Integrand size = 24, antiderivative size = 313 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\left (3 a-10 \sqrt {a} \sqrt {b}+4 b\right ) \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{64 a^{3/2} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{5/4} d}+\frac {\left (3 a+10 \sqrt {a} \sqrt {b}+4 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{64 a^{3/2} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{5/4} d}-\frac {\cos (c+d x) \left (a+b-b \cos ^2(c+d x)\right )}{8 (a-b) b d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )^2}+\frac {\cos (c+d x) \left (a^2-11 a b-2 b^2+2 b (2 a+b) \cos ^2(c+d x)\right )}{32 a (a-b)^2 b d \left (a-b+2 b \cos ^2(c+d x)-b \cos ^4(c+d x)\right )} \]
-1/8*cos(d*x+c)*(a+b-b*cos(d*x+c)^2)/(a-b)/b/d/(a-b+2*b*cos(d*x+c)^2-b*cos (d*x+c)^4)^2+1/32*cos(d*x+c)*(a^2-11*a*b-2*b^2+2*b*(2*a+b)*cos(d*x+c)^2)/a /(a-b)^2/b/d/(a-b+2*b*cos(d*x+c)^2-b*cos(d*x+c)^4)+1/64*arctan(b^(1/4)*cos (d*x+c)/(a^(1/2)-b^(1/2))^(1/2))*(3*a+4*b-10*a^(1/2)*b^(1/2))/a^(3/2)/b^(5 /4)/d/(a^(1/2)-b^(1/2))^(5/2)+1/64*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^( 1/2))^(1/2))*(3*a+4*b+10*a^(1/2)*b^(1/2))/a^(3/2)/b^(5/4)/d/(a^(1/2)+b^(1/ 2))^(5/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 6.24 (sec) , antiderivative size = 786, normalized size of antiderivative = 2.51 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\frac {\frac {32 \cos (c+d x) \left (a^2-9 a b-b^2+b (2 a+b) \cos (2 (c+d x))\right )}{a (8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x)))}-\frac {512 (a-b) \cos (c+d x) (2 a+b-b \cos (2 (c+d x)))}{(-8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x)))^2}+\frac {i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {4 a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+2 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-2 i a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+12 a^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-64 a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+10 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-6 i a^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+32 i a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-5 i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-12 a^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+64 a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-10 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+6 i a^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-32 i a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+5 i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-4 a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-2 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6+2 i a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6+i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{a}}{128 (a-b)^2 b d} \]
((32*Cos[c + d*x]*(a^2 - 9*a*b - b^2 + b*(2*a + b)*Cos[2*(c + d*x)]))/(a*( 8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c + d*x)])) - (512*(a - b)*Cos [c + d*x]*(2*a + b - b*Cos[2*(c + d*x)]))/(-8*a + 3*b - 4*b*Cos[2*(c + d*x )] + b*Cos[4*(c + d*x)])^2 + (I*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^ 4 - 4*b*#1^6 + b*#1^8 & , (4*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] + 2*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - (2*I)*a*b*Log[1 - 2*Cos [c + d*x]*#1 + #1^2] - I*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 12*a^2*Ar cTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - 64*a*b*ArcTan[Sin[c + d*x]/( Cos[c + d*x] - #1)]*#1^2 + 10*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] *#1^2 - (6*I)*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + (32*I)*a*b*Log[ 1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (5*I)*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 12*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + 64*a* b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - 10*b^2*ArcTan[Sin[c + d* x]/(Cos[c + d*x] - #1)]*#1^4 + (6*I)*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] *#1^4 - (32*I)*a*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + (5*I)*b^2*Log[ 1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 - 4*a*b*ArcTan[Sin[c + d*x]/(Cos[c + d* x] - #1)]*#1^6 - 2*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 + (2* I)*a*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^6 + I*b^2*Log[1 - 2*Cos[c + d* x]*#1 + #1^2]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) & ])/a)/(128*(a - b)^2*b*d)
Time = 0.57 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3694, 1517, 27, 1492, 27, 1480, 218, 221}
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.
