2) + 4*a^7*b + a^4*b^4 - 10*a^5*b^3 + 21*a^6*b^2 - 6*a*b^2*(a^7*b^3)^(1/2) + 5*a^2*b*(a^7*b^3)^(1/2 )))/(16384*(a^7*b^7 - 5*a^8*b^6 + 10*a^9*b^5 - 10*a^10*b^4 + 5*a^11*b^3 - a^12*b^2)))^(1/2)*1i - (((3*(49152*a^7*b + 16384*a^3*b^5 - 98304*a^4*b^4 + 196608*a^5*b^3 - 163840*a^6*b^2))/(32768*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4 *b^2)) + (tan(c + d*x)*((9*(16*a^3*(a^7*b^3)^(1/2) + b^3*(a^7*b^3)^(1/2) + 4*a^7*b + a^4*b^4 - 10*a^5*b^3 + 21*a^6*b^2 - 6*a*b^2*(a^7*b^3)^(1/2) + 5 *a^2*b*(a^7*b^3)^(1/2)))/(16384*(a^7*b^7 - 5*a^8*b^6 + 10*a^9*b^5 - 10*...