\(\displaystyle \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^5}{\left (a-b \sin (c+d x)^4\right )^3}dx\) |
\(\Big \downarrow \) 3694 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^2}{\left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )^3}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 1517 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\int \frac {2 a \left (5 b \cos ^2(c+d x)+a-7 b\right )}{\left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )^2}d\cos (c+d x)}{16 a b (a-b)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\int \frac {5 b \cos ^2(c+d x)+a-7 b}{\left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )^2}d\cos (c+d x)}{8 b (a-b)}}{d}\) |
\(\Big \downarrow \) 1492 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\frac {\cos (c+d x) \left (a^2+2 b (2 a+b) \cos ^2(c+d x)-11 a b-2 b^2\right )}{4 a (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}-\frac {\int -\frac {2 b \left (3 a^2-17 b a+2 b^2+2 b (2 a+b) \cos ^2(c+d x)\right )}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{8 a b (a-b)}}{8 b (a-b)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\frac {\int \frac {3 a^2-17 b a+2 b^2+2 b (2 a+b) \cos ^2(c+d x)}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{4 a (a-b)}+\frac {\cos (c+d x) \left (a^2+2 b (2 a+b) \cos ^2(c+d x)-11 a b-2 b^2\right )}{4 a (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}}{8 b (a-b)}}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\frac {\frac {\sqrt {b} \left (-2 \sqrt {a} \sqrt {b}+a+b\right ) \left (10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{2 \sqrt {a}}-\frac {\sqrt {b} \left (\sqrt {a}+\sqrt {b}\right )^2 \left (-10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \int \frac {1}{-b \cos ^2(c+d x)-\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {b}}d\cos (c+d x)}{2 \sqrt {a}}}{4 a (a-b)}+\frac {\cos (c+d x) \left (a^2+2 b (2 a+b) \cos ^2(c+d x)-11 a b-2 b^2\right )}{4 a (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}}{8 b (a-b)}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\frac {\frac {\sqrt {b} \left (-2 \sqrt {a} \sqrt {b}+a+b\right ) \left (10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \int \frac {1}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}-b \cos ^2(c+d x)}d\cos (c+d x)}{2 \sqrt {a}}+\frac {\left (-10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} \sqrt {\sqrt {a}-\sqrt {b}}}}{4 a (a-b)}+\frac {\cos (c+d x) \left (a^2+2 b (2 a+b) \cos ^2(c+d x)-11 a b-2 b^2\right )}{4 a (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}}{8 b (a-b)}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {\cos (c+d x) \left (a-b \cos ^2(c+d x)+b\right )}{8 b (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )^2}-\frac {\frac {\cos (c+d x) \left (a^2+2 b (2 a+b) \cos ^2(c+d x)-11 a b-2 b^2\right )}{4 a (a-b) \left (a-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)-b\right )}+\frac {\frac {\left (-10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\left (-2 \sqrt {a} \sqrt {b}+a+b\right ) \left (10 \sqrt {a} \sqrt {b}+3 a+4 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} \sqrt {\sqrt {a}+\sqrt {b}}}}{4 a (a-b)}}{8 b (a-b)}}{d}\) |
-(((Cos[c + d*x]*(a + b - b*Cos[c + d*x]^2))/(8*(a - b)*b*(a - b + 2*b*Cos [c + d*x]^2 - b*Cos[c + d*x]^4)^2) - ((((Sqrt[a] + Sqrt[b])^2*(3*a - 10*Sq rt[a]*Sqrt[b] + 4*b)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]] ])/(2*Sqrt[a]*Sqrt[Sqrt[a] - Sqrt[b]]*b^(1/4)) + ((a - 2*Sqrt[a]*Sqrt[b] + b)*(3*a + 10*Sqrt[a]*Sqrt[b] + 4*b)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[S qrt[a] + Sqrt[b]]])/(2*Sqrt[a]*Sqrt[Sqrt[a] + Sqrt[b]]*b^(1/4)))/(4*a*(a - b)) + (Cos[c + d*x]*(a^2 - 11*a*b - 2*b^2 + 2*b*(2*a + b)*Cos[c + d*x]^2) )/(4*a*(a - b)*(a - b + 2*b*Cos[c + d*x]^2 - b*Cos[c + d*x]^4)))/(8*(a - b )*b))/d)
3.3.26.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb ol] :> Simp[x*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a + b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 5.97 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {b \left (2 a +b \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{2}-19 a b -6 b^{2}\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}-14 a b -3 b^{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (3 a^{2}+15 a b +2 b^{2}\right ) \cos \left (d x +c \right )}{32 \left (a -b \right ) a b}}{{\left (a -b +2 b \left (\cos ^{2}\left (d x +c \right )\right )-b \left (\cos ^{4}\left (d x +c \right )\right )\right )}^{2}}-\frac {\frac {\left (4 a \sqrt {a b}+2 \sqrt {a b}\, b -3 a^{2}+13 a b -4 b^{2}\right ) \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (4 a \sqrt {a b}+2 \sqrt {a b}\, b +3 a^{2}-13 a b +4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(347\) |
default | \(\frac {-\frac {\frac {b \left (2 a +b \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{2}-19 a b -6 b^{2}\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{32 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}-14 a b -3 b^{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{16 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (3 a^{2}+15 a b +2 b^{2}\right ) \cos \left (d x +c \right )}{32 \left (a -b \right ) a b}}{{\left (a -b +2 b \left (\cos ^{2}\left (d x +c \right )\right )-b \left (\cos ^{4}\left (d x +c \right )\right )\right )}^{2}}-\frac {\frac {\left (4 a \sqrt {a b}+2 \sqrt {a b}\, b -3 a^{2}+13 a b -4 b^{2}\right ) \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\left (4 a \sqrt {a b}+2 \sqrt {a b}\, b +3 a^{2}-13 a b +4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}}{32 a \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(347\) |
risch | \(\text {Expression too large to display}\) | \(1459\) |
1/d*(-(1/16*b*(2*a+b)/a/(a^2-2*a*b+b^2)*cos(d*x+c)^7+1/32*(a^2-19*a*b-6*b^ 2)/a/(a^2-2*a*b+b^2)*cos(d*x+c)^5-1/16*(5*a^2-14*a*b-3*b^2)/a/(a^2-2*a*b+b ^2)*cos(d*x+c)^3+1/32*(3*a^2+15*a*b+2*b^2)/(a-b)/a/b*cos(d*x+c))/(a-b+2*b* cos(d*x+c)^2-b*cos(d*x+c)^4)^2-1/32/a/(a^2-2*a*b+b^2)*(1/2*(4*a*(a*b)^(1/2 )+2*(a*b)^(1/2)*b-3*a^2+13*a*b-4*b^2)/(a*b)^(1/2)/(((a*b)^(1/2)-b)*b)^(1/2 )*arctan(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))-1/2*(4*a*(a*b)^(1/2)+2*(a *b)^(1/2)*b+3*a^2-13*a*b+4*b^2)/(a*b)^(1/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arct anh(cos(d*x+c)*b/(((a*b)^(1/2)+b)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4524 vs. \(2 (262) = 524\).
Time = 0.97 (sec) , antiderivative size = 4524, normalized size of antiderivative = 14.45 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
-1/128*(8*(2*a*b^2 + b^3)*cos(d*x + c)^7 + 4*(a^2*b - 19*a*b^2 - 6*b^3)*co s(d*x + c)^5 - 8*(5*a^2*b - 14*a*b^2 - 3*b^3)*cos(d*x + c)^3 + ((a^3*b^3 - 2*a^2*b^4 + a*b^5)*d*cos(d*x + c)^8 - 4*(a^3*b^3 - 2*a^2*b^4 + a*b^5)*d*c os(d*x + c)^6 - 2*(a^4*b^2 - 5*a^3*b^3 + 7*a^2*b^4 - 3*a*b^5)*d*cos(d*x + c)^4 + 4*(a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*d*cos(d*x + c)^2 + (a^5 *b - 4*a^4*b^2 + 6*a^3*b^3 - 4*a^2*b^4 + a*b^5)*d)*sqrt((15*a^4 - 30*a^3*b - 229*a^2*b^2 + 116*a*b^3 - 16*b^4 + (a^8*b^2 - 5*a^7*b^3 + 10*a^6*b^4 - 10*a^5*b^5 + 5*a^4*b^6 - a^3*b^7)*d^2*sqrt((81*a^6 - 1548*a^5*b + 12814*a^ 4*b^2 - 53212*a^3*b^3 + 104361*a^2*b^4 - 48160*a*b^5 + 6400*b^6)/((a^13*b^ 5 - 10*a^12*b^6 + 45*a^11*b^7 - 120*a^10*b^8 + 210*a^9*b^9 - 252*a^8*b^10 + 210*a^7*b^11 - 120*a^6*b^12 + 45*a^5*b^13 - 10*a^4*b^14 + a^3*b^15)*d^4) ))/((a^8*b^2 - 5*a^7*b^3 + 10*a^6*b^4 - 10*a^5*b^5 + 5*a^4*b^6 - a^3*b^7)* d^2))*log((81*a^5 - 1458*a^4*b + 9389*a^3*b^2 - 24972*a^2*b^3 + 10896*a*b^ 4 - 1280*b^5)*cos(d*x + c) + ((a^10*b^4 + 10*a^9*b^5 - 69*a^8*b^6 + 160*a^ 7*b^7 - 185*a^6*b^8 + 114*a^5*b^9 - 35*a^4*b^10 + 4*a^3*b^11)*d^3*sqrt((81 *a^6 - 1548*a^5*b + 12814*a^4*b^2 - 53212*a^3*b^3 + 104361*a^2*b^4 - 48160 *a*b^5 + 6400*b^6)/((a^13*b^5 - 10*a^12*b^6 + 45*a^11*b^7 - 120*a^10*b^8 + 210*a^9*b^9 - 252*a^8*b^10 + 210*a^7*b^11 - 120*a^6*b^12 + 45*a^5*b^13 - 10*a^4*b^14 + a^3*b^15)*d^4)) - (27*a^7*b - 411*a^6*b^2 + 2383*a^5*b^3 - 5 529*a^4*b^4 + 1962*a^3*b^5 - 160*a^2*b^6)*d)*sqrt((15*a^4 - 30*a^3*b - ...
Timed out. \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
1/8*(8*(2*a*b^4 + b^5)*cos(2*d*x + 2*c)*cos(d*x + c) + 8*(2*a^2*b^3 - 24*a *b^4 - 5*b^5)*sin(3*d*x + 3*c)*sin(2*d*x + 2*c) + 8*(2*a*b^4 + b^5)*sin(2* d*x + 2*c)*sin(d*x + c) - ((2*a*b^4 + b^5)*cos(15*d*x + 15*c) + (2*a^2*b^3 - 24*a*b^4 - 5*b^5)*cos(13*d*x + 13*c) - (70*a^2*b^3 - 76*a*b^4 - 9*b^5)* cos(11*d*x + 11*c) + (96*a^3*b^2 + 164*a^2*b^3 - 54*a*b^4 - 5*b^5)*cos(9*d *x + 9*c) + (96*a^3*b^2 + 164*a^2*b^3 - 54*a*b^4 - 5*b^5)*cos(7*d*x + 7*c) - (70*a^2*b^3 - 76*a*b^4 - 9*b^5)*cos(5*d*x + 5*c) + (2*a^2*b^3 - 24*a*b^ 4 - 5*b^5)*cos(3*d*x + 3*c) + (2*a*b^4 + b^5)*cos(d*x + c))*cos(16*d*x + 1 6*c) - (2*a*b^4 + b^5 - 8*(2*a*b^4 + b^5)*cos(14*d*x + 14*c) - 4*(16*a^2*b ^3 - 6*a*b^4 - 7*b^5)*cos(12*d*x + 12*c) + 8*(32*a^2*b^3 + 2*a*b^4 - 7*b^5 )*cos(10*d*x + 10*c) + 2*(256*a^3*b^2 - 64*a^2*b^3 - 26*a*b^4 + 35*b^5)*co s(8*d*x + 8*c) + 8*(32*a^2*b^3 + 2*a*b^4 - 7*b^5)*cos(6*d*x + 6*c) - 4*(16 *a^2*b^3 - 6*a*b^4 - 7*b^5)*cos(4*d*x + 4*c) - 8*(2*a*b^4 + b^5)*cos(2*d*x + 2*c))*cos(15*d*x + 15*c) + 8*((2*a^2*b^3 - 24*a*b^4 - 5*b^5)*cos(13*d*x + 13*c) - (70*a^2*b^3 - 76*a*b^4 - 9*b^5)*cos(11*d*x + 11*c) + (96*a^3*b^ 2 + 164*a^2*b^3 - 54*a*b^4 - 5*b^5)*cos(9*d*x + 9*c) + (96*a^3*b^2 + 164*a ^2*b^3 - 54*a*b^4 - 5*b^5)*cos(7*d*x + 7*c) - (70*a^2*b^3 - 76*a*b^4 - 9*b ^5)*cos(5*d*x + 5*c) + (2*a^2*b^3 - 24*a*b^4 - 5*b^5)*cos(3*d*x + 3*c) + ( 2*a*b^4 + b^5)*cos(d*x + c))*cos(14*d*x + 14*c) - (2*a^2*b^3 - 24*a*b^4 - 5*b^5 - 4*(16*a^3*b^2 - 206*a^2*b^3 + 128*a*b^4 + 35*b^5)*cos(12*d*x + ...
\[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sin \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \]
Time = 19.51 (sec) , antiderivative size = 6362, normalized size of antiderivative = 20.33 \[ \int \frac {\sin ^5(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \]
- ((cos(c + d*x)^3*(14*a*b - 5*a^2 + 3*b^2))/(16*a*(a - b)^2) - (cos(c + d *x)^5*(19*a*b - a^2 + 6*b^2))/(32*a*(a^2 - 2*a*b + b^2)) + (b*cos(c + d*x) ^7*(2*a + b))/(16*a*(a^2 - 2*a*b + b^2)) + (cos(c + d*x)*(15*a*b + 3*a^2 + 2*b^2))/(32*a*b*(a - b)))/(d*(a^2 - 2*a*b + b^2 + cos(c + d*x)^2*(4*a*b - 4*b^2) - cos(c + d*x)^4*(2*a*b - 6*b^2) - 4*b^2*cos(c + d*x)^6 + b^2*cos( c + d*x)^8)) - (atan(((((16384*a^3*b^6 - 172032*a^4*b^5 + 319488*a^5*b^4 - 188416*a^6*b^3 + 24576*a^7*b^2)/(16384*(a^7 - 4*a^6*b + a^3*b^4 - 4*a^4*b ^3 + 6*a^5*b^2)) - (cos(c + d*x)*((80*b^3*(a^9*b^5)^(1/2) - 9*a^3*(a^9*b^5 )^(1/2) + 16*a^3*b^7 - 116*a^4*b^6 + 229*a^5*b^5 + 30*a^6*b^4 - 15*a^7*b^3 - 301*a*b^2*(a^9*b^5)^(1/2) + 86*a^2*b*(a^9*b^5)^(1/2))/(16384*(a^6*b^10 - 5*a^7*b^9 + 10*a^8*b^8 - 10*a^9*b^7 + 5*a^10*b^6 - a^11*b^5)))^(1/2)*(16 384*a^3*b^8 - 65536*a^4*b^7 + 98304*a^5*b^6 - 65536*a^6*b^5 + 16384*a^7*b^ 4))/(256*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*((80*b^3*(a^9 *b^5)^(1/2) - 9*a^3*(a^9*b^5)^(1/2) + 16*a^3*b^7 - 116*a^4*b^6 + 229*a^5*b ^5 + 30*a^6*b^4 - 15*a^7*b^3 - 301*a*b^2*(a^9*b^5)^(1/2) + 86*a^2*b*(a^9*b ^5)^(1/2))/(16384*(a^6*b^10 - 5*a^7*b^9 + 10*a^8*b^8 - 10*a^9*b^7 + 5*a^10 *b^6 - a^11*b^5)))^(1/2) + (cos(c + d*x)*(9*a^4*b - 100*a*b^4 + 16*b^5 + 2 09*a^2*b^3 - 62*a^3*b^2))/(256*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^ 4*b^2)))*((80*b^3*(a^9*b^5)^(1/2) - 9*a^3*(a^9*b^5)^(1/2) + 16*a^3*b^7 - 1 16*a^4*b^6 + 229*a^5*b^5 + 30*a^6*b^4 - 15*a^7*b^3 - 301*a*b^2*(a^9*b^5...